(Hartshorne 1.8) 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 $r-1$.
If I have another affine variety $X$ with dimension $k$ with $X \nsubseteq H$. Assume $k < r$. Then $Z = X \cup Y$ has dimension $r$. Let $W$ be an irreducible component of $Z \cap H$, then by above, $W$ has dimension $r-1$.
But what if $W$ is strictly contained in $X$, then shouldn't $W$ have dimension $k-1$ instead? So what went wrong?
Best Answer
The union $Z$ is not a variety in the sense of Hartshorne, because it is reducible. So the exercise does not apply to it, and there is no contradiction.