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

Examples

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Let A = K^* 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 $f\backslash ,$ in A to $\backslash hat\{f\}\backslash ,$ in B by the rule $\backslash hat\{f\}(t)\; =\; f(t)\; \backslash ,$. A routine check shows that the mapping $f\; \backslash rightarrow\; \backslash hat\{f\}\backslash ,$ is a homomorphism of the algebras A and B. If K is a finite field then let

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

p is a nonzero polynomial in K^*, however $p(t)\; =\; 0\backslash ,$ for all t in K, so $\backslash hat\{p\}\; =\; 0\backslash ,$ is the zero function and the algebras are not isomorphic.

If K is infinite then let $\backslash hat\{f\}\; =\; 0\backslash ,$. We want to show this implies that $f\; =\; 0\backslash ,$. Let $deg(f)\; =\; n\backslash ,$ and let $t\_0,t\_1,\backslash dots,t\_n\backslash ,$ be n + 1 distinct elements of K. Then $f(t\_i)\; =\; 0\backslash ,$ for $0\; \backslash le\; i\; \backslash le\; n$ and by Lagrange interpolation we have $f\; =\; 0\backslash ,$. Hence the mapping $f\; \backslash rightarrow\; \backslash hat\{f\}\backslash ,$ is injective and an algebra isomorphism of A and B.

Algebra