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Let $R = k[X_1, ... , X_n]$, $Y$ an affine variety in $k^n$ of dimension $r$, and $H$ a hypersurface in $k^n$, with $Y \cap H \subsetneq Y$. I am trying to understand why every irreducible component of $Y \cap H$ must have dimension $r-1$. I have looked up some solutions online, but they seem to be glossing over some details I am not comfortable with.

We can write $Y$ as the zero set of a prime ideal $\mathfrak p$ of $R$, with $r$ the Krull dimension of $R/\mathfrak p$. We can also write $H$ as $Z(p)$, where $p$ is an irreducible polynomial in $R$. Necessarily $n-1$ is the dimension of $R/(p)$, and since $Y \cap H \subsetneq Y$, we have $p \not\in \mathfrak p$.

Now $Y \cap H$ is closed in $k^n$, and if $E$ is an irreducible component of $Y \cap H$, then $E = Z(\mathfrak q)$, where $\mathfrak q$ is a prime ideal of $R$ which is minimal with respect to containing $\mathfrak p$ and $p$. I need to argue that in fact $\mathfrak q$ is minimal with respect to properly containing $\mathfrak p$. Since $R$ is catenary, it will follow that the dimension of $R/\mathfrak q$, i.e. the dimension of $E$, is exactly $r-1$.

Now suppose $\mathfrak q_0$ is a prime ideal satisfying $\mathfrak p \subseteq \mathfrak q_0 \subsetneq \mathfrak q$. Then necessarily $p \not\in \mathfrak q_0$. I want to argue that $\mathfrak p = \mathfrak q_0$, but I don't know where to go with this.

user26857
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D_S
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1 Answers1

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Thanks to basket for the suggestion. In the integral domain $R/\mathfrak p$, the element $ p + \mathfrak p$ is not zero (we supposed $Y \cap H \subsetneq Y \iff Y \not\subseteq H \iff p \not\in \mathfrak p$). If $p + \mathfrak p$ is a unit, then the ideal $\mathfrak q/\mathfrak p$ contains a unit, so it must be the whole ring $R/\mathfrak p$. This implies $\mathfrak q = R$, or in other words $E = Z(\mathfrak q) = \emptyset$. The only way this can happen is if $Y \cap H$ is the empty set. That's because irreducible components of a nonempty Noetherian topological space are themselves nonempty.

So we may assume $p + \mathfrak p$ is not a unit in $R/\mathfrak p$. Then $\mathfrak q/\mathfrak p$ is a minimal prime ideal containing $p + \mathfrak p$, so the Hauptidealsatz implies that $\mathfrak q/\mathfrak p$ has height one, i.e. there are no prime ideals strictly between $\mathfrak q$ and $\mathfrak p$.

D_S
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