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

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?

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 $$\mathfrak{m} = (X_1-a_1, \ldots , X_n - a_n)$$.
I can give a positive answer when $$k$$ is algebraically closed.
$$k^n \to \{\mathrm{maximal \ ideals \ in \ k[X_1, \dots, X_n]}\}: (a_1, \dots, a_n) \mapsto I(\{a_1, \dots, a_n\}) = (X_1-a_1, \dots, X_n-a_n)$$ is a bijection and the result readily follows.