Abstract Algebra/Ring Homomorphisms

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

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:
 * $$f(r_1r_2)=f(r_1)f(r_2),$$
 * $$f(r_1+r_2)=f(r_1)+f(r_2).$$

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: $$1_Sf(1_R) = f(1_R) = f(1_R^2) = f(1_R)f(1_R)$$. 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 $$\ker \varphi = 0$$.

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 $$g:\mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}), g(a+b\sqrt{2}) = a-b\sqrt{2}$$