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 $g(x)=(x^2-2)(x^2-3)$.

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