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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