Algebra homomorphism

A homomorphism between two algebras, A and B, over a field (or ring) K, is a map $F:A\rightarrow B$ such that for all k in K and x,y in A,

• F(kx) = kF(x)

• F(x + y) = F(x) + F(y)

• F(xy) = F(x)F(y)

If F is bijective then F is said to be an isomorphism between A and B.

A common abbreviation for "homomorphism between algebras" is "algebra homomorphism" or "algebra map". Every algebra homomorphism is a homomorphism of K-modules.

Unital algebra homomorphisms

If A and B are two unital algebras, then an algebra homomorphism $F:A\rightarrow B$ is said to be unital if it maps the unity of A to the unity of B. Often the words "algebra homomorphism" are actually used in the meaning of "unital algebra homomorphism", so non-unital algebra homomorphisms are excluded.

A unital algebra homomorphism is a ring homomorphism.

Examples

Let A = K[x] be the set of all polynomials over a field K and B be the set of all polynomial functions over K. Both A and B are algebras over K given by the standard multiplication and addition of polynomials and functions, respectively. We can map each $f\,$ in A to $\hat{f}\,$ in B by the rule $\hat{f}(t) = f(t) \,$. A routine check shows that the mapping $f \mapsto \hat{f}\,$ is a homomorphism of the algebras A and B. This homomorphism is an isomorphism if and only if K is an infinite field.

Proof. If K is a finite field then let

$p(x) = \prod\limits_{t \in K} (x-t).\,$

p is a nonzero polynomial in K[x], however $p(t) = 0\,$ for all t in K, so $\hat{p} = 0\,$ is the zero function and our homomorphism is not an isomorphism (and, actually, the algebras are not isomorphic, since the algebra of polynomials is infinite while that of polynomial functions is finite).

If K is infinite then choose a polynomial f such that $\hat{f} = 0\,$. We want to show this implies that $f = 0\,$. Let $\deg f = n\,$ and let $t_0,t_1,\dots,t_n\,$ be n + 1 distinct elements of K. Then $f(t_i) = 0\,$ for $0 \le i \le n$ and by Lagrange interpolation we have $f = 0\,$. Hence the mapping $f \mapsto \hat{f}\,$ is injective. Since this mapping is clearly surjective, it is bijective and thus an algebra isomorphism of A and B.

If A is a subalgebra of B, then for every invertible b in B the function that takes every a in A to b⁻¹ a b is an algebra homomorphism (in case $A=B$, this is called an inner automorphism of B). If A is also simple and B is a central simple algebra, then every homomorphism from A to B is given in this way by some b in B; this is the Skolem-Noether theorem.

See also

• Augmentation (algebra)