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 {atimes b=btimes 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 {ntimes 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=Abigtriangleup B:={xin Acup B: xnotin Acap B}.} This is the symmetric difference of {A} and {B.} Define {A``timesmbox{''}B=Acap 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 {Ssubseteq 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 {Ssubseteq R.} Then {S} is a subring of {R} iff the following conditions hold:

  1. {S} is closed under addition and multiplication.
  2. {Sneemptyset.}
  3. Whenever {ain S,} then {-ain 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:min{mathbb Z}}} be the set of multiples of {{mathbb Z}.} Then {n{mathbb Z}} is a subring of {{mathbb Z}.}

In fact: Suppose {Ssubseteq{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 {ntimes 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:Arightarrow B} such that for any {a,bin A}:

  1. {h(0_A)=0_B.}
  2. {h(a+_A b)=h(a)+_B h(b).}
  3. {h(atimes_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:Arightarrow B} is a homomorphism of rings and {C=h[A]:={h(a):ain 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:Arightarrow B} be the map that to each {kin A} assigns its class modulo {n.} Then {h} is a homomorphism.

Let {A={mathbb Q}[x],} let {rin{mathbb C}} and let {h:Arightarrow{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:Arightarrow 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 {Isubseteq R} is an ideal provided the following hold:

  1. {Ineemptyset.}
  2. Whenever {a,bin I,} then {a-bin I.}
  3. Whenever {ain I} and {bin R,} then {abin I} and {bain 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 {Bin R.} As an exercise, find a counterexample.

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

 

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