Ring Homomorphism - Properties

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 (multiplicative) group of units of R to the (multiplicative) 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 that f can only be the zero function if S is a trivial ring or if we don't require that f preserves the multiplicative identity.
• If both R and S are fields (and f is not the zero function), then im(f) is a subfield of S, so this constitutes a field extension.
• 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.