Here is my personal note of the book, Advanced Modern Algebra, written by Joseph J. Rotman. It's the second time I try to write a note concerning abstract algebra. Maybe this article contains many faults, and if you find some of them, please leave a comment to let me know. Please note that some of the math marks in this article do not use the same mark as in the book, because of the habit of mine.

Abstract algebra is an important part of modern mathematics. It is the study of algebra structures, such as groups, rings and fields etc. It concentrates on the common properties of algebra structures instead of certain numbers or equations. It's the basis for learning many fields of mathematics.

Change log:

• 2023.5.24: Update the section 1.7.
• 2023.5.10: Update the section 1.6.
• 2023.4.21: Finished the section 1.5 and fixed some bugs.
• 2023.4.17: Start to write the section 1.5.

1. Commutative Rings

Let's begin with commutative rings. Commutative rings are algebra structures which have good properties and close to some sets we have learned well, such as $\mathbb {R}$ and $\mathbb{Z}$.

Definition 1.0.1 $f:A\to B,g:B\to C$ and $h:A\to C$ are functions. And the diagram of them is:
$$\xymatrix{
A \ar@{->}[r]^{f} \ar@{->}[rd]_{h} & B \ar@{->}[d]^{g} \\
& C
}
$$
And a commutative diagram is a diagram such that all directed paths in the diagram with the same start and endpoints lead to the same result. If we say this diagram is commutative, it means $h=g\circ f$.

Definition 1.0.2 A binary operation on a set $R$ is a function $\cdot:R\times R\to R,(a,b)\mapsto a\cdot b$. If the binary operation is multiplication, we always write it as $ab$ instead of $a\cdot b$.

Definition 1.0.3 $R$ is a set, and there are two binary operations, addition $+$ and multiplication $\cdot$, on it, which satisfy:

• $(R, +)$ is an abelian group; that is
  • $a+(b+c)=(a+b)+c,\forall a,b,c\in R$;
  • there is an element $0\in R$ with $0+a=a,\forall a\in R$;
  • for each $a\in R$, there is $b\in R$ with $b+a=a+b=0$ (we always denote $b$ as $(-a)$);
  • $a+b=b+a$;
• Associativity: $(ab)c=a(bc)$;
• there is $1\in R$ with $1a=a1=a$ ($1$ is called an identity, and in general we always think a ring may not contain an identity, but here we think it always has);
• Distributivity: $a(b+c)=ab+ac$ and $(b+c)a=ba+ca$, $\forall a,b,c\in R$.

And then we call $(R,+,\cdot)$ a ring, usually denote it as $R$.

If $\forall a\in R^\times:=R\textbackslash\{0\}$, $a$ have an (multiplicative) inverse $b\in R^\times$, which means $ab=ba=1$, then we call $R$ is an division ring (we always denote $b$ as $a^{-1}$).

We should know that not every binary operation have associativity, such as the subtraction on $\mathbb R$. $(a-b)-c=a-(b-c)$ is not always true.

Here is a diagram of associativity:
$$\xymatrix{
R\times R\times R \ar@{->}[d]_{1\times\cdot} \ar@{->}[rr] \ar@{->}[rr] \ar@{->}[rr]^{\cdot\times 1} & & R\times R \ar@{->}[d]^{\cdot} \\
R\times R \ar@{->}[rr]^{\cdot} & & R
}
$$
The function $\cdot\times 1:R\times R\times R\to R\times R$ is defined by $(a,b,c)\mapsto (ab,c)$ while $1\times\cdot:R\times R\times R\to R\times R$ is defined by $(a,b,c)\mapsto(a,bc)$. Associativity says that this diagram is commutative, and in other words the two composite functions $R\times R\times R$ are equal.

Example 1.0.1 Here are some examples of rings.

• Denote the set of all $n\times n$ matrices $(a_{ij})$ with entries in $\mathbb R$ by $\operatorname{M}_n(\mathbb R)$. And $\operatorname M_{n}(\mathbb R)$ is a ring with binary operations matrix addition and matrix multiplication.
• Define an equivalence relation $\sim$ on $\mathbb Z$: $a\sim b$ if $a\equiv b\pmod p$. And define $\mathbb{Z}_p=\mathbb Z/\sim:=\{[a];a\in \mathbb Z\}$. Let $[a]$ be the equivalence class of $a$ under the relation $\sim$. And we define $[a]+[b]:=[a+b],[a]\cdot[b]=[ab]$ (obviously these are well-defined). Then $(\mathbb Z_p,+,\cdot)$ is a ring.

Proposition 1.0.1 Let $R$ be a ring. Here are some trivial propositions about rings which I won't prove in my note.

1. $0\cdot a=a\cdot 0=0,\forall a\in R$.
2. If $1=0$, then $R$ consists of the single element $0$. In this case, $R$ is called the zero ring or trivial ring.
3. $(-1)a=a(-1)=(-a)$.

Definition 1.0.4 Let $R$ be a ring. A subset $S$ is a subring of $R$ if

• $1\in S$;
• $\forall a,b\in S,a-b\in S$;
• $\forall a,b\in S,ab\in S$.

If $S$ is a proper subset of $R$ ($S\ne R$), $S$ is a proper subring of $R$. It's easy to show that $S$ is a ring.

Definition 1.0.5 A ring $R$ is commutative if $ab=ba,\forall a,b\in R$.

The sets $\mathbb {Z},\mathbb Q,\mathbb R$ and $\mathbb C$ are commutative rings with the usual addition and multiplication. And then all the rings in the rest of this chapter are commutative unless we say otherwise.

Proposition 1.0.2 (Binomial Theorem) Let $R$ be a commutative ring. If $a,b\in R$, then
$$(a+b)^n=\sum_{r=0}^n C_n^ra^rb^{n-r}
$$
And by using mathematical induction, it's easy to prove.

Example 1.0.2 Here is an example of a commutative ring from set theory.

If $A,B\subset X$, then their symmetric difference is $A+B=(A\cup B)\textbackslash(A\cap B)$. And define $A\cdot B=A\cap B$. Then it's not difficult to show $(P(X),+,\cdot)$ is a commutative ring. And we tell it a Boolean ring.

We can find that $0=\emptyset$ and $1=X$. Using Boolean ring we can prove the de Morgan law
$$(A\cup B)^c=A^c\cap B^c
$$
To prove it using set-theoretic methods needs too much on the meaning of the words and, or, and not. But using algebraic method is brief and clear. We can see $A\cup B=A+B+AB$ and $A^c=1+A$, and then proves
$$(A\cup B)^c=1+(A+B+AB)=(1+A)(1+B)=A^c\cap B^c\ \square
$$
Definition 1.0.6 A domain (often called an integral domain) is a commutative ring $R$ that satisfies two extra axioms:

• $1\ne 0$;
• Cancellation Law: $\forall a,b,c\in R,(ac=bc\land c\ne 0)\Rightarrow a=b$.

The familiar examples of commutative rings, $\mathbb Z,\mathbb Q,\mathbb R,$ and $\mathbb C$ are domains; the zero ring is not a domain.

Proposition 1.0.3 A nonzero commutative ring $R$ is a domain if and only if $\forall a,b\in R,ab=0\Rightarrow(a=0\lor b=0)$.

Proof.

• ($\Rightarrow$).

  Let's assume that $a\ne 0$.

  Then $ab=0=0b\Rightarrow b=0$.

• ($\Leftarrow$).

  $ab=bc\Rightarrow (a-b)c=0$.

  Because $c\ne 0$, $a-b=0$.

  So $a=b$. $\square$

Proposition 1.0.4 If a nonzero commutative ring $R$ is a division ring, it is a domain.

Corollary 1.0.1 The commutative ring $\mathbb Z_p$ is a domain if and only if $p$ is a prime.

Definition 1.0.7 Let $a$ and $b$ be elements of a commutative ring $R$. Then $a$ divides $b$ in $R$ (or $a$ is a divisor of $b$ or $b$ is a multiple of $a$), denoted by
$$a\mid b,
$$
if $\exists c\in R$ such that $b=ca$.

Definition 1.0.8 An element $u$ in a commutative ring $R$ is called a unit or invertible element if $u\mid 1$. In other words, $u$ has an inverse $u^{-1}$ in $R$.

Definition 1.0.9 If $R$ is a nonzero commutative ring, then the group of units of $R$ is $U(R):=\{\text{all units in } R\}$.

It's easy to check that $U(R)$ is a multiplicative group.

Definition 1.0.10 A nonzero commutative ring $R$ is a field, if $U(R)=R^\times$.

Proposition 1.0.5 Every field $F$ is a domain.

It's easy to prove by using Proposition-1.0.4.

Theorem 1.0.1 If $R$ is a domain, then there is a field containing $R$ as a subring.

