A variety $X$ is irreducible $\iff I(X)$ is prime

Question

EDIT: after I finished writing all this I came up with a possible solution, I decided to post it anyways (hope it's not against the rules) for two reason:

1. This may be useful to other people with the same doubt
2. To check if it's actually right

Given the proposition

A variety $X$ is irreducible if and only if $I(X)$ is a prime

where $I(X)$ is the ideal $I(X) = \{f\in > K[X_1,\ldots, X_n] | f(p) = 0 \text{ for all } p\in X\}$

Going in the first direction:

if $I$ is not prime then i can find $fg\in I$ such that $f\not\in I$ and $g\not \in I$. Define $J_1 = (I,f)$ and $J_2 = (I,g)$. It follows that $V(J_1)\subsetneq X$ and $V(J_2)\subsetneq X$ and that's pretty easy because intuitively $J_1,J_2$ contain more polinomial equations so the solutions need to satisfy a bigger system of equations, hence less solutions, meaning that $V(J_1),V(J_2)$ are smaller than $X$.

The last part i'm not sure of:

So $X = V(J_1) \cup V(J_2)$ is reducible

I agree on the fact that $X \supset V(J_1)\cup V(J_2)$ but i can't see why it must be equal to $X$. Of course since $X$ is a variety i have that $V(I(X)) = X$ but that's not what i have.

EDIT: My idea is that being $K[X_1,\ldots,X_n]$ a field it's also a domain, so $fg\in I(X)$ means that for all $x\in X$ $(fg)(x) = 0$ implies that $f(x) = 0$ or $g(x) = 0$ thus every solution found in $X$ is in $V(J_1)$ or $V(J_2)$. Is this right?

Answer 1

$fg$ vanishes on $X$ means $X \subset V(fg) = V(f) \cup V(g)$ so that $X = (X \cap V(f)) \cup (X \cap V(g))$.

$X \cap V(f) \subsetneq X$ because $f$ vanishes on the former but not on the latter.

Same for $g$ vanishing on $X \cap V(g)$ but not on $X$. Thus $X$ is reducible.

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With more details on the decompositions we obtain : for $\mathfrak{a}$ an ideal and $X = V(\mathfrak{a})$.

If $I(X)$ is not a prime ideal let $fg \in I(X), f,g\not \in I(X)$ $$Y = V(I(X),f) = X \cap V((f)),\qquad \qquad X = Y \cup\overline{X-Y}$$ where $\overline{X-Y}= V(I(X-Y))$ is the closure in Zariski topology.

• $Y \subsetneq X$ since $f \in I(Y), f \not \in I(X)$.

• That $fg$ vanishes on $X$ means $g$ vanishes on $X - (X \cap V(f)) =X-Y$. Whence $g \in I(X-Y) = I(\overline{X-Y})$ and $g \not \in I(X) \implies \overline{X-Y} \subsetneq X$.