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 if for every $r_1,r_2\in R$, the following properties hold:

f(r₁r₂) = f(r₁)f(r₂),

f(r₁ + r₂) = f(r₁) + f(r₂).

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 φ(R') is a subring of S and φ ^{− 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 φ is injective if and only if ker φ = 0.

Proof: Consider φ 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 φ is injective, and φ(x) ^{− 1} = φ(x ^{− 1}).

Proof: We know φ(1) = 1 since fields are integral domains. Let $x \in F$ be nonzero. Then φ(x ^{− 1})φ(x) = φ(x ^{− 1}x) = φ(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 φ = 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.

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