Math4302 Modern Algebra (Lecture 32)

Rings

Ideals

Let $R$ be a commutative ring.

$S\subseteq R$ is a subring if

$(S,+)\leq (R,+)$
$a,b\in S$ , then $ab\in S$ .

So $(S,+,\cdot)$ is a subring of $R$ .

Example

$(\mathbb{Z},+,\cdot)$ is a subring of $(\mathbb{R},+,\cdot)$ .

Definition of ideal

A subset $I\subset R$ is an ideal if

$(I,+)\leq (R,+)$
if $a\in I,b\in R$ , then $ab\in I$ . (Trivially $I$ is a subring of $R$ , but take $b\in R$ to $I$ if multiply with $a\in I$ .)

Example

$(\mathbb{Z},+,\cdot)$ is not an ideal of $(\mathbb{R},+,\cdot)$ .

$1\in \mathbb Z$ , $\sqrt{2}\in \mathbb R$ , $\sqrt{2}\notin \mathbb Z$ .

$(\{0\},+,\cdot)$ is an ideal of $(\mathbb{R},+,\cdot)$ .

For any non zero element, if $I$ is an ideal and $a\neq 0,a\in I$ , then $a^{-1}\in I$ , so $1\in I$ .

For every $b\in \mathbb{R}$ , $b\in I$ since $b\cdot 1\in I$ .

Therefore, there are only two idals of $(\mathbb{R},+,\cdot)$ : $(\{0\},+,\cdot)$ and $(\mathbb{R},+,\cdot)$ .

More generally, every ring with unity and $I$ is an ideal of $R$ which contains a unit, then $1\in I$ , so $I=R$ .

In particular, if $R$ is a field, then $R$ has only two ideals: $(\{0\},+,\cdot)$ and $R$ itself.

Let $R$ be the ring of functions $f:\mathbb{R}\to \mathbb{R}$ . with

(f+g)(x)=f(x)+g(x),\quad (fg)(x)=f(x)g(x)

Then $I=\{f\in R|f(2)=0\}$ is an ideal of $R$ .

What are all ideals of $\mathbb{Z}$ ?

$n\mathbb{Z}$ is an ideal of $\mathbb{Z}$ . ( $n\geq 0$ ).

clearly $n\mathbb{Z}$ is an ideal and if $I\subseteq \mathbb{Z}$ , then $(I,+)$ is a subgroup of $(\mathbb{Z},+)$ . (Cyclic, therefore $I$ must be generated by some element. $\langle n\rangle=n\mathbb{Z}$ .)

If $f:R\to S$ where $R$ is a commutative ring, $f$ is a ring homomorphism.

Then $\ker(f)$ is an ideal of $R$ .

$\ker(f)=\{a\in R|f(a)=0\}$ .

Suppose $a\in\ker(f)$ and $b\in R$ , then $f(ab)=f(a)f(b)=f(a)0=f(0)=0$ . Therefore $ab\in \ker(f)$ .

Factor ring (Quotient ring)

Suppose $R$ is commutative and $I\subseteq R$ is an ideal.

Then $R/I$ is a set of cosets of $I$ in $R$ : $a+I$ for $a\in R$ .

This is an abelian group with operation

(a+I)+(b+I)=(a+b)+I

$R/I$ has also a multiplication operation:

(a+I)(b+I)=ab+I

This is well defined

Addition is trivial.

Suppose $a+I=a'+I$ , $b+I=b'+I$ , we need to show that $ab+I=a'b'+I$ .

Since $a+I=a'+I$ , $a-a'\in I$ , and $b+I=b'+I$ , $b-b'\in I$ . So $b(a-a')\in I$ , $a'(b-b')\in I$ , (since $I$ is an ideal). So $b(a-a')+a'(b-b')=ab-a'b'\in I$ , so $ab+I=a'b'+I$ .

So $R/I$ is a ring.

Fundamental Theorem of ring homomorphisms

Let $R$ be a commutative ring and $f:R\to S$ be a ring homomorphism.

Then $R/\ker(f)\simeq \operatorname{Im}(f)$

Warning

$\operatorname{Im}(f)$ is a subring of $S$ but not necessarily an ideal.

Consider $f:\mathbb Z\to \mathbb R$ $f(a)=a$ , $\mathbb Z$ is not an ideal of $\mathbb R$ .

The proof should be similar to the proof of the Fundamental Theorem of group homomorphisms, so we skip this part.

Example

$\mathbb{Z}\to \mathbb{Z}_n$ by $a\mapsto [a]$ . This a ring homomorphism. $\mathbb{Z}/n\mathbb{Z}\simeq \mathbb{Z}_n$ as rings.

Definition of Maximal Ideals

Assume $R$ is a commutative ring with a unity.

If $I\subseteq R$ is an ideal, then $R/I$ is also a commutative ring with a unity ( $1+I$ ).

(a+I)\cdot (b+I)=(ab)+I=(ba)+I=(b+I)\cdot (a+I)

An ideal $M\subsetneq R$ is called a maximal ideal if $M\subseteq I\subseteq R$ for ideal $I$ in $R$ implies $I=M$ or $I=R$ .

Example

The maximal ideal of $\mathbb{Z}$ .

$n\mathbb{Z}\subset m\mathbb{Z}\iff n\in m\mathbb \iff m|n$ .

So $n\mathbb{Z}=m\mathbb{Z}\iff m=\pm n$ .

So if $p|n$ then $n\mathbb{Z}\subseteq p\mathbb{Z}$ .

If $n>0$ $p\neq n$ then $n\mathbb{Z}\subsetneq p\mathbb{Z}$ .

Therefore $p\mathbb{Z}$ is the maximal ideal of $\mathbb{Z}$ , where $p$ is a prime number.

Theorem of maximal ideals

$M\subsetneq R$ is a maximal ideal if and only if $R/M$ is a field.

Proof next lecture.