## Ring Homomorphisms (Pt. II)

**Point of Post: **This post is a continuation of this one.

*Unital Rings*

We now discuss what happens for homomorphisms if we assume things about the unitalness of the domain or codomain rings. For example, we make the observation that

**Theorem: ***Let be a homomorphism where is unital. Then, is unital with unity given by .*

**Proof: **For any we have that and from where the conclusion follows.

Note that this multiplicative identity for need not be the multiplicative identity for if is unital. Indeed, consider and . Both are unital rings (the first when given coordinatewise multiplication and addition) with units and respectively. Consider though the obvious map . This is a ring homomorphism and while is indeed a multiplicative identity for yet it is, of course, not the identity of . This example also illustrates another nice principle. Suppose for a second our target ring is unital with non-zero unity but the image has some different unit. What we shall see now is that it has to be a zero-divisor. Indeed:

**Theorem: ***Let and be unital rings with unities and respectively. If is a non-zero homomorphism and then is a zero divisor.*

**Proof: **We merely note that . Now, since is non-zero we know that (exercise) and thus if the above shows that and are both zero divisors.

**Corollary: ***If and are unital rings with having no zero divisors, and if is a non-zero homomorphism then is unital.*

*Remark: *An *idempotent * of a ring is one such that (compare to linear projections and projections into the group algebra). The above more generically shows (if one looks hard enough) that an idempotent of a unital ring is either the identity or a zero divisor.

We now define a morphism such that a *unital homomorphism.* What we’d now like to note is a fascinating observation about unital homomorphisms and units. Namely:

**Theorem: ***Let and be unital rings with unities and respectively. Suppose then that is a unital homomorphism then with for each .*

**Proof: **Let note then that and and thus . since was arbitrary the conclusion follows. .

From this we get the following corollary:

**Corollary: ***Let and be unital rings. Then, every unital homomorphism induces a group homomorphism .*

Moreover we see that this carries isomorphisms to isomorphisms. In particular

**Theorem: ***Let and be unital rings and a unital isomorphism, then is an isomorphism.*

**Proof: **We have by simple observation that is a group monomorphism and so it suffices to prove that is surjective. To see this let be arbitrary. Since is surjective there exists such that . Similarly, there exists such that and so and since is injective we may conclude that . By the same method we easily prove that and so . Thus, we may conclude that and so . Since was arbitrary we may conclude that is surjective and thus an isomorphism as desired.

Another way to think about this last theorem(for people who know the lingo) is by noticing that:

**Theorem: ***Let and where is a unital ring with unity for . Then, . Moreover, one has that for any unital ring .*

**Proof: **We merely note that for every one has that but since we have that and so but since we have that and so . Since was arbitrary the first claim follows. The second claim is immediately obvious.

Or rephrasing this we see we have a functor with and . Thus, from basic category theory we have that if implies or .

From all of this we of course have that

**Theorem: ***Let and be two unital rings, if then .*

We end by showing that “being a division ring”, “being an integral domain”, and “being a field” are isomorphism invariants:

**Theorem: ***Let and be isomorphic unital rings. If is an integral domain, division ring, or a field then is an integral domain, division ring, or a field respectively.*

**Proof: **Suppose that is an integral domain, if then and so , thus or and so either or . Since we have proven that the image of a commutative ring is commutative the conclusion follows.

Suppose next that is a division ring, we know then for any that and so by assumption and so by previous theorem .

Lastly, we combine this previous result with the fact that homomorphic images of commutative rings are commutative to conclude.

**References:**

1. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

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