Abstract Algebra/Ring Homomorphisms

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

Homomorphisms[edit]

Definition[edit]

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[edit]

  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.

Isomorphisms[edit]

Definition[edit]

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[edit]

  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}