# 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

* 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.
* If we define $g:a\mapsto {\begin{pmatrix}a&0\\0&a\end{pmatrix}}$, then we can see that $g$ is a unital homomorphism.
* 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

* The function $g$ defined above is an isomorphism between $\mathbb {Z}$ and the set of integer scalar matrices of size 2, $S=\left\{\lambda I_{2}|\lambda \in \mathbb {Z} \right\}$.
* 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.
* 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:
* $f:\mathbb {C} \to \mathbb {C} ,f(a+bi)=a-bi$
* Define the set $\mathbb {Q} ({\sqrt {2}})=\left\{a+b{\sqrt {2}}|a,b\in \mathbb {Q} \right\}$, and let $g:\mathbb {Q} ({\sqrt {2}})\to \mathbb {Q} ({\sqrt {2}}),g(a+b{\sqrt {2}})=a-b{\sqrt {2}}$