$2 \otimes_{R} 2 + x \otimes_{R} x$ is not a simple tensor in $I \otimes_R I$ This question is related to Properties of the element $2 \otimes_{R} x - x \otimes_{R} 2$.
Another exercise of Dummit-Foote is to show that

Let $I = (2, x)$ be the ideal generated by $2$ and $x$ in the ring $R = \mathbb{Z}[x]$ . Show that the element $2 \otimes_R 2 + x \otimes_R x$ in $I \otimes_R I$ is not a simple tensor, i.e., cannot be written as $a \otimes_R b$ for some $a, b \in I $. 

I know how to show that a tensor is not simple in tensor products like $M \otimes_R N$ where $M$ and $N$ are free $R$-modules, but here we can't use this fact; so how to do it ?

The bilinear map $\varphi:I\times I\to R$ given by $\varphi(u,v)=uv$ gives rise to a homomorphism $I\otimes I\to R$. If $2\otimes 2+X\otimes X=u\otimes v$ with $u,v\in I$, then $X^2+4=uv$. Assume that $u,v$ are monic of degree one, that is, $u(X)=X+2a$ and $v(X)=X+2b$. From $a+b=0$ and $ab=1$ we get a contradiction. If $\deg u=0$, then $u=\pm1$ which is not in $I$.