Moreover, such a field $F$ can be chosen so that, for each $f\in F$. there are $a,b\in R$ with $b\ne 0$ and $f=ab^{-1}$.

Proof.

Define a relation $\equiv$ on $R\times R^\times$ by $(a,b)\equiv (c,d)$ if $ad=bc$.

Then, let's prove $\equiv$ is an equivalence relation.

• Reflexivity: $ab\equiv ab\Rightarrow (a,b)=(a,b)$
• Symmetry: $(a,b)\equiv(c,d)\Rightarrow ad=bc\Rightarrow bc=ad\Rightarrow (c,d)\equiv (a,b)$
• Transitivity: $((a,b)\equiv (c,d)\land(c,d)\equiv(e,f))\Rightarrow (ad=bc\land cf=de)\Rightarrow bde=bcf=adf\Rightarrow (be)d=(af)d$ and since $d\in R^\times$ and $R$ is a domain, $be=af\Rightarrow (a,b)\equiv (e,f)$

Denote the equivalence class of $(a,b)$ by $[a,b]$, define $F=R/\equiv$ and equip $F$ with the following addition and multiplication (if we pretend that $[a,b]$ is the fraction $a/b$, then these are just the familiar formulas):
$$[a,b]+[c,d]=[ad+bc,bd] \text{ and } [a,b][c,d]=[ac,bd].
$$
It's easy to show that addition and multiplication are well-defined and $F$ is a commutative ring. We can find that the family $R'=\{[a,1];a\in R\}$ is a subring of $F$ and we identify $a\in R$ with $[a,1]\in R'$. To see that $F$ is a field, observe that if $[a,b]\ne [0,1]$, then $a\ne 0$, and the inverse of $[a,b]$ is $[b,a]$.

Finally if $b\ne 0$, then $[1,b]=[b,1]^{-1}$, and so $[a,b]=[a,1][b,1]^{-1}$. $\square$

Definition 1.0.11 The field $F$ constructed from $R$ in Theorem-1.0.1 is called the fraction field of $R$; we denote it by
$$\operatorname{Frac}(R),
$$
and we denote $[a,b]\in \operatorname{Frac}(R)$ by $a/b$; in particular, the elements $[a,1]$ of $F$ are denoted by $a/1$ or, more simply, by $a$.

We can find that the fraction field of $\mathbb Z$ is $\mathbb Q$.

Definition 1.0.12 A subfield of a field $K$ is a subring $k$ of $K$ that is also a field.

It's easy to see that a subset $k$ of a field $K$ is a subfield if and only $k$ is a subring that is closed under inverses; that is, if $a\in k^\times$, then $a^{-1}\in k^\times$. It is also routine to see that any intersection of subfields of $K$ is itself a subfield of $K$.

1.1 Polynomials

Definition 1.1.1 If $R$ is a commutative ring, then a formal power series over $R$ is a sequence of elements $s_i\in R$ for all $i\ge 0$, called the coefficients of $\sigma$:
$$\sigma = (s_0,s_1,s_2,\dots,s_i,\dots).
$$
We can consider $\sigma$ as a function from $\mathbb N$ to $R$. Denote $R[[x]]$ as the set of all formal power series over $R$.

Let $\sigma=(s_0,s_1,s_2,\dots,s_i,\dots)$ and $\tau=(t_0,t_1,t_2,\dots,t_i,\dots)$ be formal power series.

We define two binary operations, addition and multiplication, on $R[[x]]$.
$$\sigma +\tau := (s_0+t_0,s_1+t_1,\dots,s_i+t_i,\dots);\\
\sigma\cdot \tau := (s_0t_0,s_0t_1+s_1t_0,\dots,\sum_{j=0}^is_jt_{i-j},\dots).
$$

Definition 1.1.2 A polynomial over a commutative ring $R$ is a formal power series $\sigma$ over $R$ for which there exists an integer $n\ge 0$ with $\sigma(i)=0,\forall i>n$.

A polynomial has only finitely many nonzero coefficients. The zero polynomial, denoted by $\sigma = 0$, is the sequence $\sigma = (0,0,0,\dots)$.

Definition 1.1.3 If $\sigma=(s_0,s_1,\dots,s_n,0,0,\dots)$ is a nonzero polynomial, then there is $n\ge 0$ with $s_n\ne 0$ and $s_i=0,\forall i>n$. We call $s_n$ the leading coefficient of $\sigma$, we call $n$ the degree of $\sigma$, and we denote the degree by
$$n = \deg(\sigma)
$$
If the leading coefficient $s_n=1$, the $\sigma$ is called monic.

Specifically, the zero polynomial $0$ does not have a degree because it has no nonzero coefficients.

Denote $R[x]$ as the set of all polynomials over $R$. It's trivial that $R[x]\subset R[[x]]$.

Proposition 1.1.1 If $R$ is a commutative ring, then $R[[x]]$ is a commutative ring that contains $R[x]$ and $R'$ as subrings, where $R'=\{(r,0,0,\dots);r\in R\}$.

Lemma 1.1.1 Let $R$ be a commutative ring and let $\sigma,\tau\in R$ be nonzero polynomials.

• Either $\sigma\tau=0$ or $\deg(\sigma\tau)\le \deg(\sigma)+\deg(\tau)$.
• If $R$ is a domain, then $\sigma\tau\ne 0$ and $\deg(\sigma\tau)=\deg(\sigma)+\deg(\tau)$.
• If $R$ is a domain, $\sigma,\tau\ne 0$ and $\tau\mid\sigma$ in $R[x]$, then $\deg(\tau)\le \deg(\sigma)$.
• If $R$ is a domain, then $R[x]$ is a domain.

Definition 1.1.4 Let $R$ be a commutative ring. The indeterminate $x\in R[x]$ is
$$x=(0,1,0,0,\dots).
$$
After defining indeterminate, we can denote a formal power series as $s_0+s_1x+s_2x^2+\cdots+s_ix^i+\cdots$.

Now we can describe the usual role of $x$ in $f(x)$ as a variable. Each polynomial $f(x)=s_0+s_1x+s_2x^2+\cdots+s_nx^n\in R[x]$ defines a polynomial function
$$f^\flat:R\to R
$$
by evaluation: If $a\in R$, $f^\flat(a)=s_0+s_1a+s_2a^2+\cdots+s_na^n\in R$. It should be realized that polynomial and polynomial function are distinct objects.

Sometimes, we write $f^\flat$ as $f$.

Definition 1.1.5 Let $K$ be a field. The fraction field $\operatorname{Frac}(K[x])$ of $K[x]$, denoted by $K(x)$, is called the field of rational functions over $K$.

Proposition 1.1.2 If $K$ is a field, then the elements of $K(x)$ have the form $f(x)/g(x)$, where $f(x),g(x)\in K[x]$ and $g(x)\ne 0$.

We usually call $R[x]$ the ring of all polynomials over $R$ in one variable, but also there exist polynomials over $R$ in more than one variables.

Definition 1.1.6 Let $R$ be a commutative ring, $R[x_1,x_2,\dots,x_n]$ is the ring of polynomials over $R$ in $n$ variables. When $n\ge 2$,
$$R[x_1,x_2,\dots,x_n]:=(R[x_1,x_2,\dots,x_{n-1}])[x_n].
$$
Moreover, when $K$ is a field, we can describe $\operatorname{Frac}(K[x_1,x_2,\dots,x_n])$ as all rational functions in $n$ variables
$$K(x_1,x_2,\dots,x_n).
$$

1.2 Homomorphisms

Homomorphisms allow us to compare rings.

Definition 1.2.1 If $A$ and $R$ are (not necessarily commutative) rings, a (ring) homomorphism is a function $\varphi :A\to R$ such that

• $\varphi(1)=1$,
• $\varphi (a+a')=\varphi(a)+\varphi(a')$,
• $\varphi(aa')=\varphi(a)\varphi(a')$.

A ring homomorphism that is also a bijection is called an isomorphism. Rings $A$ and $R$ are called isomorphic, denoted by
$$A\cong R,
$$
if there is an isomorphism $\varphi:A\to R$.

Theorem 1.2.1 Let $R$ and $S$ be commutative rings, and let $\varphi:R\to S$ be a homomorphism, and let $\varphi:R\to S$ be a homomorphism. If $s_1,\dots,s_n\in S$, then there exists a unique homomorphism
$$\Phi :R[x_1,\dots,x_n]\to S
$$
with $\Phi(x_i)=s_i$ for all $i$ and $\Phi (r)=\varphi(r)$ for all $r\in R$.

By using mathematical induction, it's easy to prove.

Definition 1.2.2 If $R$ is a commutative ring and $a\in R$, then evaluation at $a$ is the function $e_a:R[x]\to R$, defined by $e_a(f(x))=f^\flat(a)$; that is, $e_a(\sum_{i}r_ix^i)=\sum_i r_ia^i$.

