I'm doing this exercise from Dummit-Foote:
Let $I = (2, x)$ be the ideal generated by $2$ and $x$ in the ring $R = \mathbb{Z}[x]$. Show that the nonzero element $2 \otimes_{R} x - x \otimes_{R} 2$ in $I \otimes_{R} I$ is a torsion element. Show infact that $2 \otimes_{R} x - x \otimes_{R} 2$ is annihilated by both $2$ and $x$ and that the submodule of $I \otimes_{R} I$ generated by $2 \otimes_{R} x - x \otimes_{R} 2$ is isomorphic to $R/ I$.
I have proved that $2 \otimes_{R} x - x \otimes_{R} 2 \neq 0$, it is annhilated by $2$ and $x$.
My question is: if $$q \in R \ \ \ \text{and} \ \ q \cdot ( 2 \otimes_{R} x - x \otimes_{R} 2 ) = 0$$
why $ q \in I = (2, x) $ ?
First of all, it should be clear that since $2$ and $x$ annihilate $2\otimes x - x \otimes 2$, then clearly all of $I$ annihilates $2\otimes x - x \otimes 2$. Moreover, the set of elements which annihilate $2\otimes x - x \otimes 2$ is clearly an ideal of $R$.
Now, consider $R/I$. The way to think about this is that we are basically killing everything in $R$ which is a multiple of $x$ or a multiple of $2$. The only things left are $0$ and $1$! Hence, $I$ is a maximal ideal of $R$. Since $2\otimes x - x \otimes 2 \neq 0$, the annihilator ideal cannot be all of $R$, so the only option is that it is actually $I$ itself.