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I do not get why every maximal ideal of $K[X_1,...,X_n]$ is of the form $n=(X_1-a_1,...,X_n-a_n)$. (where I assume $K$ algebraically closed field)

Taking the quotient I deduce that actually $n$ is maximal. However given a maximal ideal $m$ the textbook suggests me to consider the homomorphism $f:K[X_1,...,X_n]\to K[X_1,...,X_n]/m $ given by $X_i\to a_i$.

How do I continue? Why $n$ is contained in $m$? I noticed that $f(n)=0$ since every generator of $n$ goest to $0$, but I can't see the thesis. I think it should be a small detail.

Any hints?

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  • $\begingroup$ Do you know Hilbert's nullstellensatz? $\endgroup$ – Kenny Wong Mar 1 '17 at 21:57
  • $\begingroup$ Which version? Please state the version you find useful $\endgroup$ – Richard Mar 1 '17 at 21:58
  • $\begingroup$ The thing you state IS one of the versions of the nullstellensatz. $\endgroup$ – Kenny Wong Mar 1 '17 at 21:59
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I'm going to rewrite your ideals $n,m$ as $\frak n,m$ to make the notation clearer.

Okay, so we have a maximal ideal $\frak m$, and we want to show that it is of the form $(X_1-a_1,\dots,X_n-a_n)$ for some $a_i\in K$.

Now, since $\frak m$ is maximal, $K[X_1,\dots,X_n]/\mathfrak m$ is a field, and in particular it is isomorphic to $K$ since $K$ is algebraically closed.

(Note in the last paragraph I am actually glossing over an important detail: the field $K[X_1,\dots,X_n]/\mathfrak m$ is finitely generated as a $K$-algebra, which by a version of the Nullstellensatz implies that it is finitely generated as a $K$-module, i.e. it is a finite extension of $K$.)

Now, from this isomorphism we can associate to each residue $\overline X_i$ an element $a_i\in K$. We then claim that $\mathfrak m$ is equal to $\mathfrak n=(X_1-a_1,\dots,X_n-a_n)$. To see this, since we know $\mathfrak n$ is maximal, it suffices to show $\mathfrak n\subseteq\mathfrak m$. To do this, we just need to show that $X_i-a_i\in\mathfrak m$ for each $i$, since these elements generate $\mathfrak n$. However, this is easy since if we consider the composition

$$K[X_1,\dots,X_n]\to K[X_1,\dots,X_n]/\mathfrak m\overset{\sim}{\to}K,$$

this is really just a map sending $X_i\to a_i$ and fixing elements of $K$. Clearly then $X_i-a_i$ is mapped to zero, i.e. $X_i-a_i$ is in the kernel, which is equal to $\mathfrak m$.

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  • $\begingroup$ @Richard yes that's exactly it! $\endgroup$ – Alex Mathers Mar 1 '17 at 22:10
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I assume you know about Zariski's Lemma:

Let $K$ be a field and let $A$ be a finitely generated $K$-algebra. If $A$ is a field, then it is a finite extension of $K$.

Applying this in your situation to $A=K[X_1,\ldots,X_n]/\mathfrak{m}$ shows that the canonical map $K\to A$ is an isomorphism since $K$ is algebraically closed. In particular, you may choose a preimage $a_i\in K$ of $X_i + \mathfrak{m}$ for each $i=1,\ldots,n$. But then the projection $\pi \colon K[X_1,\ldots,X_n]\to A$ sends $X_i-a_i$ to $0$, i.e. $(X_1-a_1,\ldots,X_n-a_n)\subset \mathfrak{m}$, which implies that $\mathfrak{m}=(X_1-a_1,\ldots,X_n-a_n)$.

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