Corollary 1.2.1 If $R$ and $S$ are commutative rings and $\varphi: R\to S$ is a homomorphism, then there is a homomorphism $\varphi_*:R[x]\to S[x]$ given by
$$\varphi_*:r_0+r_1x+r_2x^2+\cdots\mapsto \varphi(r_0)+\varphi(r_1)x+\varphi(r_2)x^2+\cdots.
$$
Moreover, $\varphi_*$ is a isomorphism if $\varphi$ is.

Proposition 1.2.1 Let $\varphi:A\to R$ be a homomorphism.

• $\varphi(a^n)=\varphi(a)^n,\forall a\in A,n\in \mathbb{N}$.
• If $a\in A$ is a unit, then $\varphi(a)$ is a unit and $\varphi(a^{-1})=\varphi(a)^{-1}$, and so $\varphi(U(A))\subset U(R)$. Moreover, if $\varphi$ is an isomorphism, then $U(A)\cong U(R)$.

Definition 1.2.3 If $\varphi:A\to R$ is a homomorphism, then its kernel is
$$\ker \varphi = \{a\in A;\varphi(a)=0\}
$$
and its image is
$$\operatorname{im}\varphi = \{\varphi(a);a\in A\}.
$$
Definition 1.2.4 An ideal in a commutative ring $R$ is a subset $I$ of $R$ such that

• $0\in I$
• if $a,b\in I$, then $a+b\in I$,
• if $a\in I$ and $r\in R$, then $ra\in I$.

The ring $R$ itself and $(0)$, the subset consisting $0$ alone, are always ideals in a commutative ring $R$. An ideal $I\ne R$ is called a proper ideal.

Proposition 1.2.2 If $\varphi:A\to R$ is a homomorphism, then $\ker \varphi$ is an ideal in $A$ and $\operatorname{im} \varphi$ is a subring of $R$. Moreover, if $A$ and $R$ are not zero rings, then $\ker\varphi$ is a proper ideal.

Proposition 1.2.3 A homomorphism $\varphi:A\to R$ is an injection if and only if $\ker \varphi=(0)$.

Definition 1.2.5 If $b_1,b_2,\dots,b_n$ lie in $R$, then the set of all linear combinations
$$I=\{r_1b_1+r_2b_2+\dots+r_nb_n;r_i\in R,\text{for all }i\}
$$
is an ideal in $R$. We write $I=(b_1,b_2,\dots,b_n)$ in this case, and we call $I$ the ideal generated by $b_1,b_2,\dots,b_n$. In particular, if $n=1$, then
$$I=(b)=\{rb;r\in R\}
$$
is an ideal in $R$. The ideal $(b)$ (often denoted by $Rb$), consisting of all the multiples of $b$, is called principal ideal generated by $b$.

Theorem 1.2.2 Every ideal $I$ in $\mathbb Z$ is a principal ideal; that is, there is $d\in\mathbb Z$ with $I=(d)$.

Proposition 1.2.4 Let $R$ be a commutative ring and let $a,b\in R$. If $a\mid b$ and $b\mid a$, then $(a)=(b)$.

Definition 1.2.6 Elements $a$ and $b$ in a commutative ring $R$ are associates if there exists a unit $u\in R$ with $b=ua$.

Proposition 1.2.5 Let $R$ be a domain and let $a,b\in R$.

• $a\mid b$ and $b\mid a$ if and only if $a$ and $b$ are associates.
• The principal ideals $(a)$ and $(b)$ are equal if and only if $a$ and $b$ are associates.

1.3 Quotient Rings

Definition 1.3.1 Let $I$ be an ideal in a commutative ring $R$. If $a\in R$, then the coset $a+I$ is the subset
$$a+I=\{a+i;i\in I\}.
$$
The coset $a+I$ is often called $a\bmod I$. The family of all cosets is denoted by $R/I$:
$$R/I=\{a+I;a\in R\}
$$
The relation $\equiv$ on $R$, defined by $a\equiv b$ if $a-b\in I$, is called congruence mod $I$; it is an equivalence relation on $R$, and its equivalence classes are the cosets.

Proposition 1.3.1 Let $I$ be an ideal in a commutative ring $R$. If $a,b\in R$, then $a+I=b+I$ if and only if $a-b\in I$. In particular, $a+I=I$ if and only if $a\in I$.

Definition 1.3.2 Let $R$ be a commutative ring and $I$ be an ideal in $R$. Define addition $+:R/I\times R/I\to R/I$ by
$$+:(a+I,b+I)\mapsto a+b+I,
$$
and multiplication $\cdot:R/I\times R/I\to R/I $ by
$$\cdot:(a+I,b+I)\mapsto ab+I
$$
It's easy to prove addition and multiplication $R/I\times R/I\to R/I$ are well-defined.

Theorem 1.3.1 If $I$ is an ideal in a commutative ring $R$, then $R/I$ is a commutative ring.

Definition 1.3.2 The commutative ring $R/I$ equipped with addition and multiplication is called quotient ring of $R$ modulo $I$; it is usually pronounced $R\bmod I$.

Definition 1.3.3 Let $I$ be an ideal in a commutative ring $R$. The natural function is the function $\pi:R\to R/I,a\mapsto a+I$.

Proposition 1.3.2 If $I$ is an ideal in a commutative ring $R$, then the natural map $\pi: R\to R/I$ is a surjective homomorphism and $\ker \pi = I$.

Corollary 1.3.1 Given an ideal $I$ in a commutative ring $R$, there exists a commutative ring $A$ and a (surjective) homomorphism $\varphi:R\to A$ with $I=\ker \varphi$.

If we set $A=R/I$, then the natural map $\pi:R\to R/I$ is a homomorphism with $I=\ker \pi$.

Theorem 1.3.2 (First Isomorphism Theorem) Let $R$ and $A$ be commutative rings. If $\varphi:R\to A$ is a homomorphism, then $\ker \varphi$ is an ideal in $R$, and $\operatorname{im}\varphi$ is a subring of $A$, and
$$R/\ker\varphi\cong \operatorname{im}\varphi
$$
Let $I=\ker \varphi$.

In the commutative diagram below, $\pi:R\to R/I$ is the natural map, $i:\operatorname{im}\varphi\to A$ is the inclusion, and define $\tilde \varphi:R/I\to A,a+I\mapsto \varphi(a)$ ($a+I=b+I\Leftrightarrow b-a\in I\Leftrightarrow \varphi(a)=\varphi(a)+\varphi(b-a)=\varphi(b)$ show that $\tilde\varphi$ is well-defined).
$$% https://darknmt.github.io/res/xypic-editor/#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
\xymatrix{
R \ar@{->}[d]_{\pi} \ar@{->}[r]^{\varphi} & A \\
R/I \ar@{->}[r]_{\tilde\varphi} & \operatorname{im} \varphi \ar@{->}[u]_{i}
}
$$
Definition 1.3.4 If $K$ is a field, the intersection of all the subfields of $K$ is called the prime field of $K$.

For example, the prime field of $\mathbb C$ is $\mathbb Q$, because every subfield of $\mathbb C$ contains $\mathbb Q$.

Definition 1.3.5 If $X$ is a subset of a field, define $\left<X\right>$, the subfield of generated by $X$, to be the intersection of all the subfields containing $X$.

Proposition 1.3.3 Let $K$ be a field with identity $\mathscr l$, and let $\chi:\mathbb Z\to K,n\mapsto n\mathscr l$ be a homomorphism.

1. Either $\operatorname{im}\chi\cong \mathbb Z$ or $\operatorname{im}\chi \cong \mathbb Z_p$ for some prime $p$.
2. The prime field of $K$ is isomorphism to $\mathbb Q$ or to $\mathbb Z_p$ for some prime $p$.

Proof.

1. Since every ideal in $\mathbb Z$ is principal, $\ker \chi = (m)$ for some integer $m\ge 0$.

   The First Isomorphism Theorem gives $\operatorname{im}\chi\cong \mathbb Z/(m)$.

   If $m=0$, $\operatorname{im}\chi\cong\mathbb Z/(0)\cong\mathbb Z$.

   If $m\ne 0$, $\operatorname{im}\chi\cong\mathbb Z/(m)=\mathbb Z_m$. And then we will prove $m$ is a prime.

   Let's suppose $m$ isn't a prime. Then there are two integers $a,b>0$ such that $ab=m$.

   Therefore $\chi(a)\chi(b)=\chi(ab)=0$, which contradicts the fact that $K$ is a domain.

2. If $\operatorname{im} \chi\cong \mathbb Z$. There is a field $Q\cong \operatorname{Frac}(\mathbb Z)=\mathbb Q$ with $\operatorname{im}\chi\subset Q\subset K$. Now $Q$ is the prime field of $K$, for it is the subfield generated by $\mathscr l$.

