Algebra homomorphism

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A homomorphism between two algebras over a field K, A and B, 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 injective then F is said to be an isomorphism between A and B.

[edit] 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 \rightarrow \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 \rightarrow \hat{f}\, is injective and an algebra isomorphism of A and B.

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