Algebra homomorphism

A homomorphism between two algebras over a field "K", "A" and "B", is a map $F:A\; ightarrow\; B$ 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".

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 $f\; ightarrow\; hat\{f\},$ is a homomorphism of the algebras "A" and "B". If "K" is a finite field then let

:$p(x)\; =\; Pi\_\{t\; in\; K\}\; (x-t).,$

"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 the algebras are not isomorphic.

If "K" is infinite then let $hat\{f\}\; =\; 0,$. 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 $f\; ightarrow\; hat\{f\},$ is injective. Since the mapping is clearly surjective, "F" 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 which takes "a" in "A" to "b"^-1 "a" "b" is an algebra homomorphism, 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.