Show that if $X$ is an integral affine scheme, then every open subscheme is dense

Question

So we're doing Scheme theory in my undergraduate Algebraic Geometry course, and I'm totally lost on this question. First, we define the following notions,

Say that an open subscheme $U \subseteq X$ is dense if the only open subscheme $V\subseteq X$ satisfying $U\cap V=\emptyset$ is $V=\emptyset$, the empty (sub)scheme.

Say that $X$ is integral if its function algebra, $\mathcal{O}(X)$ is is an integral domain.

Now we want to show that $X$ integral $\implies$ every non-empty, open subscheme is dense.

The first bit is just to re-interpret this question in the language of ideals, so hopefully the following is an accurate translation of the statement:

Let $U=D_I$ and $V=D_J$ (where the notation $D_I := X\backslash Z_I$ , $Z_I$ a closed subscheme). Since we have proven that $D_{I}\cap D_{J}=D_{IJ}$, and that we know that the empty scheme is the zero locus $Z_{\langle 1 \rangle}$, it suffices to show that if $\mathcal{O}(X)=A$ is integral, $I,J \subset A$, $I\neq \langle 1 \rangle$ then

$$\sqrt{IJ}=\langle 1 \rangle \quad \text{iff} \quad J=\langle 1 \rangle$$

that is, the radical of $IJ$ generates the unit ideal if and only if $J$ itself is the unit ideal.

Is this the correct ideal-theoretic interpretation of this question? If so, do you have any hints on how to approach this? I'm trying a contrapositive approach, supposing $\sqrt{IJ} = \sqrt{I}\sqrt{J} = 1$, with neither equal to the unit ideal, and assuming $A$ to be not an integral domain. Given the first assumption, there exists $f_i \in \sqrt{I}$ and $g_j \in \sqrt{J}$ such that

$$\big(\sum_{i=1}^N \alpha_i f_i \big)\big(\sum_{i=1}^M \beta_j g_j \big) =1 $$

where $\alpha_i,\beta_j \in A$. It's not at all clear to me how to use this to gain a contradiction with the fact that $A$ is integral, say, by showing there is a nilpotent.

Any help would be appreciated.

Edit: Given that this class is approaching algebraic geometry in an unusual way, I've additionally included our definition of "open subschemes

First, let $X$ be an affine scheme (ie. a representable functor from the category of R-Algebras to Sets). Say $Z_T$ is a closed subfunctor/subscheme of $X$ if $T \subseteq \mathcal{O}(X)$, the coordinate ring of $X$. Then we say a subscheme $U$ is open if it is the complement of a closed subscheme, $U = X\backslash Z_T$. To formally define the complement, we introduce the notation of a hypersurface complement, $D_I$, given by

$$D_I(C) = \{x \in X(C)\ | \ \langle\{f_C(x) \ | \ f \in I \}\rangle \cong \langle 1 \rangle \}.$$

(where $C$ is an R-Algebra).

Answer 1

Firstly, note that to prove the claim, it suffices to show that any two nonempty open subschemes intersect. (This makes the argument nice and symmetric).

Recall (or prove) the following facts about the subfunctors $D(I)$:

1. $D(0)$ Is the empty subfunctor of $X$, and $D(k[X]) = X$.
2. $D(I) \subseteq D(J)$ if and only if $\sqrt{I} \subseteq \sqrt{J}$.
3. $D(I) = D(\sqrt{I})$.

In particular, $D(I)$ is the empty subfunctor iff $\sqrt{I} \subseteq \sqrt{0}$, which is iff $I \subseteq \sqrt{0}$. So the correct ideal-theoretic interpretation of the question is this:

• Let $I$, $J$ be ideals not contained in the nilradical ($D(I)$ and $D(J)$ nonempty). Then the ideal $IJ$ is not contained in the nilradical ($D(I) \cap D(J)$ is nonempty).

The proof of that last statement is where the integrality hypothesis must be used.