Algebra homomorphism

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A homomorphism between two algebras over a field K, A and B, is a map <math>F:A\rightarrow B</math> 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.

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 <math>f\,</math> in A to <math>\hat{f}\,</math> in B by the rule <math>\hat{f}(t) = f(t) \, </math>. A routine check shows that the mapping <math>f \rightarrow \hat{f}\,</math> is a homomorphism of the algebras A and B. If K is a finite field then let

<math>p(x) = \Pi_{t \in K} (x-t).\,</math>

p is a nonzero polynomial in K[x], however <math>p(t) = 0\,</math> for all t in K, so <math>\hat{p} = 0\,</math> is the zero function and the algebras are not isomorphic.

If K is infinite then let <math>\hat{f} = 0\,</math>. We want to show this implies that <math>f = 0\,</math>. Let <math>deg(f) = n\,</math> and let <math>t_0,t_1,\dots,t_n\,</math> be n + 1 distinct elements of K. Then <math>f(t_i) = 0\,</math> for <math>0 \le i \le n</math> and by Lagrange interpolation we have <math>f = 0\,</math>. Hence the mapping <math>f \rightarrow \hat{f}\,</math> is injective and an algebra isomorphism of A and B.