   In case $\operatorname{im}\chi\cong \mathbb Z_p$, then $\operatorname{im}\chi$ must be the prime field of $K$, for it is a field which is obviously the subfield generated by $\mathscr l$. $\square$

Definition 1.3.6 A field $K$ has characteristic $0$ if its prime field is isomorphic to $\mathbb Q$; it has characteristic $p$ if its prime field is isomorphic to $\mathbb Z_p$ for some prime $p$.

Proposition 1.3.4 If $K$ is a finite field, then $|K|=p^n$ for some prime $p$ and some $n\ge 1$.

Proof.

The prime field of $K$ is isomorphic to $\mathbb Z_p$ for some prime $p$.

We can regard $K$ as a vector space over $\mathbb Z_p$. As $K$ is finite, it is obviously finite-dimensional.

If $\operatorname{dim}_{\mathbb Z_p}(K)=n$, then $|K|=p^n$. $\square$

Proposition 1.3.5 Let $I$ be an ideal in a commutative ring $R$. If $J$ is an ideal in $R$ containing $I$, define the subset $J/I$ of $R/I$ by
$$J/I=\{a+I;a\in I\}.
$$

1. $\pi^{-1}(J/I)=J$, where $\pi:R\to R/I$ is the natural map.
2. $J/I$ is an ideal in $R/I$.
3. If $I\subset J\subset J'$ are ideals in $R$, $J/I\subset J'/I$. Moreover, if $J\ne J'$, then $J/I\ne J'/I$.

Proof.

1. • First, we prove $\pi^{-1}(J/I)\subset J$.

     $\forall a\in \pi^{-1}(J/I),\pi(a)=a+I\in J/I$.

     Then there is $b\in J,b+I=a+I\Rightarrow a-b\in I\subset J$.

     Therefore $a\in J$.

   • Second, we prove $J\subset \pi^{-1}(J/I)$.

     $\forall a\in J,\pi(a)=a+I\in J/I\Rightarrow a\in \pi^{-1}(J/I)$.

2. $\forall a+I\in J/I,\forall b+I\in R/I,(a+I)(b+I)=ab+I$.

   Since $ab\in J\Rightarrow ab+I\in J/I$, $J/I$ is an ideal.

3. $J\subset J'\Rightarrow J/I=\pi(J)\subset \pi(J')=J'/I$.

   Then $J=\pi^{-1}(J/I)\ne \pi^{-1}(J'/I)=J'\Rightarrow J/I\ne J'/I$. $\square$

Proposition 1.3.6 A commutative ring $R$ is a field if and only if $R$ only have trivial ideals, $(0)$ and $R$.

Proof.

• ($\Rightarrow$).

  Let $I$ be an ideal of $R$, with $I\ne (0)$.

  Then there is $a\in I\textbackslash \{0\}$ with $a$ is a unit.

  Because $\forall b\in R,b=(ba^{-1})a\in I$, $I=R$.

• ($\Leftarrow$).

  If $R$ isn't a field, then there is $a\in R$ with $a$ isn't a unit.

  Then $(a)$ is an ideal with $1\notin (a)$, which lead to a contradiction. $\square$

1.4 From Arithmetic to Polynomials

Theorem 1.4.1 (Division Algorithm) If $K$ is a field and $f(x),g(x)\in K[x]$ with $f\ne 0$, then there are unique polynomials $q(x),r(X)\in K[x]$ with
$$g(x)=q(x)f(x)+r(x),
$$
where either $r=0$ or $\deg r(x)<\deg f(x)$.

Definition 1.4.1 If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then the polynomials $q(x)$ and $r(x)$ occurring in the Division Algorithm are called the quotient and the remainder after dividing $g$ by $f$.

Corollary 1.4.1 Let $R$ be a commutative ring, and let $f(x)\in R[x]$ be a monic polynomial. If $g(x)\in R[x]$, then there exist $q(x),r(x)\in R[x]$ with
$$g(x)=q(x)f(x)+r(x),
$$
where either $r=0$ or $\deg r(x)<\deg f(x)$.

Theorem 1.4.2 If $K$ is a field, then every ideal $I$ in $K[x]$ is a principal ideal; that is, there is $d\in I$ with $I=(d)$, then $d$ can be chosen to be a monic polynomial. Moreover, if $I\ne (0)$, then $d$ can be chosen to be a monic polynomial.

Definition 1.4.2 If $f(x)\in R[x]$, where $R$ is a ring, then a root of $f(x)$ in $R$ is an element $a\in R$ with $f(a)=0$.

Lemma 1.4.1 Let $f(x)\in K[x]$, where $K$ is a field, and let $u\in K$. Then there is $q(x)\in K[x]$ with
$$f(x)=q(x)(x-u)+f(u).
$$
Proposition 1.4.1 If $f(x)\in K[x]$, where $K$ is a field, then $a$ is a root of $f(x)$ in $K$ if and only if $x-a\mid f(x)$.

Proposition 1.4.2 Let $K$ be a field and let $f(x)\in K[x]$. If $f$ has degree $n$, then $f$ has at most $n$ roots in $K$.

Corollary 1.4.2 Every $n$th root of unity in $\mathbb C$ is equal to
$$e^{2\pi ik/n}=\cos(2\pi k/n)+i\sin(2\pi k/n),
$$
where $k=0,1,2,\dots,n-1$.

Corollary 1.4.3 Let $K$ be an infinite field and let $f(x)$ and $g(x)$ be polynomials in $K[x]$. If $f$ and $g$ determine the same polynomial function (that is, $f(a)=g(a)$ for all $a\in K$), then $f=g$.

Proof.

Let $h(x)=f(x)-g(x)$. Then if $h\ne 0$, $h$ has some degree.

However, the roots of $h$ are too much, which contradicts the Proposition 1.4.2. $\square$

Corollary 1.4.4 Let $K$ be a (possibly finite) field, let $f(x),g(x)\in K[x]$, and let $\deg f(x)\le\deg g(x)=n$. If $f(a)=g(a)$ for $n+1$ elements $a\in K$, then $f=g$.

Proof.

Let $h(x)=f(x)-g(x)$. Then if $h\ne 0$, $\deg h(x)\le n$.

That $h(x)$ have $n+1$ roots contradicts the Proposition 1.4.2. $\square$

Proposition 1.4.3 Let $f(X),g(X)\in K[X]=K[x_1,\dots,x_n]$, where $K$ is an infinite field.

1. If $f(X)$ is nonzero, then there are $a_1,\dots,a_n\in K$ with $f(a_1,\dots,a_n)\ne 0$.
2. If $f(a_1,\dots,a_n)=g(a_1,\dots,a_n)$ for all $(a_1,\dots,a_n)\in K^n$, then $f=g$.

Theorem 1.4.3 Let $K$ be a field. If $G$ is a finite subgroup of the multiplicative group $K^\times$, then $G$ is cyclic. In particular, if $K$ is finite, then $K^\times$ is cyclic.

Proof.

$G$ is an abelian group and $x^m-1$ have less than $m$ roots in $K$ (also in $G$).

For 命题1.1.7 in 抽象代数笔记, $G$ is cyclic. $\square$

Definition 1.4.3 If $K$ is a finite field, a generator of the cyclic group $K^\times$ is called a primitive element of $K$.

Definition 1.4.4 If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then a common divisor is a polynomial $c(x)\in K[x]$ with $c\mid f$ and $c\mid g$. If $f$ and $g$ in $K[x]$ are not both $0$, define their greatest common divisor abbreviated $\gcd$, to be the monic common divisor having largest degree. If $f=0=g$, define $\gcd(f,g)=0$.

Theorem 1.4.4 If $K$ is a field and $f(x),g(x)\in K[x]$, then their gcd $d(x)$ is a linear combination of $f$ and $g$; that is there are $s(x),t(x)\in K[x]$ with
$$d=sf+tg.
$$
Proof.

The set $(f,g)$ of all linear combinations of $f$ and $g$ is an ideal in $K[x]$.

Since every ideal in $K[x]$ is a principle ideal, there is $d(x)\in K[x]$ such that $(d)=(f,g)$.

Then there are $s(x),t(x)\in K[x]$ with $d=sf+tg$.

Because $f(x),g(x)\in (d)$, $d$ is a common divisor of $f$ and $g$.

If $q(x)$ is a common divisor of $f$ and $g$, then there are $f_1(x),g_1(x)\in K[x]$ with $f=f_1q,g=g_1q$.

$d=sf+tg=(sf_1+tg_1)q$ shows that $q$ is a divisor of $d$, and $\deg q(x)\le\deg d(x)$.

Therefore, $d$ is the gcd of $f$ and $g$. $\square$

Corollary 1.4.5 Let $K$ be a field and let $f(x),g(x)\in K[x]$.

