We know that if $k$ is algebraically closed, then each maximal ideals of $k[x_1, x_2, \ldots , x_n]$ are of the form $(x_1 - a_1, x_2 - a_2, \ldots, x_n - a_n),$ where $a_1, a_2, \ldots , a_n \in k$ (Hilbert's Nullstellensatz Theorem). In the case when $k$ is not algebraically closed is it correct to say that a maximal ideal $m$ of $k[x_1, x_2, \ldots, x_n]$ has residue field $k$ if and only if $m = (x_1 - a_1, x_2 - a_2, \ldots, x_n - a_n)$ for some $a_1, a_2, \ldots, a_n \in k.$

Thank you.

  • $\begingroup$ The converse is correct. We can define surjective ring homomorphism from $k[x_{1}, x_{2}, ..., x_{n}]$ to $k$ by $x_{1} $ to $a_{1},..,$ $x_{n}$ to $a_{n}$ and identity on $k,$ then the kernel is $m = (x_{1} - a, ..., x_{n} - a_{n}).$ Being surjective by ring homomorphism the residue field is $k.$ $\endgroup$
    – CAA
    Jun 14, 2015 at 22:37
  • 2
    $\begingroup$ If we're going to be so careful then maybe it's good to note that you really mean that $k[x]/\mathfrak{m}$ is $k$ in the "natural" way. We know that it's possible to fields to be isomorphic to proper subfields. $\endgroup$
    – Hoot
    Jun 15, 2015 at 5:33
  • $\begingroup$ @ Hoot Thank you. Are you talking about something similar Lüroth's theorem which is related with the isomorphic to subfield? $\endgroup$
    – CAA
    Jun 15, 2015 at 7:13

1 Answer 1


Yes. Set $R=k[x_1...x_n]$ and suppose $R/m\cong k$. Then we have a homomorphism $R\to k$ given by composing $R\to R/m$ with this isomorphism, and each $x_i$ is sent to some element $a_i$ of $k$. This completely determines the kernel of $R\to k$, for instance since $R$ is a free algebra, and so realizes $m$ in the desired form.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .