Maximal ideal of $K[x_1,\cdots,x_n]$ such that the quotient field equals to $K$

Question

I was wondering whether a maximal ideals of $K[x_1,\cdots,x_n]$ such that the quotient field equals to $K$ must be the form of $(x_1-a_1,\cdots,x_n-a_n)$? Here $K$ is not necessary to be algebraically closed.

I tried to consider the Zariski's lemma, if $\mathfrak m$ is a maximal ideal of finitely generated $K$-algebra $A$, then $A/\mathfrak m$ is a finite extension of $K$. But I don't know the degree $[A/\mathfrak m:K]=1$ means what?

Answer 1

It depends on what you mean by "equal". A field can be isomorphic to a non-trivial finite extension of itself, e.g. $\mathbb{C}(Y^2) \cong \mathbb{C}(Y)$, so the residue field being isomorphic to $K$ is not a strong enough condition. Concretely, a counterexample is given by $(X^2 - Y^2) \subset \mathbb{C}(Y^2)[X]$, a maximal ideal with quotient field $\mathbb{C}(Y)$.

However, if the quotient $K[X_1, \ldots , X_n] / \mathfrak{m}$ is actually isomorphic to $K$ as a $K$-algebra, then, if $a_i$ denotes the image of $X_i$ in $K$, it is easy to see that $X_i - a_i$ lies in $\mathfrak{m}$ for each $i$, and therefore $\mathfrak{m} = (X_1-a_1, \ldots , X_n - a_n)$.

Answer 2

I can give a positive answer when $k$ is algebraically closed.

From the Hilbert Nulstellensatz, we know that

$$k^n \to \{\mathrm{maximal \ ideals \ in \ k[X_1, \dots, X_n]}\}: (a_1, \dots, a_n) \mapsto I(\{a_1, \dots, a_n\}) = (X_1-a_1, \dots, X_n-a_n)$$ is a bijection and the result readily follows.

See also: basic question on maximal ideals in $K[X_1,...,X_n]$