1. A monic common divisor $d(x)$ is the gcd if and only if $d$ is divisible by every common divisor; that is, if $h(x)$ is a common divisor, then $h\mid d$.
2. $f$ and $g$ have a unique gcd.

Definition 1.4.5 An element $p$ in a domain $R$ is irreducible if $p$ is neither $0$ nor a unit and, in every factorization $p=uv$ in $R$, either $u$ or $v$ is a unit.

For example, a prime $p\in \mathbb Z$ is irreducible element, as is $-p$.

Proposition 1.4.4 If $K$ is a field, then a polynomial $p(x)\in K[x]$ is irreducible if and only if $\deg(p)=n\ge 1$ and there is no factorization in $K[x]$ for the form $p(x)=g(x)h(x)$ in which both factors have degree smaller than $n$.

Corollary 1.4.6 Let $K$ be a field and let $f(x)\in K[x]$ be a quadratic or cubic polynomial. Then $f$ is irreducible in $K[x]$ if and only if $f$ has no roots in $K$.

Theorem 1.4.5 (Gauss's Lemma) Let $f(x)\in \mathbb Z[x]$. If $f(x)=G(x)H(x)$ in $\mathbb Q[x]$, where $\deg(G),\deg(H)<\deg(f)$, then $f(x)=g(x)h(h)$ in $\mathbb Z[x]$, where $\deg(g)=\deg(G)$ and $\deg(h)=\deg(H)$.

Proof.

Clearing denominators, there are positive integers $n',n''$ such that $g(x)=n'G(x)$ and $h(x)=n''H(x)$.

Setting $n=n'n''$, we have
$$nf(x)=n'G(x)n''H(x)=g(x)h(x)\text{ in }\mathbb Z[x].
$$
If $p$ is a prime divisor of $n$, consider the map $\mathbb Z[x]\to \mathbb Z_p[x]$, denoted by $g\to \overline g$, which reduces all coefficients mod $p$. The equation becomes
$$0=\overline g(x)\overline h(x).
$$
But $\mathbb Z_p[x]$ is a domain, and so at least one of these factors is $0$.

Let's assume that $\overline g(x)=0$; that is all the coefficients of $g(x)$ are multiples of $p$.

Therefore, we may write $g(x)=pg'(x)$, where all the coefficients of $g'(x)$ lie in $\mathbb Z$. If $n=pm$, then
$$pmf(x)=pg'(x)h(x)\text{ in } \mathbb Z.
$$
Cancel $p$, and continue canceling primes until we reach a factorization $f(x)=g^*(x)h^*(x)$ in $\mathbb Z[x]$. $\square$

The contrapositive version of Gauss's Lemma is more convenient to use. If $f(x)\in \mathbb Z[x]$ has no factorization in $\mathbb Z[x]$ as a product of two polynomials, each having degree smaller than $\deg(f)$, then $f$ is irreducible in $\mathbb Q[x]$.

Lemma 1.4.2 Let $K$ be a field, let $p(x),f(x)\in K[x]$, and let $d(x)=\gcd(p,f)$. If $p$ is a monic irreducible polynomial, then
$$d(x)=\begin{cases}
1,&p\nmid f,\\
p(x),&p\mid f.
\end{cases}
$$

Proof. Since $d\mid p$, we have $d=1$ or $d=p$. $\square$

Theorem 1.4.6 (Euclid's Lemma) Let $K$ be a field and let $f(x),g(x)\in K[x]$. If $p(x)$ is an irreducible polynomial in $K[x]$, and $p\mid fg$, then either
$$p\mid f\text{ or }p\mid g.
$$
More generally, if $p\mid f_1(x)\cdots f_n(x)$, then $p\mid f_i$ for some $i$.

Proof.

Assume that $p\mid fg$ but that $p\nmid f$.

Since $p$ is irreducible, $\gcd(p,f)=1$, and so $1=sp+sf$ for some polynomials $s$ and $t$.

Therefore $g=spg+sfg$. Because $p\mid fg$, $fg=hp$ for some polynomial $h$.

Then $g=spg+shp=(sg+sh)p$, so $p\mid g$. $\square$

Definition 1.4.6 Two polynomials $f(x),g(x)\in K[x]$, where $K$ is a field, are called relatively prime if their gcd is $1$.

Corollary 1.4.7 Let $f(x),g(x),h(x)\in K[x]$, where $K$ is a field, and let $h$ and $f$ be relatively prime. If $h\mid fg$, then $h\mid g$.

Definition 1.4.7 If $K$ is a field, then a retaional function $f(x)/g(x)\in K(x)$ is in lowest terms if $f(x)$ and $g(x)$ are relatively prime.

Theorem 1.4.7 (Euclidean Algorithms) If $K$ is a field $f(x),g(x)\in K[x]$, then there are algorithms for computing $\gcd(f,g)$, as well as for finding a pair of polynomials $s(x)$ and $t(x)$ with
$$\gcd(f,g)=sf+tg.
$$
Corollary 1.4.8 Let $k$ be a subfield of a field $K$, so that $k[x]$ is a subring of $K[x]$. If $f(x),g(x)\in k[x]$, then their gcd in $k[x]$ is equal to their gcd in $K[x]$.

Corollary 1.4.9 If $f(x),g(x)\in \mathbb R[x]$ have no common root in $\mathbb C$, then $f,g$ are relatively prime in $\mathbb R[x]$.

Theorem 1.4.8 (Unique Factorization) If $K$ is a field, then every polynomial $f(x)\in K[x]$ of degree $\ge1$ is a product of a nonzero constant and monic irreducibles. Moreover, if $f(x)$ has two such factorizations,
$$f(x)=ap_1(x)\cdots p_m(x) \text{ and } f(x)=bq_1(x)\cdots q_n(x)
$$
that is, $a$ and $b$ are nonzero constants and the $p$'s and $q$'s are monic irreducibles, then $a=b,m=n$ and the $q$'s may be reindexed so that $q_i=p_i$ for all $i$.

Definition 1.4.8 Let $f(x)\in K[x]$, where $K$ is a field. A prime factorization of $f(x)$ is
$$f(x)=ap_1(x)^{e_1}\cdots p_m(x)^{e_m},
$$
where $a$ is a nonzero constant, the $p_i$ are distinct monic irreducible polynomials, and $e_i\ge 0$ for all $i$.

Let $K$ be a field, and assume that there are $a,r_1,\dots,r_n\in K$ with
$$f(x)=a\prod_{i=1}^n(x-r_i);
$$
we say that $f$ splits over $k$. If $r_1,\dots,r_s$, where $s\le n$, are the distinct roots of $f(x)$, then a prime factorization of $f(x)$ is
$$f(x)=a(x-r_1)^{e_1}(x-r_2)^{e_2}\cdots(x-r_s)^{e_s}.
$$
We call $e_j$ the multiplicity of the root $r_j$.

Definition 1.4.9 If $f$ and $g$ are elements in a commutative ring $R$, then a common multiple is an element $m\in R$ with $f\mid m$ and $g\mid m$. If $f$ and $g$ in $R$ are not both $0$, define their last common multiple, abbreviated $\operatorname{lcm}(f,g)$, to be a common mutiple $c$ of them with $c\mid m$ for every common multiple $m$. If $f=0=g$, define their $\operatorname{lcm}(f,g)=0$, If $R=K[x]$, we require lcm's to be monic.

Proposition 1.4.5 If $K$ is a field and $f(x),g(x)\in K[x]$ have prime factorizations $f(x)=a_0p_1^{a_1}\cdots p_n^{a_n}$ and $g(x)=b_0p_1^{b_1}\cdots p_n^{b_n}$ in $K[x]$, then

1. $f\mid g$ if and only if $a_i\le b_i,\forall i$.

2. If $m_i=\min\{a_i,b_i\}$ and $M_i=\max\{a_i,b_i\}$, then
   $$
   \gcd(f,g)=p_1^{m_1}\cdots p_n^{m_n}\text{ and }\operatorname{lcm}(f,g)=p_1^{M_1}\cdots p_n^{M_n}.
   $$

Corollary 1.4.10 If $K$ is a field and $f(x),g(x)\in K[x]$ are monic polynomials, then
$$\gcd(f,g)\operatorname{lcm}(f,g)=fg.
$$

1.5 Maximal Ideals and Prime Ideals

Definition 1.5.1 An ideal $I$ in a commutative ring $R$ is called a maximal ideal if $I$ is a proper ideal for which there is no proper ideal $J$ with $I\subsetneq J$.

Proposition 1.5.1 A proper ideal $I$ in a commutative ring $R$ is a maximal ideal if and only if $R/I$ is a field.

Proof.

• ($\Rightarrow$).

  Since $I$ is a maximal ideal, then $I/I=(0)$ is a maximal ideal in $R/I$ for Proposition1.3.5.

  Then because a commutative ring only having trivial ideals is a field, $R/I$ is a field.

• ($\Leftarrow$).

