Algebra homomorphism

A homomorphism between two algebras over a field K, A and B, is a map Failed to parse (Missing texvc executable; please see math/README to configure.): 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.

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 Failed to parse (Missing texvc executable; please see math/README to configure.): f\,

in A to Failed to parse (Missing texvc executable; please see math/README to configure.): \hat{f}\,
in B by the rule Failed to parse (Missing texvc executable; please see math/README to configure.): \hat{f}(t) = f(t) \, 

. A routine check shows that the mapping Failed to parse (Missing texvc executable; please see math/README to configure.): f \rightarrow \hat{f}\,

is a homomorphism of the algebras A and B.  If K is a finite field then let 

Failed to parse (Missing texvc executable; please see math/README to configure.): p(x) = \Pi_{t \in K} (x-t).\,

p is a nonzero polynomial in K[x], however Failed to parse (Missing texvc executable; please see math/README to configure.): p(t) = 0\,

for all t in K, so Failed to parse (Missing texvc executable; please see math/README to configure.): \hat{p} = 0\,
is the zero function and the algebras are not isomorphic.  

If K is infinite then let Failed to parse (Missing texvc executable; please see math/README to configure.): \hat{f} = 0\, . We want to show this implies that Failed to parse (Missing texvc executable; please see math/README to configure.): f = 0\, . Let Failed to parse (Missing texvc executable; please see math/README to configure.): \deg f = n\,

and let Failed to parse (Missing texvc executable; please see math/README to configure.): t_0,t_1,\dots,t_n\,
be n + 1 distinct elements of K.  Then Failed to parse (Missing texvc executable; please see math/README to configure.): f(t_i) = 0\,
for Failed to parse (Missing texvc executable; please see math/README to configure.): 0 \le i \le n
and by Lagrange interpolation we have Failed to parse (Missing texvc executable; please see math/README to configure.): f = 0\,

. Hence the mapping Failed to parse (Missing texvc executable; please see math/README to configure.): f \rightarrow \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.it:Omomorfismo algebrale