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In exercise 1.8 of chap I in Hartshorne algebraic geometry,

Let $Y$ be an affine variety of dimension $r$ in $\mathbf A^n$. Let $H$ be a hypersurface in $\mathbf A^n$, and assume that $Y \nsubseteq H$. Then every irreducible component of $Y \cap H$ has dimension $r-1$.

I refered to a solution. In this solution, why $f$ is not a unit in $B$?

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  • $\begingroup$ It could well be a unit if the two do not meet, e.g., $n = 1$, $Y = Z(x)$, $f = x - 1$. He is probably assuming that the two do meet so that there is something to prove; if you win a Fields Medal I think you're allowed to be a little sketchy here. $\endgroup$
    – Hoot
    Jun 6 '15 at 21:17
  • $\begingroup$ @Hoot My comment under the answer points out very clearly what condition we need for getting an invertible element. $\endgroup$
    – user26857
    Jun 6 '15 at 21:29
  • $\begingroup$ @user26857 Yes, definitely; I hadn't read the comments to Ayman's answer, I must admit. I just wanted to point out that this is very easy for the OP to verify. $\endgroup$
    – Hoot
    Jun 6 '15 at 21:32
  • $\begingroup$ Some of you understood why the statement is true if the intersection $Y\cap H=\emptyset$ ? I read on somewhere that the dimension of the empty set is $-1$ by definition. Is it true? How does it imply the thesis? I'm considering only basic theory (since this exercise is on page 8) $\endgroup$
    – Richard
    Mar 4 '17 at 14:02
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Let $H = Z(f)$ where $f$ is irreducible. Let $Y = Z(\mathfrak a)$ where $\mathfrak a$ is a prime ideal. Let $\overline f$ be the image of $f$ in the integral domain $B = A / \mathfrak a$. Every irreducible component of $Y \cap H$ corresponds to a minimal prime ideal of $B$ that contains $\overline f$. If $\overline f$ is a unit, no such prime ideal exists. Thus $Y \cap H = \varnothing$ and the given statement is vacuously true.

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    $\begingroup$ It seems that in the end $f$ is not invertible modulo $\mathfrak a$ follows from $Y\cap H\ne\emptyset$, not from $Y\nsubseteq H$ which gives only $f\ne0$ modulo $\mathfrak a$. $\endgroup$
    – user26857
    Jun 6 '15 at 16:39
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    $\begingroup$ @user26857 Indeed. What's left now is a straightforward application of Krull’s Hauptidealsatz. $\endgroup$ Jun 6 '15 at 17:14
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    $\begingroup$ Sorry why if the intersection is empty then the statement is true? $\endgroup$
    – Richard
    Mar 4 '17 at 12:29
  • $\begingroup$ Can you elaborate on this answer and prove that if $W$ is an irreducible component of $Y \cap H$, then $I(W)$ is minimal among all prime ideals of $B$ that contain $\overline f$? This is the part I have trouble with. $\endgroup$
    – JDZ
    Oct 25 '18 at 22:34
  • $\begingroup$ @Richard If their intersection is empty then by the Nullstellensatz the vanishing ideal is the whole ring $k[x_1, ..., x_n]$. So there is nothing contradictory about $f$ being a unit here. $\endgroup$ Oct 21 '20 at 19:14

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