  Since $R/I$ is a field, $I/I$ is a maximal ideal in $R/I$.

  Because of Proposition1.3.5, $I$ is a maximal ideal in $R$. $\square$

Example 1.5.1

• If $p$ is a prime number, then $(p)$ is a maximal ideal in $\mathbb Z$, for $\mathbb Z_p$ is a field.
• If $K$ is a field, then $(x)$ is a maximal ideal in $K[x]$, for $K[x]/(x)\cong K$.
• $(x^2+1)$ is a maximal ideal in $\mathbb R[x]$, for $\mathbb R[x]/(x^2+1)\cong \mathbb C$.

Proposition 1.5.2 If $K$ is a field, then $I=(x_1-a_1,\dots,x_n-a_n)$ is a maximal ideal in $K[x_1,\dots,x_n]$ whenever $a_1,\dots,a_n\in K$.

Definition 1.5.2 An ideal $I$ in a commutative ring $R$ is called a prime ideal if $I$ is a proper ideal such that $ab\in I\Rightarrow (a\in I\lor b\in I)$.

Proposition 1.5.3 If $I$ is a proper ideal in a commutative ring $R$, then $I$ is a prime ideal if and only if $R/I$ is a domain.

Proof.

• ($\Rightarrow$).

  If $(a+I)(b+I)=0+I$, then $ab\in I\Rightarrow (a\in I\lor b\in I)\Rightarrow (a+I=0+I\lor b+I=0+I)$.

• ($\Leftarrow$).

  If $ab\in I$, then $(a+I)(b+I)=0+I\Rightarrow (a+I=0+I\lor b+I=0+I)\Rightarrow (a\in I\lor b\in I)$. $\square$

Corollary 1.5.1 Every maximal ideal is a prime ideal.

Definition 1.5.3 If $I$ and $J$ are ideals in a commutative ring, then
$$IJ:=\{\text{all finite sums}\sum_{\mathscr l}a_{\mathscr l}b_\mathscr l;a_{\mathscr l}\in I,b_\mathscr l\in J\}
$$
It's easy to see that $IJ$ is an ideal in $R$.

Proposition 1.5.4 Let $P$ be a prime ideal in a commutative ring $R$. If $I$ and $J$ are ideals with $IJ\subset P$, then $I\subset P$ or $J\subset P$.

Proof.

If $I\not\subset P$ and $J\not\subset P$, then there are $a\in I$ and $b\in J$ with $a,b\notin P$.

But $ab\in IJ\subset P$, contradicting $P$ being prime. $\square$

Proposition 1.5.5 If $K$ is a field and $I=(f)$, where $f(x)$ is a nonzero polynomial in $K[x]$, then the following are equivalent:

1. $f$ is irreducible;

2. $K[x]/I$ is a field;

3. $K[x]/I$ is a domain.

Proof.

• (1) $\Rightarrow$ (2).

  If there is an ideal $J$ on $K[x]$, with $I\subset J$.

  Since every ideal in $K[x]$ is principle, there is $d(x)\in K[x]$ such that $J=(d)$.

  Then because $f\in(f)\subset (d)$, there is $g(x)\in K[x]$ with $f=gd$.

  Then $d$ is a unit or $g$ is a unit for $f$ is irreducible.

  Therefore, $(d)=K[x]$ or $(d)=(f)$. $I$ is a maximal ideal.

• (2) $\Rightarrow$ (3).

  Every field is a domain.

• (3) $\Rightarrow$ (1).

  Since $K[x]/I$ is a domain, $I$ is a prime ideal in $K[x]$.

  If some polynomials $g(x),h(x)\in K[x]$ such that $f=gh$, then $gh=f\in I\Rightarrow (g\in I\lor h\in I)$.

  Since $\deg(g)\le \deg(f)$ and $\deg(h)\le \deg(f)$, $g$ is a unit or $h$ is a unit. $\square$

Proposition 1.5.6 Let $k$ be a field, let $p(x)$ be a monic irreducible polynomial in $k[x]$ of degree $d$, let $K=k[x]/I$, where $I=(p)$, and let $\beta=x+I\in K$. Then:

1. $K$ is a field and $k'=\{a+I;a\in k\}$ is a subfield of $K$ isomorphic to $k$.
2. $\beta$ is a root of $p$ in $K$.
3. If $g(x)\in k[x]$ and $\beta$ is a root of $g$ in $K$, then $p\mid g$ in $k[x]$.
4. $p$ is the unique monic irreducible polynomial in $k[x]$ having $\beta$ as a root.
5. The list $1,\beta,\beta^2,\dots,\beta^{d-1}$ is a basis of $K$ as a vector space over $k$, and so $\dim_{k}(K)=d$.

Proof.

1. Trivial.

2. Let $p(x)=a_0+a_1x+\cdots+a_{d-1}x^{d-1}+x^d$, where $a_i\in k$ for all $i$. In $K=k[x]/I$, we have
   $$ \begin{align}
   p(\beta)&=(a_0+I)+(a_1+I)\beta+\cdots+(1+I)\beta^d\\
   &=(a_0+I)+(a_1+I)(x+I)+\cdots+(1+I)(x+I)^d\\
   &=(a_0+I)+(a_1x+I)+\cdots+(1x^d+I)\\
   &=a_0+a_1x+\cdots+x^d+I\\
   &=p(x)+I=0+I.
   \end{align}
   $$

3. We have $g(x)=h(x)p(x)+r(x)$ in $k[x]$ with $r=0\lor\deg(r)<\deg(p)$ for the Division Algorithm.

   Then $0+I=g(\beta)=h(\beta)p(\beta)+r(\beta)=r(\beta)$; that is $r(x)\in I\Rightarrow r=0$.

   Therefore $p\mid g$.

4. (3) $\Rightarrow$ (4) is trivial.

5. If $a_0+a_1\beta+\cdots+a_{d-1}\beta^{d-1}=0$, letting $g(x)=a_0+a_1x+\cdots+a_{d-1}x^{d-1}$, then $g=0$ for (3). $\square$

Definition 1.5.4 If $K$ is a field containing $k$ as a subfield, then $K$ is called an extension field of $k$, and we denote an extension field by
$$K/k.
$$
An extension field $K/k$ is a finite extension if $K$ is a finite-dimensional vector space over $k$. The dimension of $K$, denoted by
$$[K:k],
$$
is called the degree of $K/k$.

Definition 1.5.5 Let $K/k$ be an extension field. An element $a\in K$ is algebraic over $k$, if there is some nonzero polynomial $f(x)\in k[x]$ having $\alpha$ as a root; otherwise, $\alpha$ is transcendental over $k$. An extension field $K/k$ is algebraic if every $\alpha\in K$ is algebraic over $k$.

Proposition 1.5.7 If $K/k$ is a finite extension field, then $K/k$ is an algebraic extension.

Proof.

Let $n=\dim_k(K)$, for all $\beta\in K$, then $1,\beta,\dots,\beta^n$ are dependent.

Therefore, there are $a_0,a_1,\dots,a_n\in k$ such that $a_0+a_1\beta+\dots+a_n\beta^n=0$. $\square$

Definition 1.5.6 If $K/k$ is an extension field and $\alpha\in K$, then
$$k(\alpha)
$$
is the intersection of all those subfields of $K$ containing $k$ and $\alpha$; we call $k(\alpha)$ the subfield of $K$ obtained by adjoining $\alpha$ to $k$.

More generally, if $A$ is a (possibly infinite) subset of $K$, define $k(A)$ to be the intersection of all the subfields of $K$ containing $k\cup A$; we call $k(A)$ the subfield of $K$ obtained by adjoining $A$ to $k$. In particular, if $A=\{z_1,\dots,z_n\}$ is a finite subset, then we may denote $k(A)$ by $k(z_1,\dots,z_n)$.

Theorem 1.5.1

1. If $K/k$ is an extension field and $\alpha\in K$ is algebraic over $k$, then there is a unique monic irreducible polynomial $p(x)\in k[x]$ having $\alpha$ as a root. Moreover, if $I=(p)$, then $k[x]/I\cong k(\alpha)$; indeed, there exists an isomorphism
   $$
   \varphi:k[x]/I\to k(\alpha)
   $$
   with $\varphi(x+I)=\alpha$ and $\varphi(c+I)=c,\forall c\in k$.

2. If $\alpha'\in K$ is another root of $p(x)$, then there is an isomorphism
   $$
   \theta:k(\alpha)\to k(\alpha')
   $$
   with $\theta(\alpha)=\alpha'$ and $\theta(c)=c,\forall c\in k$.

