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$\def\frm{\mathfrak{m}} $Let $k$ be a field. Let $\frm\subset k[x_1,\dots,x_n]$ be a maximal ideal of the polynomial ring. We get a field extension $$ \label{eq}\tag{1} k\hookrightarrow k[x_1,\dots,x_n]/\frm. $$ If $k$ is algebraically closed, from the Nullstellensatz one deduces that $$ \label{max_id}\tag{2} \frm=(x_1-a_1,\dots,x_n-a_n),\;\text{for some }a_i\in k. $$ In particular, the field extension \eqref{eq} is of degree 1. Now suppose $k$ to be arbitrary again (possibly not algebraically closed). My question is: if the field extension \eqref{eq} is of degree 1 (i.e., if \eqref{eq} is bijective), then does this imply \eqref{max_id}? Even more strongly: must $\frm$ be of the form \eqref{max_id} if $k[x_1,\dots,x_n]/\frm$ is isomorphic to $k$ as a ring? (Note that for a field extension $k\hookrightarrow K$ it can happen that $K\cong k$ as rings but $[K:k]>1$, as the field endomorphism $t\in k(t)\mapsto t^2\in k(t)$ shows.)

To be honest, I don't know in what kind of fields and ideals I should look for counterexamples. The only thing to try that comes to my mind is $k=\mathbb{R}$, $n=1$, $\frm=(x^2+1)$. But this supposes no counterexample to my question.

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  • $\begingroup$ @math54321 Thanks. Do you want to post this as an answer? One question: What do you mean by a "rational maximal ideal"? $\endgroup$ Mar 9, 2023 at 8:49
  • $\begingroup$ "Rational maximal ideal" = maximal ideal at a rational point, i.e. $(x_i - a_i)$ $\endgroup$
    – math54321
    Mar 9, 2023 at 16:17

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Yes to the first question, no to the second. The point is that you want a $k$-algebra isomorphism $k[x_1, \ldots, x_n]/\mathfrak{m} = k$, i.e. your inclusion (1) is the identity. Given this, you get a surjection of $k$-algebras $\pi : k[x_1, \ldots, x_n] \twoheadrightarrow k$ whose kernel is $\mathfrak{m}$, so if $a_i := \pi(x_i)$, then $\mathfrak{m} = (x_i - a_i)$. (Note that $\pi(a_i) = a_i$ precisely because $\pi$ is a $k$-algebra map.)

If you only have a ring isomorphism $k[x_1, \ldots, x_n]/\mathfrak{m} \cong k$ this need not be the case, and you already gave a counterexample: if $t$ is an indeterminate over $k$, then $k(t)$ is a finite $k(t^2)$-algebra, e.g. $k(t) \cong k(t^2)[x]/(x^2 - t^2)$, where $(x^2 - t^2)$ is maximal in $k(t^2)[x]$, and is not of the form $(x - a)$ for any $a \in k(t^2)$.

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  • $\begingroup$ Interestingly, this argument is valid for an arbitrary number of indeterminates: if $k\to k[x_i\mid i\in I]/\mathfrak{m}$ is an isomorphism of $k$-algebras, then $\mathfrak{m}=(x_i-a_i\mid i\in I)$, where $a_i$ is the preimage of the coset of $x_i$ along the isomorphism. $\endgroup$ Aug 26, 2023 at 9:38

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