Ideal property is not upper join-closed for alternating rings

Statement
It is possible to have an alternating ring $$R$$ (i.e., an abelian group with an alternating biadditive map to itself), a subring $$I$$ of $$R$$, and intermediate subrings $$A,B$$ of $$R$$ such that $$I$$ is an ideal in both $$A$$ and $$B$$ but not in the subring generated by $$A$$ and $$B$$.

Related facts

 * Ideal property is upper join-closed for Lie rings
 * Normality is upper join-closed
 * Normality is not upper join-closed for algebra loops

Proof
Consider a four-dimensional vector space $$R$$ over any field, with basis $$x,y,z,w$$. To define the product, it suffices to describe what it does on the basis elements, subject to the condition that the product of each basis element with itself is zero and the product of any two distinct basis elements is the negative of the product in the reverse order. We define the product as follows:

$$x \cdot y = 0, x \cdot z = 0, x \cdot w = w, y \cdot z = w, y \cdot w = w, z \cdot w = w$$.

Define $$I$$ to be the one-dimensional subspace spanned by $$x$$. Let $$A$$ be the subspace spanned by $$x$$ and $$y$$ and $$B$$ be the subspace spanned by $$x$$ and $$z$$. Then:


 * $$I$$ is an ideal in $$A$$: In fact, the product on $$A$$ is identically zero.
 * $$I$$ is an ideal in $$B$$: In fact, the product on $$B$$ is identically zero.
 * $$I$$ is not an ideal in the subring generated by $$A$$ and $$B$$: The subring generated by $$A$$ and $$B$$ contains $$w = y \cdot z$$. But $$x \cdot w = w \notin I$$, so $$I$$ is not an ideal in the subring generated.