Definition 1.5.7 If $K/k$ is an extension field and $\alpha\in K$ is algebraic over $k$, then the unique monic irreducible polynomial $p(x)\in k[x]$ having $\alpha$ as a root is called the minimal polynomial of $\alpha$ over $k$; it is denoted by
$$\operatorname{irr}(\alpha,k)=p(x).
$$
Theorem 1.5.2 Let $k\subset E\subset K$ be fields, with $E$ a finite extension field of $k$ and $K$ a finite extension field of $E$. Then $K$ is a finite extension field of $k$ and
$$[K:k]=[K:E][E:k].
$$
Proof.

Let $n=[K:E],m=[E:k]$.

Then let $\alpha_1,\dots,\alpha_n$ be a base of vector space $K$ over $E$, and $\beta_1,\dots,\beta_m$ be a base of vector space $E$ over $k$.

For all $\alpha\in K$, there are $b_1,\dots,b_n\in E$ with $b_1\alpha_1+\cdots+b_n\alpha_n=\alpha$.

And for all $i(1\le i\le n)$, there are $a_{i1},\dots,a_{im}$ such that $b_i=a_{i1}\beta_1+\cdots +a_{im}\beta_m$.

Therefore $\alpha=\sum_{i,j} a_{ij}\alpha_i\beta_j$, which means $\{\alpha_i\beta_j;1\le i\le n,1\le j\le m\}$ is a base of vector space $K$ over $k$.

Then let's prove $\{\alpha_i\beta_j;1\le i\le n,1\le j\le m\}$ is independent.

Suppose that $\sum_{i,j}a_{ij}\alpha_i\beta_j=0$, and then $\sum_{i} (\sum_j a_{ij}\beta_j)\alpha_i=0$.

Since $\alpha_1,\dots,\alpha_n$ is a base, $\sum_ja_{ij}\beta_j=0$ for all $i$.

Because $\beta_1,\dots,\beta_m$ is also a base, $a_{ij}=0$ for all $i,j$. $\square$

Example 1.5.2 Let $f(x)=x^4-10x^2+1\in \mathbb Q[x]$. If $\beta$ is a root of $f$, then the quadratic formula gives $\beta^2=5\pm 2\sqrt 6$. But the identity $a+2\sqrt{ab}+b=(\sqrt a+\sqrt b)^2$ gives $\beta=\pm (\sqrt 2+\sqrt 3)$. Similarly, $5-2\sqrt{6}=(\sqrt 2-\sqrt 3)^2$, so that the roots of $f$ are
$$\sqrt 2+\sqrt 3,-\sqrt{2}-\sqrt 3,\sqrt 2-\sqrt 3,-\sqrt 2+\sqrt 3.
$$
We claim that $f$ is irreducible in $\mathbb Q[x]$. If $g$ is a quadratic factor of $f$ in $\mathbb Q[x]$, then
$$g(x)=(x-a\sqrt 2-b\sqrt 3)(x-c\sqrt 2-d\sqrt 3),
$$
there $a,b,c,d\in \{1,-1\}$. Multiplying,
$$g(x)=x^2-\left((a+c)\sqrt 2+(b+d)\sqrt 3\right)x+2ac+3bd+(ad+bc)\sqrt 6.
$$
We check easily that $(a+c)\sqrt 2+(b+d)\sqrt 3$ is rational if and only if $a+c=0=b+d$; but these equations force $ad+bc\ne 0$, and so the constant term of $g$ is not rational. Therefore, $g\notin \mathbb Q[x]$, and so $f$ is irreducible in $\mathbb Q[x]$. If $\beta = \sqrt 2+\sqrt 3$, then $f(x)=\operatorname{irr}(\beta,\mathbb Q)$.

Consider the field $E=\mathbb Q(\beta)=\mathbb Q(\sqrt 2+\sqrt 3).$ There is a tower of fields $\mathbb Q\subset E\subset F$, where $F=\mathbb Q(\sqrt 2,\sqrt 3)$, and so
$$[F:\mathbb Q]=[F:E][E:\mathbb Q].
$$
Since $E=\mathbb Q$ and $\beta$ is a root of an irreducible polynomial of degree $4$, namely, $f$, we have $[E:\mathbb Q]=4$.

On the other hand,
$$[F:\mathbb Q]=[F:\mathbb Q(\sqrt 2)][\mathbb Q(\sqrt 2):\mathbb Q].
$$
Now $[\mathbb Q(\sqrt2):\mathbb Q]=2$, because $\sqrt 2$ is a root of the irreducible quadratic $x^2-2$ in $\mathbb Q[x]$. We claim that $[F:\mathbb Q(\sqrt 2)]\le 2$. The field $F$ arises by adjoining $\sqrt 3$ to $\mathbb Q(\sqrt 2)$; either $\sqrt 3\in \mathbb Q(\sqrt 2)$, in which case the degree is $1$, or $x^2-3$ is irreducible in $\mathbb Q(\sqrt 2)[x]$, in which case the degree is $2$ (in fact, the degree is $2$). It follows that $[F:\mathbb Q]\le 4$, and so the equation $[F:\mathbb Q]=[F:E][E:\mathbb Q]$ gives $[F:E]=1$; that is $F=E$.

Let us note that $F$ arises from $\mathbb Q$ by adjoining all the roots of $f$, but it also arises from $\mathbb Q$ by adjoining all the roots of the reducible polynomial $g(x)=(x^2-2)(x^2-3)$.

1.6 Finite Fields

Theorem 1.6.1 (Kronecker) If $k$ is a field and $f(x)\in k[x]$, there exists an extension field $K/k$ with $f$ a product of linear polynomials in $K[x]$.

Proof.

The proof is by induction on $\deg(f)$.

• If $\deg(f)=1$, then $f$ is linear and we can choose $K=k$.

• If $\deg(f)>1$, write $f=pg$, where $p(x),g(x)\in k[x]$ and $p$ is irreducible.

  Let $F=k/(p(x))$, then $F$ containing $k$ and a root $z$ of $p$.

  Hence, in $F[x]$, there is $h(x)$ with $p=(x-z)h$, and so $f=(x-z)hg$.

  By induction, there is a field $K$ containing $F$ (and hence $k$) so that $hg$, and hence $f$, is a product of linear factors in $K[x]$. $\square$

Definition 1.6.1 If $K/k$ is an extension field and $f(x)\in k[x]$ is nonconstant, then $f$ splits over $K$ if $f(x)=a(x-z_1)\cdots(x-z_n)$, where $z_1,\dots,z_n\in K$ and $a\in k$. An extension field $E/k$ is called a splitting field of $f$ over $k$ if splits over $E$, but $f$ does not split over any proper subfield of $E$.

Corollary 1.6.1 If $k$ is a field and $f(x)\in k[x]$, then a splitting field of $f$ over $k$ exists.

Proof.

By Kronecker's Theorem, there is an extension field $K/k$ such that $f$ splits in $K[x]$; say, $f(x)=a(x-\alpha_1)\cdots(x-\alpha_n)$. The subfield $E=k(\alpha_1,\dots,\alpha_n)$ of $K$ is a splitting field of $f$ over $k$. $\square$

A splitting field of $f(x)\in k[x]$ is a smallest extension field $E/k$ containing all the roots of $f$. We say "a" splitting field instead of "the" splitting field because it is not obvious whether an two splitting fields of $f$ over $k$ are isomorphic (they are). Analysis of this technical point will not only prove uniqueness of splitting fields, it will enable us to prove that any two finite fields with the same number of elements are isomorphic.

Example 1.6.1 Let $f(x)=x^n-1\in k[x]$ for some field $k$, and let $E/k$ be a splitting field. Because every finite subgroup of the multiplicative group $k^\times$ is cyclic, the set of all $n$th roots of identity in $E$ is a cyclic group. Then let $\omega$ be a generator of it. It follows that $k(\omega)=E$ is a splitting field of $f$.

Proposition 1.6.1 Let $p$ be prime, then let $k$ be a field. If $f(x)=x^p-c\in k[x]$ and $\alpha$ is a $p$th root of $c$ (in some splitting field), then either $f$ is irreducible in $k[x]$ or $c$ has a $p$th root in $k$. In either case, if $k$ contains the $p$th roots of identity, then $k(\alpha)$ is a splitting field of $f$.

Theorem 1.6.2 (Galois) If $p$ is a prime and $n$ is a positive integer, then there exists a field having exactly $p^n$ elements.

Proof.

Write $q=p^n$, and consider the polynomial $g(x)=x^q-x\in \mathbb Z_p[x]$.

By Kronecker's Theorem, there is an extension field $K/\mathbb Z_p$ with $g$ a product of linear factors in $K[x]$. Define
$$E=\{\alpha\in K;g(\alpha)=0\};
$$
that is, $E$ is the set of all the roots of $g$. Since the derivative $g'(x)=qx^{q-1}-1=p^nx^{q-1}-1=-1$, we have $\gcd(g,g')=1$. Therefore, all the roots of $g$ are distinct; that is, $E$ has exactly $q=p^n$ elements.

