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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.

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 $K[X_1,...,X_n]$