# Abstract Algebra/Ring Homomorphisms

Just as with groups, we can study homomorphisms to understand the similarities between different rings.

## Homomorphisms

### Definition

Let R and S be two rings. Then a function $f:R\to S$  is called a ring homomorphism or simply homomorphism if for every $r_{1},r_{2}\in R$ , the following properties hold:

$f(r_{1}r_{2})=f(r_{1})f(r_{2}),$
$f(r_{1}+r_{2})=f(r_{1})+f(r_{2}).$

In other words, f is a ring homomorphism if it preserves additive and multiplicative structure.

Furthermore, if R and S are rings with unity and $f(1_{R})=1_{S}$ , then f is called a unital ring homomorphism.

### Examples

1. Let $f:\mathbb {Z} \to M_{2}(\mathbb {Z} )$  be the function mapping $a\mapsto {\begin{pmatrix}a&0\\0&0\end{pmatrix}}$ . Then one can easily check that $f$  is a homomorphism, but not a unital ring homomorphism.
2. If we define $g:a\mapsto {\begin{pmatrix}a&0\\0&a\end{pmatrix}}$ , then we can see that $g$  is a unital homomorphism.
3. The zero homomorphism is the homomorphism which maps ever element to the zero element of its codomain.

Theorem: Let $R$  and $S$  be integral domains, and let $f:R\to S$  be a nonzero homomorphism. Then $f$  is unital.

Proof: $1_{S}f(1_{R})=f(1_{R})=f(1_{R}^{2})=f(1_{R})f(1_{R})$ . But then by cancellation, $f(1_{R})=1_{S}$ .

In fact, we could have weakened our requirement for R a small amount (How?).

Theorem: Let $R,S$  be rings and $\varphi :R\to S$  a homomorphism. Let $R'$  be a subring of $R$  and $S'$  a subring of $S$ . Then $\varphi (R')$  is a subring of $S$  and $\varphi ^{-1}(S')$  is a subring of $R$ . That is, the kernel and image of a homomorphism are subrings.

Proof: Proof omitted.

Theorem: Let $R,S$  be rings and $\varphi :R\to S$  be a homomorphism. Then $\varphi$  is injective if and only if $\ker \varphi =0$ .

Proof: Consider $\varphi$  as a group homomorphism of the additive group of $R$ .

Theorem: Let $F,E$  be ﬁelds, and $\varphi :F\to E$  be a nonzero homomorphism. Then $\varphi$  is injective, and $\varphi (x)^{-1}=\varphi (x^{-1})$ .

Proof: We know $\varphi (1)=1$  since fields are integral domains. Let $x\in F$  be nonzero. Then $\varphi (x^{-1})\varphi (x)=\varphi (x^{-1}x)=\varphi (1)=1$ . So $\varphi (x)^{-1}=\varphi (x^{-1})$ . So $\varphi (x)\neq 0$  (recall you were asked to prove units are nonzero as an exercise). So $\ker \varphi =0$ .

## Isomorphisms

### Definition

Let $R,S$  be rings. An isomorphism between $R$  and $S$  is an invertible homomorphism. If an isomorphism exists, $R$  and $S$  are said to be isomorphic, denoted $R\cong S$ . Just as with groups, an isomorphism tells us that two objects are algebraically the same.

### Examples

1. The function $g$  defined above is an isomorphism between $\mathbb {Z}$  and the set of integer scalar matrices of size 2, $S=\left\{\lambda I_{2}|\lambda \in \mathbb {Z} \right\}$ .
2. Similarly, the function $\varphi :\mathbb {C} \to M_{2}(\mathbb {R} )$  mapping $z\mapsto {\begin{pmatrix}a&-b\\b&a\end{pmatrix}}$  where $z=a+bi$  is an isomorphism. This is called the matrix representation of a complex number.
3. The Fourier transform ${\mathcal {F}}:L^{1}\to L^{1}$  defined by ${\mathcal {F}}(f)=\int _{\mathbb {R} }f(t)e^{-i\omega t}dt$  is an isomorphism mapping integrable functions with pointwise multiplication to integrable functions with convolution multiplication.

Exercise: An isomorphism from a ring to itself is called an automorphism. Prove that the following functions are automorphisms:

1. $f:\mathbb {C} \to \mathbb {C} ,f(a+bi)=a-bi$
2. Define the set $\mathbb {Q} ({\sqrt {2}})=\left\{a+b{\sqrt {2}}|a,b\in \mathbb {Q} \right\}$ , and let $g:\mathbb {Q} ({\sqrt {2}})\to \mathbb {Q} ({\sqrt {2}}),g(a+b{\sqrt {2}})=a-b{\sqrt {2}}$