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Let $k$ be a field, $E$ be a finitely generated $k$-algebra (suppose it is generated by $x_1, x_2,...,x_n$). This means that every element of K can be written as a polynomial in $x_1, x_2,...,x_n$ with coefficients in $k$, so the evaluation homomorphism of the polynomial ring $k[X_1, X_2,...,X_n]$ to $E$ is surjective, and therefore $E$ is isomorphic to a quotient of $k[X_1, X_2,...,X_n]$ . However, in Proposition $7.9$ of Atiyah-Mcdonald, $E$ is identified precisely as $k[x_1, x_2,...,x_n]$, so the evaluation homomorphism is in fact an isomorphism, without the need of quotienting by the kernel. My question is, is this always the case? If not, when are the two constructions canonically isomorphic?

JBuck
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    This is not the case: consider that $\mathbb{C}$ is a finitely-generated $\mathbb{R}$ algebra. But because it is dimension 2 as an $\mathbb{R}$-vector space, it is not a polynomial ring. – Mark Saving Dec 02 '21 at 17:00

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Hint: Try to understand what it means to be in the kernel of the evaluation homomorphism. What condition must the $x_i$ satisfy if the kernel is trivial?

  • Be algebraically independent. So I guess, if E is a field, then an element in the kernel (an equation of algebraic dependence on the generators) gives us a generator $x_i$ that can be written as a polynomial of the other generators, and thus the generator list can be reduced by inductively taking out those elements until we arrive at an algebraically independent set of generators such that the kernel is now trivial? Sorry for the late reply, I really had to think this out. – JBuck Dec 04 '21 at 15:02
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    In general, by the Noether normalisation theorem, one can always find algebraically independent elements $y_1, ..., y_m$ s.t. $k \subset k[y_1, ..., y_m] \subset E$ and $k[y_1, ..., y_m] \subset E$ is finite (as a module). In fact one can choose $m$ as the dimension of $E$ (as a ring). In general though, the extension $k[y_1, ..., y_m] \subset E$ is not trivial. – Johannes Oertel Dec 04 '21 at 20:58