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
is called a ring homomorphism or simply homomorphism if for every
, the following properties hold:
![{\displaystyle f(r_{1}r_{2})=f(r_{1})f(r_{2}),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/253ecb4bc8f85a108899f15f2a072fb711ca868b)
![{\displaystyle f(r_{1}+r_{2})=f(r_{1})+f(r_{2}).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63492b628483a6ad32ffe26960ee82907b40cb80)
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
, then f is called a unital ring homomorphism.
- Let
be the function mapping
. Then one can easily check that
is a homomorphism, but not a unital ring homomorphism.
- If we define
, then we can see that
is a unital homomorphism.
- The zero homomorphism is the homomorphism which maps ever element to the zero element of its codomain.
Theorem: Let
and
be integral domains, and let
be a nonzero homomorphism. Then
is unital.
Proof:
. But then by cancellation,
.
In fact, we could have weakened our requirement for R a small amount (How?).
Theorem: Let
be rings and
a homomorphism. Let
be a subring of
and
a subring of
. Then
is a subring of
and
is a subring of
. That is, the kernel and image of a homomorphism are subrings.
Proof: Proof omitted.
Theorem: Let
be rings and
be a homomorphism. Then
is injective if and only if
.
Proof: Consider
as a group homomorphism of the additive group of
.
Theorem: Let
be fields, and
be a nonzero homomorphism. Then
is injective, and
.
Proof: We know
since fields are integral domains. Let
be nonzero. Then
. So
. So
(recall you were asked to prove units are nonzero as an exercise). So
.
Let
be rings. An isomorphism between
and
is an invertible homomorphism. If an isomorphism exists,
and
are said to be isomorphic, denoted
. Just as with groups, an isomorphism tells us that two objects are algebraically the same.
- The function
defined above is an isomorphism between
and the set of integer scalar matrices of size 2,
.
- Similarly, the function
mapping
where
is an isomorphism. This is called the matrix representation of a complex number.
- The Fourier transform
defined by
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:
![{\displaystyle f:\mathbb {C} \to \mathbb {C} ,f(a+bi)=a-bi}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2cc8ead31f5699fa8fec456e749e602dcabfedc0)
- Define the set
, and let ![{\displaystyle g:\mathbb {Q} ({\sqrt {2}})\to \mathbb {Q} ({\sqrt {2}}),g(a+b{\sqrt {2}})=a-b{\sqrt {2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf64606332974c1e88121e0cee2ce3d13666ebc2)