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Question:

$k$ is an algebraically closed field. Let $f \in k[x_1, \ldots, x_n]$ be an irreducible polynomial. Show that $Z(yf-1)\subseteq \textbf{A}^{n+1}$, with coordinates $x_1, \ldots, x_n, y$, is irreducible.

Attempt:

I tried to use the general approach that the set is irreducible iff $(yf-1)$ is a prime ideal iff the coordinate ring is an integral domain, but there has been no concrete progress. I think the main difficulty is I don't know where to use the condition $f$ is irreducible.

Thanks for help.

user26857
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ViciousPaws
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    Do we really need $f$ irreducible? – user26857 Apr 13 '16 at 16:37
  • Thanks for the answer. I thought $f$ being irreducible is to ensure it's never zero, but in that case $yf \neq 1$. Probably the condition is for the later part (about a morphism induced by projection $\textbf{A}^{n+1} \to \textbf{A}^{n}$). – ViciousPaws Apr 13 '16 at 16:58

2 Answers2

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As suggested by user26857, let's use the fact that the localisation $R_{(f)}$ is isomorphic to $R[y]/(fy - 1)$. If $R_{(f)}$ is an integral domain, then the ideal $(fy - 1)$ is prime.

I leave it to you to prove that $f$ irreducible $\Longrightarrow R_{(f)}$ is an integral domain.

EDIT

$R$ is an integral domain, hence $R_{(f)}$ is an integral domain (no need that $f$ is irreducible for that).

Olivier Roche
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Hint. $R[y]/(ay-1)\simeq R[a^{-1}]$. (For more details see Localization in a ring.)

Community
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user26857
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