Abstract Algebra/Ring Homomorphisms

Template:TOC right 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 $a\mapsto \left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)$. Then one can easily check that $f$ is a homomorphism, but not a unital ring homomorphism.
2. If we define $g:a\mapsto \left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)$, 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 $\phi :R\to S$ a homomorphism. Let $R^{\prime }$ be a subring of $R$ and $S^{\prime }$ a subring of $S$. Then $\phi (R^{\prime })$ is a subring of $S$ and ${\phi }^{-1}(S^{\prime })$ 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 $\phi :R\to S$ be a homomorphism. Then $\phi$ is injective if and only if $\ker \phi =0$.

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

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

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

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, $S=\{\lambda I_2|\lambda \in \mathbb{Z}\}$.
2. Similarly, the function $\phi :ℂ\to M_2(ℝ)$ mapping $z\mapsto \left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ where $z=a+bi$ is an isomorphism. This is called the matrix representation of a complex number.
3. The Fourier transform $ℱ:L^1\to L^1$ defined by $ℱ(f)=\underset{ℝ}{\int }f(t)e^{-i\omega t}dt$ 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. $f:ℂ\to ℂ,f(a+bi)=a-bi$
2. Define the set $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}|a,b\in \mathbb{Q}\}$, and let $g:\mathbb{Q}(\sqrt{2})\to \mathbb{Q}(\sqrt{2}),g(a+b\sqrt{2})=a-b\sqrt{2}$