# Abstract Nonsense

## 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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References:

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

June 18, 2011 -

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