minimal prime ideals over the union of two prime ideals

Question

When two subvarieties intersect properly ($X_1\cap X_2$), it should end up with a new subvariety($X_3$=$X_1\cap X_2$). I do not know how to keep track of the intersection operation from the algebraic view. I feel I can say little about $I(X_3)$.

If $p_1$ and $p_2$ are prime ideals in a commutative ring $R$, what do we know about minimal prime ideal over (ideal generated by elements in $p_1$ and $p_2$)?

I hope someone could give some examples like <$p_1$, $p_2$> is not prime...to help me get more organic feeling of what may happen.

Thanks!

Answer 1

For convenience, we discuss subvarieties of $\mathbb{A}^n$.

Say $X_1=V(I_1),X_2=V(I_2)$, so $X_i$ consists of $n$-tuples on which every polynomial in $I_i$ vanishes. The intersection consists of all tuples on which every element in both $I_1$ and $I_2$ vanishes, thus $$X_1\cap X_2=V(I_1+I_2)=V(\sqrt{I_1+I_2}).$$This intersection is a(n irreducible) variety if and only if $\sqrt{I_1+I_2}$ is prime.

Now, for example, let $n=2$, $X_1=V(x^2-y-1),\quad X_2=V(y)$. It is easy to see that $$I_1+I_2=(x^2-1,y),$$which is radical but not prime. Geometrically, the intersection consists of two points, namely $(1,0)$ and $(-1,0)$, thus it is not connected and certainly reducible.

Answer 2

The intersection of two irreducible varieties need not be irreducible: e.g. two plane curves typically intersect in finitely many points. Algebraically, if $f_1, f_2 \in k[x,y]$ are (irreducible) polynomials, then $(f_1, f_2)$ typically has height $2$, in which case every minimal prime of $(f_1, f_2)$ will be a maximal ideal in $k[x,y]$.

Explicitly: in $\mathbb{C}[x,y]$, if $f_1 = y^2 - x^3$, $f_2 = x - 1$, then any prime containing $(f_1, f_2)$ must contain $x - 1$ as well as $(y^2 - x^3) + (x^3 - 1) = y^2 - 1$, hence either $y+1$ or $y-1$, so $(f_1, f_2)$ has two minimal primes, namely $(x-1,y-1)$ and $(x-1,y+1)$. This corresponds to the two intersection points of the curves $y^2 = x^3$ and $x = 1$.

If you require the intersection to be proper, then since $p_1, p_2$ are prime by assumption, $p_1 + p_2$ will have strictly larger codimension than both $p_1$ and $p_2$.