Abstract Algebra/Ring Homomorphisms

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

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

${\displaystyle f(r_{1}r_{2})=f(r_{1})f(r_{2}),}$

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

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 ${\displaystyle f(1_{R})=1_{S}}$, then f is called a unital ring homomorphism.

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

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

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

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

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

Proof: Proof omitted.

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

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

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

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

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

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