Abstract Algebra – 1.0 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:
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:
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 $\forall a,b\in R,ab=0\Rightarrow(a=0\lor b=0)$.


  • ($\Rightarrow$).

    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 $f=ab^{-1}$.


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
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 $K$.


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