Maximal ideals in polynomial rings

Question

Let $K$ be a field. Let $\mathfrak{m}$ be an ideal of the polynomial ring $K[x_1,\ldots,x_n]$ and suppose the quotient $\frac{K[x_1,\ldots,x_n]}{\mathfrak{m}}$ to be isomorphic to $K$ itself. I want to prove that $\mathfrak{m}$ is of the form $$\mathfrak{m}=(x_1-a_1,\ldots,x_n-a_n)$$ for some $a_1,\ldots,a_n$ in $K$.

All textbooks I consulted omit the proof of this fact, since it is "obvious". Ok, the fact in itself is obvious for me too (since the quotient must be $K$, then all variables $x_i's$ are to be deleted...) but I would like to write down a precise proof. Any help is welcome.

Answer 1

You're missing a hypothesis. Questions like this are often implicitly asked not in the category of rings, but in the category of $K$-rings. In particular, a $K$-ring is a ring $R$ equipped with the extra structure of an action of $K$ on $R$ (equivalently, a homomorphism $K \to R$). A homomorphism of $K$-rings must respect this extra structure.

Without this hypothesis, we have counterexamples. $F(t)$ is an algebraic extension of $F(t^2)$ that is isomorphic to $F(t^2)$. So we have

$$ F(t)[x] / (x^2 - t) \cong F(t) $$

where the isomorphism sends $t \to t^2$ and $x \to t$.

Answer 2

If the quotient map $k[x_1, \ldots, x_n] \to k$ sends $x_i$ to $a_i$ then $x_i - a_i$ is contained in the kernel.

Answer 3

Take the isomorphism $\varphi: k[x_1,\ldots,x_n]/\mathfrak{m} \to k$ and consider the images of $\overline{x_1}, \ldots, \overline{x_n}$: Set $a_i := \varphi(\overline{x_i})$. Since $k$ is a field, we also have that $\varphi(\overline{a_i}) = a_i$, thus $x_i - a_i \in \mathfrak{m}$.