$k$-rational points in affine varieties (characterization of closed points $x$ of affine $k$-varieties with $[\kappa(x):k]=1$)

$$\def\frm{\mathfrak{m}}$$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.

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

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 $$a \in k(t^2)$$.
• 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. Aug 26, 2023 at 9:38