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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.

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It might help if you explained what part of the proof you are having trouble supplying detail for. –  Hurkyl Feb 26 '13 at 9:21

3 Answers 3

up vote 4 down vote accepted

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 $x \to t$.

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ok, i add the hypothesis: $K$ algebraically closed –  Federica Maggioni Feb 26 '13 at 10:01
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@FedericaMaggioni When $K$ is algebraically closed there is no need to assume that $\frac{K[x_1,\ldots,x_n]}{\mathfrak{m}}\simeq K$ in order to find this form of $\mathfrak m$! It is enough to know that $\mathfrak m$ is maximal and then use Hilbert Nullstellensatz. –  user26857 Feb 26 '13 at 10:13
    
wonderful, i did my question because i'm trying to prove nullstellensatz :-))) maybe i should have said this before, excuse me –  Federica Maggioni Feb 26 '13 at 10:44
    
@FedericaMaggioni Then I think that you misunderstood your textbooks: if want to prove such a thing like you have posted have to assume that the isomorphism is a $K$-isomorphism. –  user26857 Feb 26 '13 at 11:52
    
yes, now it's clear, thank you –  Federica Maggioni Feb 26 '13 at 14:57

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 $x_i - a_i \in \mathfrak{m}$.

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did you mean $a_i:=\phi(\overline{x_i})$? –  Federica Maggioni Feb 26 '13 at 8:58
    
Yes; I'll edit the answer accordingly. –  Johannes Kloos Feb 26 '13 at 10:01
    
ok, now the only thing i don't understand is: "since $k$ is a field, we also have $\phi(\overline{a_i})=a_i$". Could you explain this to me? –  Federica Maggioni Feb 26 '13 at 10:04
    
Actually, this requires that the isomorphism is a $K$-algebra isomorphism (see Hurkyl's answer). I assume that this is the case here. –  Johannes Kloos Feb 26 '13 at 10:25

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.

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