# A prime ideal with the algebraic set reducible

In "Algebraic Curves" by Fulton, section 1.7, page 11, there is the following corollary of the nullstellensatz:

Corollary 2: If $I$ is a prime ideal, then $V(I)$ is irreducible. There is a one-to-one correspondence between prime ideals and irreducible algebraic sets. The maximal ideals correspond to points.

Here $V(I) = \{x \in k^n | f(x) = 0 \text{ for } f \in I\}$. In the first exercise after this section, we are supposed to show this does not hold over a field which is not algebraically closed. I understand this as finding a counterexample for each of the three statements, and for the latter two, $(X^2+1) \subset \mathbb{R}[X]$ works. Specifically, this is a maximal ideal such that $V(I)$ is empty, and the ideal $(X^2+2)$ also gives us the same algebraic set.

With the usual definition where irreducible sets are supposed to be non-empty, this would have worked for the first statement as well. However, Fulton's definition of irreducible sets seems to include the empty case, and he considers the whole ring to be a prime ideal as well.

In any case, does there exist a prime ideal $I$ in some $k[X_1,\dots,X_n]$, such that $V(I)$ is (non-empty and) reducible? I can see that $n>2$, since the prime ideals in $k[X]$ are generated by irreducibles (or $0$), and $V(I)$ in this case is either a single point, empty or the whole space.

For an example, preferably $k$ should be some familiar field, say $\mathbb{Q}$ or $\mathbb{R}$, and ideally I would like an example that could be modified to work over any $k$ not algebraically closed, like the previous example (just replace $X^2+1$ by an irreducible without any roots).

Alternatively, I would like a proof that over any field $k$, if $I$ is a prime in $k[X_1,\dots,X_n]$, then $V(I)$ is (either empty or) irreducible.

$Z$ is irreducible if and only $I(Z)$ is prime. This is not hard to prove, and holds for every field ( use that a field has no zero divisors).
Therefore, $V(I)$ is irreducible if and only if $I(V(I))$ is prime.
Now for $k$ algebraically closed we have $I(V(I))= \sqrt{I}$. So, over algebraically closed fields, $V(I)$ is irreducible if and only if $\sqrt{I}$ is prime. Now, if $I$ is prime then clearly $\sqrt{I}=I$ is also prime, and we conclude $V(I)$ irreducible.
However, for fields that are not algebraically closed fields, it may happen that $I$ is prime, but $V(I)$ is reducible (because $I(V(I))$ will be in general bigger than $\sqrt{I}$, and there is no guarantee that $I(V(I))= I$ for $I$ prime ideals). For instance, $k=\mathbb{R}$, $I$ the principal ideal generated by the irreducible polynomial $p(x,y)=x^2(x-1)^2 + y^2$. The zero set $V(I)= V(p)= \{(0,0),(1,0)\}$ is reducible.
• Adding a quick way to see that the polynomial $x^2(x-1)^2 + y^2$ is irreducible. Any non-trivial factorization must have two factors linear in $y$, monic without loss of generality. Now matching coefficients with $(y+f(x))(y+g(x))$, we have $f(x)g(x) = x^2(x-1)^2$ and $f = -g$, which is impossible by unique factorization. – ronno Jan 29 '15 at 5:07
• @ronno: Excellent point. It is still reducible over $\mathbb{C}$, but we can use the idea to get say $x^2(x-1)^2 (x^2+ 1) + y^2$, now irreducible over any field of char $\ne 2$. – Orest Bucicovschi Jan 29 '15 at 5:22