prove that an element in tensor product is non zero 
Let $I=(x,5)$ be an ideal in $\mathbb Z[x]$ where $\mathbb Z$ is the integers. Show that $5\otimes x-x\otimes 5\ne 0$ as an element in $I\otimes _{\mathbb Z[x]} I$.

I want to construct a bilinear form which is non symmetric on $I\times I$. If we can construct such a map, then by universal property, $5\otimes x-x\otimes 5$ is nonzero.
But I find it hard to construct such a non-symmetric bilinear map.
 A: Consider the function:
\begin{align}
&\Bbb Z[x]^2\times\Bbb Z[x]^2\to\Bbb Z[x]/I&
&((a(x),b(x)),(c(x),d(x)))\mapsto\begin{vmatrix}a(x)&c(x)\\b(x)&d(x)\end{vmatrix}+I
\end{align}
This is clearly bilinear, hence induces a $\Bbb Z[x]$-linear map
$$\delta:\Bbb Z[x]^2\otimes_{\Bbb Z[x]}\Bbb Z[x]^2\to\Bbb Z[x]/I$$
Now consider the exact sequence of $\Bbb Z[x]$-modules:
$$0\xrightarrow{}\Bbb Z[x]\xrightarrow{\left[\begin{smallmatrix}-5\\x\end{smallmatrix}\right]}\Bbb Z[x]^2\xrightarrow{[x,5]}I\xrightarrow{} 0$$
which gives a $\Bbb Z[x]$-module isomorphism $I\cong\Bbb Z[x]^2/\operatorname{Im}\left[\begin{smallmatrix}-5\\x\end{smallmatrix}\right]$.
By functoriality of tensor product this induces a surjective $\Bbb Z[x]$-linear map $\xi:\Bbb Z[x]^2\otimes_{\Bbb Z[x]}\Bbb Z[x]^2\to I\otimes_{\Bbb Z[x]}I$.
By alternanting properties of determinants, $\delta$ factors through $\xi$ giving rise to a $\Bbb Z[x]$-linear map
$$\psi:I\otimes_{\Bbb Z[x]}I\to\Bbb Z[x]/I$$
Now $\xi((1,0)\otimes(0,1))=x\otimes 5$ and $\xi((0,1)\otimes(1,0))=5\otimes x$, consequently
\begin{align*}
\psi(x\otimes 5)&=\begin{vmatrix}1&0\\0&1\end{vmatrix}=1&
\psi(5\otimes x)&=\begin{vmatrix}0&1\\1&0\end{vmatrix}=-1
\end{align*}
thus proving $x\otimes 5\neq 5\otimes x$.
