Let an ideal $I = (2,x)$ be generated by 2 and $x$ in the ring $R = \mathbb{Z}[x]$. It seems both 1 and $-1$ are not in the ideal, but I don't know why. I would appreciate a simple explanation for this.
Note: I have an ideal $I = (2,x)$ generated by 2 and $x$ in the ring $R = \mathbb{Z}[x]$. I am trying to prove that the element $2⊗2+x⊗x$ is not a simple tensor, i.e. it cannot be written as $a⊗b$, for some $a$ and $b$ in $I$. The question I posted is the last piece of the puzzle for me.
It is certainly true that $-1$ and $1$ are not in $I = (2,x)$. Suppose $\pm1 \in I = (2,x)$, then by definition $$\pm1 = 2p(x) + xq(x)\text{ for }p(x), q(x) \in \mathbb{Z}[x]$$ Substituting $x=0$ you get $\pm1 = 2p(0)$. Since $p(0) \in \mathbb{Z}$ this is a contradiction. Therefore neither $-1$ nor $1$ are in $I$.