The map given by evaluation at is a ring homomorphism and hence its image, which you denote , is a subring ofE. All you need to show is that it is closed under taking multiplicative inverses. Note for use in a moment that a subring of a field is an integral domain. The fact that is algebraic is equivalent to saying that is a finite-dimensionalF-vector space. Let be a non-zero element of . Multiplication by is anF-linear endomorphism and has trivial kernel (integral domain), hence is surjective by finiteness of the dimension and the rank--nullity formula. So 1 is in the image, hence a multiple of , which is thus invertible.