Chapter 1, Exercise 1.8: Intersecting with a hypersurface decreases dimension by one

Let $Y$ be an affine variety of dimension $r$ in $\mathbb A^n$. Let $H$ be a hypersurface in $\mathbb A^n$, and assume that $Y\not\subseteq H$. Then every irreducible component of $Y\cap H$ has dimension $r-1$. (See (7.1) for a generalization.)

Note: It is implicit, but not stated, in this problem that $Y\cap H$ is nonempty. In the case $Y\cap H=\emptyset$ it is not even meaningful to talk about irreducible components since $\emptyset$ is not irreducible and as such has no decomposition.

This problem, probably the first truly nontrivial result given in an exercise thus far, works out most naturally algebraically because of the quotient decomposition of dimension. It is notable further for giving us an interesting application of Krull's Hauptidealsatz (Hartshorne's Theorem 1.11A).

First let us fix some notation: since $H$ is a hypersurface write $I(H)=(f)$, and let the irreducible components of $Y\cap H$ be denoted $G_i$.

Now, we will work in the affine coordinate ring $A(Y)=A/I(Y)$ of the variety $Y$. In particular, we claim that the canonical image of $f$ (the generator of $I(H)$) in this ring is neither a zero-divisor nor a unit. For the first statement, since $Y\not\subseteq H$ taking ideals gives $(f)\not\subseteq I(Y)$, so $f\notin I(Y)$. Thus, under the canonical map, $f$ is not sent to the additive identity of $A(Y)$. Now, we use the assertion that $Y\cap H\neq\emptyset$ to prove that $f$ is not a unit. If the image $\bar f$ was a unit in $A(Y)$ and we had a prime ideal $\mathfrak q$ containing $\bar f$, then $\mathfrak q=(1)$. Thus $\mathfrak q$ is not minimal - in particular, no minimal prime ideals in $A(Y)$ contain $\bar f$. We will show momentarily that this implies $Y\cap H=\emptyset$.

Our next goal is to study the algebraic representation of the $G_i$ under the correspondence. Clearly $I(G_i)$ is a prime ideal in $A$, but under the quotient map $A\to A(Y)$ the image $\mathfrak p_i$ is a prime ideal. Note that it contains $\bar f$, since $f\in I(G_i)$ and inclusion is preserved under quotient maps. It is minimal over $(\bar f)$ by the irreducibility of $G_i$, and in particular we see $\bar f$ is a unit. By the Hauptidealsatz, $\operatorname{height}\mathfrak p_i=1$ and by Theorem 1.8Ab) we see $$\dim A(Y)/\mathfrak p_i=\dim A(Y)-\operatorname{height}\mathfrak p_i=r-1.$$ Since $A(Y)$ is the coordinate ring of $G_i$ we see $\dim G_i=r-1$ as desired.