I am introducing my self to algebraic geometry, and by now I am reading Algebraic Geometry by Robin Hartshorne. In the exercise $2.16$, I have to prove that the intersection of two varieties need not be a variety.
For that, we consider de quadrics $Q_1 \equiv x^2-yw=0$ and $Q_2 \equiv xy-zw=0$. I proved that the intersection $Q_1 \cap Q_2$ is the union of the twisted cubic curve $y^2=xz$ and the line $x=w=0$. But at this point, I don't know why this union is not a variety.
Maybe by the fact that the union is not irreducible?
Best Answer
Yes. That's exactly why. Your algebraic set has two irreducible components, and this is what stops it from being a variety.