## 305 -6. Rings, ideals, homomorphisms

It will be important to understand the subfields of a given field; this is a key step in figuring out whether a field ${{\mathbb Q}^{p(x)}}$ is an extension by radicals or not. We need some “machinery” before we can develop this understanding.

Recall:

Definition 1 A ring is a set ${R}$ together with two binary operations ${+,\times}$ on ${R}$ such that:

1. ${+}$ is commutative.
2. There is an additive identity ${0.}$
3. Any ${a}$ has an additive inverse ${-a.}$
4. ${+}$ is associative.
5. ${\times}$ is associative.
6. ${\times}$ distributes over ${+,}$ both on the right and on the left.

So ${{\mathbb Z}}$ is a ring with the usual operations, as is each ${{\mathbb Z}_n.}$ Any field is a ring. These are all examples of commutative rings, since ${a\times b=b\times a}$ holds in them. They are also rings with identity, since there is a multiplicative identity ${1}$ in all of them.

If ${{\mathbb F}}$ is a field, ${{\mathbb F}[x]}$ is a commutative ring with identity.

Here is an exercise, to check your understanding: Suppose ${R}$ is a ring. Is ${R[x]}$ a ring?

If ${{\mathcal M}_n({\mathbb F})}$ denotes the collection of ${n\times n}$ matrices with coefficients in the field ${{\mathbb F},}$ then ${{\mathcal M}_n({\mathbb F})}$ is a ring. In general, it is not commutative. Here is another exercise: Is there some ${n}$ such that ${{\mathcal M}_n({\mathbb F})}$ is commutative?

If ${X}$ is a set and ${{\mathcal P}(X)}$ is its power set, i.e., the collection of all subsets of ${X,}$ then we can turn ${{\mathcal P}(X)}$ into a commutative ring with identity: For ${A,B}$ subsets of ${X,}$ define ${A+\mbox{''}B=A\bigtriangleup B:=\{x\in A\cup B: x\notin A\cap B\}.}$ This is the symmetric difference of ${A}$ and ${B.}$ Define ${A\times\mbox{''}B=A\cap B.}$ Check that this is indeed a commutative ring with identity. What is the ${0}$ element? What is the ${1}$ element?

Definition 2 If ${R}$ is a ring, a subring of ${R}$ is a subset ${S\subseteq R}$ which is a ring with the operations inherited from ${R.}$

Just as with subfields, there is an easy way of verifying whether a given subset of a ring is a subring.

Proposition 3 Let ${R}$ be a ring and let ${S\subseteq R.}$ Then ${S}$ is a subring of ${R}$ iff the following conditions hold:

1. ${S}$ is closed under addition and multiplication.
2. ${S\ne\emptyset.}$
3. Whenever ${a\in S,}$ then ${-a\in S.}$ ${\Box}$

The proof of the proposition is an easy modification of the corresponding argument for subfields.

Here are some examples:

For any ring ${R,}$ both ${R}$ and ${\{0\}}$ are subrings of ${R.}$

For any integer ${n,}$ let ${n{\mathbb Z}=\{nm:m\in{\mathbb Z}\}}$ be the set of multiples of ${{\mathbb Z}.}$ Then ${n{\mathbb Z}}$ is a subring of ${{\mathbb Z}.}$

In fact: Suppose ${S\subseteq{\mathbb Z}}$ is a subring. Then ${S=n{\mathbb Z}}$ for some ${n.}$ Indeed, either ${S=\{0\}}$ or else ${S}$ must contain a positive element. Using the division algorithm, one easily checks that if ${n}$ is the smallest positive element of ${S,}$ then ${S=n{\mathbb Z}.}$

Let ${R={\mathcal M}_n({\mathbb C}).}$ An ${n\times n}$ matrix ${A}$ has eigenvector ${{\mathbf v}}$ iff ${A{\mathbf v}}$ is a scalar multiple of ${{\mathbf v},}$ i.e., for some complex number ${\lambda,}$ ${A{\mathbf v}=\lambda{\mathbf v}.}$ Fix a column vector ${{\mathbf v}}$ with ${n}$ entries, and let ${S}$ be the collection of matrices in ${R}$ with eigenvector ${{\mathbf v}.}$ Then ${S}$ is a subring of ${R.}$ Make sure you verify that this is indeed the case.

