Abstract Nonsense

Crushing one theorem at a time

Ring Homomorphisms (Pt. II)

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

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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

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Theorem: Let f:R\to R' be a homomorphism where R is unital. Then, \text{im }f is unital with unity given by f(1_R).

Proof: For any f(a)\in\text{im }f we have that f(a)f(1_R)=f(a1_R)=f(a) and f(1_R)f(a)=f(1_Ra)=f(a) from where the conclusion follows. \blacksquare

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Note that this multiplicative identity for \text{im }f need not be the multiplicative identity for R' if R' is unital. Indeed, consider \mathbb{Z}\times\mathbb{Z} and \mathbb{Z}. Both are unital rings (the first when given coordinatewise multiplication and addition) with units (1,1) and 1 respectively. Consider though the obvious map f:\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}:z\mapsto (z,0). This is a ring homomorphism and while f(1)=(1,0) is indeed a multiplicative identity for \text{im }f=\mathbb{Z}\times\{0\} yet it is, of course, not the identity (1,1) of \mathbb{Z}\times\mathbb{Z}. 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:

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Theorem: Let R and R' be unital rings with unities 1_R and 1_{R'} respectively. If f:R\to R' is a non-zero homomorphism and f(1_R)\ne 1_{R'} then f(1_R) is a zero divisor.

Proof: We merely note that (f(1_R)-1_{R'})f(1_R)=f(1_R)^2-1_{R'}f(1_R)=f(1_R^2)-f(1_R)=f(1_R)-f(1_R)=0. Now, since f is non-zero we know that f(1_R)\ne 0 (exercise) and thus if f(1_R)-1_{R'}\ne 0 the above shows that f(1_R) and f(1_R)-1_{R'} are both zero divisors. \blacksquare

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Corollary: If R and R' are unital rings with R' having no zero divisors, and if  f:R\to R' is a non-zero homomorphism then f is unital.

Remark: An idempotent x of a ring is one such that x^2=x (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.

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We now define a morphism R\to R' such that 1_R\mapsto 1_{R'}unital homomorphism. What we’d now like to note is a fascinating observation about  unital homomorphisms and units. Namely:

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Theorem: Let R and S be unital rings with unities 1_R and 1_S respectively. Suppose then that f:R\to S is a unital homomorphism then \text{im }R^\times\subseteq S^\times with f(u)^{-1}=f\left(u^{-1}\right) for each u\in R^\times.

Proof: Let u\in R^\times note then that f\left(u^{-1}\right)f(u)=f\left(u^{-1}u\right)=f(1_R)=1_S and f\left(u^{-1}\right)f(u)=f\left(u^{-1}u\right)=f(1_R)=1_S and thus u\in S^\times. since u was arbitrary the conclusion follows. \blacksquare.

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From this we get the following corollary:

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Corollary: Let R and S be unital rings. Then, every unital homomorphism f:R\to S induces a group homomorphism f_\ast:U(R)\to U(S).

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

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Theorem: Let R and S be unital rings and f:R\to S a unital isomorphism, then f_\ast:U(R)\to U(S) is an isomorphism.

Proof: We have by simple observation that f_\ast is a group monomorphism and so it suffices to prove that f_\ast is surjective. To see this let w\in S^\times be arbitrary. Since f is surjective there exists u\ R such that f(u)=w. Similarly, there exists v\in R such that f(v)=w^{-1} and so f(uv)=f(u)f(v)=ww^{-1}=1_S and since f is injective we may conclude that uv=1_R. By the same method we easily prove that f(vu)=f(v)v(u)=w^{-1}w=1_S and so vu=1_R. Thus, we may conclude that u\in R^\times and so w=f(u)=f_\ast(u)\in\text{im }f_\ast. Since w was arbitrary we may conclude that f_\ast is surjective and thus an isomorphism as desired. \blacksquare

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

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Theorem: Let f\in\text{Hom}(R_1,R_2) and g\in\text{Hom}(R_2,R_3) where R_i is a unital ring with unity 1_{R_i} for i=1,2,3. Then, (g\circ f)_\ast=g_\ast\circ f_\ast. Moreover, one has that \left(\text{id}_{R}\right)_\ast=\text{id}_{U(R)} for any unital ring R.

Proof: We merely note that for every x\in U(R) one has that (g\circ f)_\ast(x)=(g\circ f)(x)=g(f(x)) but since x\in U(R_1) we have that f(x)=f_\ast(x) and so (g\circ f)_\ast(x)=g(f_\ast(x)) but since f_\ast(x)\in U(R_2) we have that g(f_\ast(x))=g_\ast(f_\ast(x))=(g_\ast\circ f_\ast)(x) and so (g\circ f)_\ast(x)=(g_\ast\circ f_\ast)(x). Since x was arbitrary the first claim follows. The second claim is immediately obvious. \blacksquare

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Or rephrasing this we see we have a functor F:\bold{Ring}\to\bold{Grp} with F(R)=U(R) and F(f)=f_\ast. Thus, from basic category theory we have that if f\in\text{Iso}(R,S) implies F(f)\in\text{Iso}(F(R),F(S)) or f_\ast\in\text{Iso}(U(R),U(S)).

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From all of this we of course have that

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Theorem: Let R and S be two unital rings, if U(R)\not\cong U(S) then R\not\cong S.

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

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

Proof: Suppose that R is an integral domain, if f(a)f(b)=0 then f(ab)=0 and so ab=0, thus a=0 or b=0 and so either f(a)=0 or f(b)=0. Since we have proven that the image of a commutative ring is commutative the conclusion follows.

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Suppose next that R is a division ring, we know then for any f(a)\ne 0 that a\ne 0 and so by assumption a\in R^\times and so by previous theorem f(a)\in S^\times.

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

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


June 18, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , , , , , , ,


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