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This is exercise 4.2.1 of Smith's "Algebraic Geometry" - I have tried to look for a previous question, but cannot find any (probably because it is very easy, yet I am stuck on what I think is the final step).

The setup: $X$ is an affine variety, $f$ a function in $\mathbb{C}[X]$ such that $f \not \equiv 0$ on $X$. We want $f$ to be invertible in $\mathbb{C}[X]$.

What I have so far is that the distinguished open subset $D(f)$ defined to be $X\setminus V(f)$ is equal to $X$ itself. If we let $X = V(F_1, \cdots, F_r)$, then we have $X \cong V(F_1, \cdots, F_r, zf-1) \subseteq \mathbb{A}^{n+1}$ where $z$ denotes the $n+1$-th coordinate.

Hence we can express $zf-1$ as some combination of the $F_1, \cdots, F_r$, i.e. $zf-1 = 0$ in the quotient ring $\mathbb{C}[X]$, and so $f$ is invertible. My only concern is that $z$ is not itself a polynomial in $x_1, \cdots, x_n$, so I don't think this final statement actually makes any sense. On the other hand, I believe this is the correct direction to be working in.

The 'easy' way out is to note that $\mathbb{C}[X] \cong \mathbb{C}[X][\frac{1}{f}]$ in this setup, after which it is clear that $1/f \in \mathbb{C}[X]$. But I want something preferably a bit more explicit, though I'm not sure if such an answer even exists.

mi.f.zh
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  • If $f\not \in \Bbb{C}[X]^\times$ then $\Bbb{C}[X]/(f)$ is a unital commutative noetherian ring, it has a maximal ideal which is a zero of $f$. – reuns Dec 08 '19 at 12:55
  • I'm not sure I follow this argument exactly. Why then does $\mathbb{C}[X]/(f)$ become Noetherian? – mi.f.zh Dec 08 '19 at 12:58
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    $\Bbb{C}[X]=\Bbb{C}[t_1,\ldots,t_K]/J$. To produce a maximal ideal of $R=\Bbb{C}[X]/(f)$ the idea is that for each $k$, there is some non-zero $a\in \Bbb{C}[t_k]$ which is not a unit in $R$ because otherwise $R$ would contain a copy of $\Bbb{C}(t_k)$ which is not finitely generated, a contradiction. Replace $R$ by $R/(a)$ and repeat with the next $k$. The obtained ring is now a finite dimensional $\Bbb{C}$ vector space, quotienting by non-units terminates in at most $\dim$ steps, giving a maximal ideal which is a zero of $f$. – reuns Dec 08 '19 at 13:50
  • That makes sense. Does the outline given in my question also make sense/is it correct? – mi.f.zh Dec 08 '19 at 14:26

1 Answers1

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The answer I have now uses the Nullstellensatz, courtesy of my advisor. Denote by $(f)$ the ideal

Since $V(f)=\emptyset$, by the Nullstellensatz we have $ \sqrt{(f)} = I(V((f))) = \Bbb{C}[X]$, and so $1 \in \sqrt{(f)}$, giving $1 \in (f)$. Hence we can find $g \in \Bbb{C}[X]$ so that $fg=1$, and we are done.

While this does solve the problem, a nicer answer would be one that uses the content given in the section in Smith, namely using the basis of the Zariski topology.

mi.f.zh
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