Ring homomorphism

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In ring theory or abstract algebra, a ring homomorphism is a function between two rings which respects the operations of addition and multiplication.

More precisely, if R and S are rings, then a ring homomorphism is a function f : R → S such that

f(a + b) = f(a) + f(b) for all a and b in R
f(ab) = f(a) f(b) for all a and b in R
f(1) = 1

Naturally, if one does not require rings to have a multiplicative identity then the last condition is dropped.

The composition of two ring homomorphisms is a ring homomorphism. It follows that the class of all rings forms a category with ring homomorphisms as the morphisms (cf. the category of rings).

1 Properties
2 Examples
3 Types of ring homomorphisms
4 See also

[edit] Properties

Directly from these definitions, one can deduce:

f(0) = 0
f(−a) = −f(a)
If a has a multiplicative inverse in R, then f(a) has a multiplicative inverse in S and we have f(a⁻¹) = (f(a))⁻¹. Therefore, f induces a group homomorphism from the group of units of R to the group of units of S.
The kernel of f, defined as ker(f) = {a in R : f(a) = 0} is an ideal in R. Every ideal in a commutative ring R arises from some ring homomorphism in this way. For rings with identity the kernel of a ring homomorphism is a subring without identity.
The homomorphism f is injective if and only if the ker(f) = {0}.
The image of f, im(f), is a subring of S.
If f is bijective, then its inverse f⁻¹ is also a ring homomorphism. f is called an isomorphism in this case, and the rings R and S are called isomorphic. From the standpoint of ring theory, isomorphic rings cannot be distinguished.
If there exists a ring homomorphism f : R → S then the characteristic of S divides the characteristic of R. This can sometimes be used to show that between certain rings R and S, no ring homomorphisms R → S can exist.
If R_p is the smallest subring contained in R and S_p is the smallest subring contained in S, then every ring homomorphism f : R → S induces a ring homomorphism f_p : R_p → S_p.
If R is a field, then f is either injective or f is the zero function. (Note, however, that if f preserves the multiplicative identity, then it cannot be the zero function.)
If both R and S are fields, then im(f) is a subfield of S (if f is not the zero function).
If R and S are commutative and S has no zero divisors, then ker(f) is a prime ideal of R.
If R and S are commutative, S is a field, and f is surjective, then ker(f) is a maximal ideal of R.
For every ring R, there is a unique ring homomorphism Z → R. This says that the ring of integers is an initial object in the category of rings.

[edit] Examples

The function f : Z → Z_n, defined by f(a) = [a]_n = a mod n is a surjective ring homomorphism with kernel nZ (see modular arithmetic).
There is no ring homomorphism Z_n → Z for n > 1.
If R[X] denotes the ring of all polynomials in the variable X with coefficients in the real numbers R, and C denotes the complex numbers, then the function f : R[X] → C defined by f(p) = p(i) (substitute the imaginary unit i for the variable X in the polynomial p) is a surjective ring homomorphism. The kernel of f consists of all polynomials in R[X] which are divisible by X² + 1.
If f : R → S is a ring homomorphism between the commutative rings R and S, then f induces a ring homomorphism between the matrix rings M_n(R) → M_n(S).

[edit] Types of ring homomorphisms

A bijective ring homomorphism is called a ring isomorphism.
A ring homomorphism whose domain is the same as its range is called a ring endomorphism.

Injective ring homomorphisms are identical to monomorphisms in the category of rings: If f:R→S is a monomorphism which is not injective, then it sends some r₁ and r₂ to the same element of S. Consider the two maps g₁ and g₂ from Z[x] to R which map x to r₁ and r₂, respectively; f o g₁ and f o g₂ are identical, but since f is a monomorphism this is impossible.

However, surjective ring homomorphisms are vastly different from epimorphisms in the category of rings. For example, the inclusion Z ⊆ Q is a ring epimorphism, but not a surjection. However, they are exactly the same as the strong epimorphisms.