Definition 4 Given rings ${(A,+_A,\times_A,0_A)}$ and ${(B,+_B,\times_B,0_B),}$ a homomorphism between ${A}$ and ${B}$ is a function ${h:A\rightarrow B}$ such that for any ${a,b\in A}$:

1. ${h(0_A)=0_B.}$
2. ${h(a+_A b)=h(a)+_B h(b).}$
3. ${h(a\times_A b)=h(a)\times_B h(b).}$

Informally, ${h}$ translates the operations of ${A}$ into the operations of ${B}$ in a “coherent” way.

Proposition 5 If ${h:A\rightarrow B}$ is a homomorphism of rings and ${C=h[A]:=\{h(a):a\in A\},}$ then ${C}$ is a subring of ${B.}$ ${\Box}$

This is easy. It is a good idea to check first that ${h(-a)=-h(a)}$ for any ${a.}$

Here are some examples:

Let ${A={\mathbb Z}}$ and ${B={\mathbb Z}_n}$ and let ${h:A\rightarrow B}$ be the map that to each ${k\in A}$ assigns its class modulo ${n.}$ Then ${h}$ is a homomorphism.

Let ${A={\mathbb Q}[x],}$ let ${r\in{\mathbb C}}$ and let ${h:A\rightarrow{\mathbb C}}$ be the evaluation by ${r:}$ ${h(p(x))=p(r).}$ Then ${h}$ is a homomorphism. For example, if ${r=\sqrt2,}$ then ${h[A]={\mathbb Q}(\sqrt2).}$ This is not only a subring of ${{\mathbb C},}$ it is in fact a field.

Proposition 6 Suppose ${h:A\rightarrow B}$ is a ring homomorphism. Then ${h}$ is 1-1 iff the only ${a}$ such that ${h(a)=0_B}$ is ${a=0_A,}$ in symbols, ${h^{-1}(0_B)=\{0_A\}.}$ ${\Box}$

For example, if ${A={\mathbb Q}[x],}$ ${B={\mathbb C}}$ and ${h(p(x))=p(\pi),}$ then ${h}$ is 1-1, because ${\pi}$ is transcendental, i.e., there is no nonzero polynomial in ${A}$ such that ${p(\pi)=0.}$

We just saw that the image of a homomorphism is a subring of the target ring. Homomorphisms also provide us with subrings of the source ring. These are special subrings:

Definition 7 Let ${R}$ be a ring. A subset ${I\subseteq R}$ is an ideal provided the following hold:

1. ${I\ne\emptyset.}$
2. Whenever ${a,b\in I,}$ then ${a-b\in I.}$
3. Whenever ${a\in I}$ and ${b\in R,}$ then ${ab\in I}$ and ${ba\in I.}$

Lemma 8 If ${I}$ is an ideal of a ring ${R}$ then ${I}$ is a subring of ${R.}$ ${\Box}$

Here is an example: Let ${R={\mathbb Q}[x],}$ let ${p(x)\in R}$ and let ${I}$ be the set of all polynomial multiples of ${p,}$ ${I=\{p(x)q(x):q(x)\in R\}.}$ Then ${I}$ is an ideal. It is the ideal generated by ${p(x),}$ and we write ${I=(p(x)).}$

On the other hand, if ${R={\mathcal M}_2({\mathbb C}),}$ ${{\mathbf v}}$ is a column vector with 2 entries and ${S}$ is the collection of matrices in ${R}$ with eigenvector ${{\mathbf v},}$ then this is a subring of ${R}$ but not necessarily an ideal, since there is no reason why ${A{\mathbf v}=\lambda{\mathbf v}}$ should imply that ${AB{\mathbf v}}$ is a multiple of ${{\mathbf v}}$ for an arbitrary ${B\in R.}$ As an exercise, find a counterexample.

Proposition 9 Let ${h:A\rightarrow B}$ be a ring homomorphism. Let ${I=h^{-1}(0_B):=\{a\in A: h(a)=0_B\}.}$ Then ${I}$ is an ideal of ${A.}$ ${\Box}$

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