It will be important to understand the subfields of a given field; this is a key step in figuring out whether a field is an extension by radicals or not. We need some “machinery” before we can develop this understanding.

Recall:

Definition 1Aringis a set together with two binary operations on such that:

- is commutative.
- There is an additive identity
- Any has an additive inverse
- is associative.
- is associative.
- distributes over both on the right and on the left.

So is a ring with the usual operations, as is each Any field is a ring. These are all examples of **commutative** rings, since holds in them. They are also rings **with identity**, since there is a multiplicative identity in all of them.

If is a field, is a commutative ring with identity.

Here is an exercise, to check your understanding: Suppose is a ring. Is a ring?

If denotes the collection of matrices with coefficients in the field then is a ring. In general, it is not commutative. Here is another exercise: Is there some such that is commutative?

If is a set and is its power set, i.e., the collection of all subsets of then we can turn into a commutative ring with identity: For subsets of define This is the **symmetric difference** of and Define Check that this is indeed a commutative ring with identity. What is the element? What is the element?

Definition 2If is a ring, asubringof is a subset which is a ring with the operations inherited from

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

Proposition 3Let be a ring and let Then is a subring of iff the following conditions hold:

- is closed under addition and multiplication.
- Whenever then

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

Here are some examples:

For any ring both and are subrings of Note that the latter is a ring *without* identity, or what is typically called a **rng**.

For any integer let be the set of integer multiples of Then is a subring of (again, without identity).

In fact: Suppose is a subring. Then for some Indeed, either or else must contain a positive element. Using the division algorithm, one easily checks that if is the smallest positive element of then

Let An matrix has **eigenvector ** iff is a scalar multiple of i.e., for some complex number (In the context of linear algebra, in addition one usually insists that where the latter is the vector all of whose entries are 0.)

Fix a column vector with entries, and let be the collection of matrices in with eigenvector Then is a subring of Make sure you verify that this is indeed the case.

Definition 4Given rings and ahomomorphismbetween and is a function such that for any :

Informally, *translates* the operations of into the operations of in a “coherent” way.

Proposition 5If is a homomorphism of rings and then is a subring of

This is easy. It is a good idea to check first that for any

Here are some examples:

Let and and let be the map that to each assigns its class modulo Then is a homomorphism.

Let let and let be the evaluation by Then is a homomorphism. For example, if then This is not only a subring of it is in fact a field.

Proposition 6Suppose is a ring homomorphism. Then is 1-1 iff the only such that is in symbols,

For example, if and then is 1-1, because is transcendental, i.e., there is no nonzero polynomial in such that

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 7Let be a ring. A subset is anidealprovided the following hold:

- Whenever then
- Whenever and then and

Lemma 8If is an ideal of a ring then is a subring of

Here is an example: Let let and let be the set of all polynomial multiples of Then is an ideal. It is the ideal **generated** by and we write

On the other hand, if is a column vector with 2 entries and is the collection of matrices in with eigenvector then this is a subring of but not necessarily an ideal, since there is no reason why should imply that is a multiple of for an arbitrary As an exercise, find a counterexample.

Proposition 9Let be a ring homomorphism. Let Then is an ideal of

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