The theorem will follow if $E$ is a subfield of $K$. Of course, $1\in E$. If $a,b\in E$, then $a^q=a$ and $b^q=b$. Therefore, $(ab)^q=a^qb^q=ab$, and $ab\in E$. $(a-b)^q=a^q-b^q=a-b$, so that $a-b\in E$. Finally, if $a\ne 0$, $a^{q-1}=1$ gives $a^{q-2}$ is the inverse of $a$. $\square$

After this, we write $\mathbb F_{p^n}$ as a finite field having $p^n$ elements.

Corollary 1.6.2 For every prime $p$ and every integer $n\ge 1$, there exists an irreducible polynomial $g(x)\in \mathbb F_p[x]$ of degree $n$. In fact, if $\alpha$ is a primitive element of $\mathbb F_{p^n}$, then its minimal polynomial $g(x)=\operatorname {irr}(\alpha,\mathbb F_p)$ has degree $n$.

Proof.

Let $E/\mathbb F_p$ be an extension field with $p^n$ elements, and let $\alpha \in E$ be a primitive element.

Let $g(x)=\operatorname{irr}(\alpha,\mathbb F_p)\in \mathbb F_p[x]$, then $g(x)$ is irreducible.

If $\deg(g)=d$, the Proposition 1.5.6 gives $[\mathbb F_p[x]/(g):\mathbb F_p]=d$; and $\mathbb F_p[x]/(g)\cong \mathbb F_p(\alpha)=E$.

Therefoer $d=[E:\mathbb F_p]=n$. $\square$

Lemma 1.6.1 Let $\varphi:k\to k'$ be an isomorphism of fields, and let $\varphi_*:k[x]\to k'[x]$ be the ring isomorphism of the Corollary 1.2.1:
$$\varphi_*:g(x)=a_0+a_1x+\cdots+a_nx^n\mapsto g'(x)=\varphi(a_0)+\varphi(a_1)x+\cdots+\varphi(a_n)x^n.
$$
Let $f(x)\in K[x]$ and $f'(x)=\varphi_*(f)\in k'[x]$, If $E$ is a splitting field of $f$ over $k$ and $E'$ is a splitting field of $f'$ over $k'$, then there is an isomorphism $\Phi:E\to E'$ extending $\varphi$:
$$%https://darknmt.github.io/res/xypic-editor/#eyJub2RlcyI6W3sicG9zaXRpb24iOlswLDBdLCJ2YWx1ZSI6IkUifSx7InBvc2l0aW9uIjpbMSwwXSwidmFsdWUiOiJFJyJ9LHsicG9zaXRpb24iOlswLDFdLCJ2YWx1ZSI6ImsifSx7InBvc2l0aW9uIjpbMSwxXSwidmFsdWUiOiJrJyJ9XSwiZWRnZXMiOlt7ImZyb20iOjIsInRvIjozLCJ2YWx1ZSI6IlxcdmFycGhpIiwibGFiZWxQb3NpdGlvbiI6InJpZ2h0In0seyJmcm9tIjowLCJ0byI6MiwiaGVhZCI6Im5vbmUifSx7ImZyb20iOjEsInRvIjozLCJoZWFkIjoibm9uZSJ9LHsiZnJvbSI6MCwidG8iOjEsImxpbmUiOiJkYXNoZWQiLCJ2YWx1ZSI6IlxcUGhpIn1dfQ==
\xymatrix{
E \ar@{-}[d] \ar@{-->}[r]^{\Phi} & E' \ar@{-}[d] \\
k \ar@{->}[r]_{\varphi} & k'
}
$$
Theorem 1.6.3 If $k$ is a field and $f(x)\in k[x]$, then any two splitting fields of $f$ over $k$ are isomorphic via an isomorphism that fixes $k$ pointwise.

Corollary 1.6.3 (Moore) Any two finite fields having exactly $p^n$ elements are isomorphic.

Proof.

If $E$ is a field with $q=p^n$ elements, then Lagrange's Theorem applied to the multiplicative group $E^\times$ shows that $a^{q-1}=1,\forall a\in E^\times$.

So ever element of $E$ is a root of $f(x)=x^q-x\in \mathbb F_p[x]$, and so $E$ is a splitting field of $f$ over $\mathbb F_p$. $\square$

Finite fields are often called Galois fields in honor of their discoverer. In light of Corollary 1.6.3, we nay speak of the field with $q$ elements, where $q=p^n$ is a power of a prime $p$, and we denote it by
$$\mathbb F_q.
$$

1.7 Irreducibility

Theorem 1.7.1 If $f(x)=a_0+a_1x+\cdots+a_nx^n\in \mathbb Z[x]\subset \mathbb Q[x]$, then every rational root of $f$ has the form $b/c$, where $b\mid a_0$ and $c\mid a_n$. In particular, if $f$ is monic, then every rational root of $f$ is an integer.

Theorem 1.7.2 Let $f(x)=a_0+a_1x+\cdots+a_{n-1}x^{n-1}+x^n\in \mathbb Z[x]$ be monic, and let $p$ be a prime. If $\overline{f}(x)=[a_0]+[a_1]x+\cdots+[a_{n-1}]x^{n-1}+x^n$ is irreducible in $\mathbb F_p[x]$, then $f$ is irreducible in $\mathbb Q[x]$.

Definition 1.7.1 If $n\ge 1$ is a positive integer, then an $n$th root of unity in a field $k$ is an element $\zeta\in k$ with $\zeta^k=1$. If $\zeta$ is an $n$th root of unity and $n$ is the smallest positive integer for which $\zeta^n=1$, we say that $\zeta$ is a primitive $n$th root of unity.

Definition 1.7.2 If $d$ is a positive integer, then the $d$th cyclotomic polynomial is defined by
$$\Phi_d(x)=\prod(x-\zeta),
$$
where $\zeta$ ranges over all the primitive $d$th roots of unity.

Proposition 1.7.1 Let $n$ be a positive integer and regard $x^n-1\in \mathbb Z[x]$. Then

1. $$
   x^n-1=\prod{d\mid n}\Phi{d}(x),
   $$

   where $d$ ranges over all the positive divisors $d$ of $n$.

2. $\Phi_n (x)$ is a monic polynomial in $\mathbb Z[x]$ and $\deg(\Phi_n)=\phi(n)$, the Euler $\phi$-function.

3. For every integer $MARKDOWN_HASH1d0fe3ad28ed27546808efeec5c626a9MARKDOWNHASH$, we have
   $$
   n=\sum{d\mid n}\phi(d).
   $$

Proof.

1. For each divisor $d$ of $n$, collect all terms in the equation $x^n-1=\prod(x-\zeta)$ with $\zeta$ a primitive $d$th root of unity.

2. We prove that $\Phi_n(x)\in \mathbb Z[x]$ by induction on $n\ge1$. The base step is true, for $\Phi_1(x)=x-1\in \mathbb Z[x]$.

   For the inductive step, let $f(x)=\prod_{d\mid n,d<n}\Phi_d(x)$, so that
   $$
   x^n-1=f(x)\Phi_n(x).
   $$
   By induction, each $\Phi_d(x)$ is a monic polynomial in $\mathbb Z[x]$, and so $f$ is a monic polynomial in $\mathbb Z[x]$.

   Since $f$ and $x^n-1$ are monic, $\Phi_n$ is a monic polynomial in $\mathbb Z[x]$.

3. Immediate from parts (1) and (2):
   $$
   n=\deg(x^n-1)=\deg(\prod_d \Phi_d)=\sum_d \deg(\Phi_d)=\sum_d \phi(d)\ \square
   $$

Corollary 1.7.1 If $q$ is a positive integer and $d$ is a divisor of an integer $n$ with $d<n$, then $\Phi_n(q)$ is a divisor of both $q^n-1$ and $(q^n-1)/(q^d-1)$.

Theorem 1.7.3 (Eisenstein Criterion) Let $f(x)=a_0+a_1x+\cdots+a_nx^n\in \mathbb Z[x]$. If there is a prime $p$ dividing $a_i$ for all $i<n$ but with $p\nmid a_n$ and $p\nmid a_0^2$, then $f$ is irreducible in $\mathbb Q[x]$.

Theorem 1.7.4 (Gauss) For every prime $p$, the $p$th cyclotomic polynomial $\Phi_p(x)$ is irreducible in $\mathbb Q[x]$.

Proof.

Since $\Phi_p(x)=(x^p-1)/(x-1)$, we have
$$\Phi(x+1)=((x+1)^p-1)/x=x^{p-1}+C_p^1x^{p-2}+C_p^2x^{p-3}+\cdots+p.
$$
Since $p$ is prime, we have $p\mid C_p^i$ for all $i\ (0<i<p)$; hence, Eisenstein Criterion applies, and $\Phi(x+1)$ is irreducible in $\mathbb Q[x]$. $\square$