Super simple question about an example of the module of differentials of a polynomial ring This will no doubt be a very simple question, but I am clearly missing something. Assume all rings are commutative with identity. Let $B$ be an $A$-algebra and denote the map
$$
\mu: B \otimes_{A} B \longrightarrow B
$$
determined by multiplication. Let $I$ denote the kernel of $\mu$. It is a well known fact that the module of differentials of $B$ over $A$ can be given by
$$
\Omega_{B/A} = I/I^{2}
$$
Now suppose $k$ is a field and we construct the polynomial ring $B = k[x]$. Then we have the map
$$
\mu: k[x] \otimes_{k} k[x] \longrightarrow k[x]
$$
determined just by polynomial multiplication. But $k[x]$ is an integral domain so the kernel of $\mu$ is surely only those tensor products $f \otimes g$ with either $f$ or $g$ being $0$. But it is also not true that $\Omega_{B/k} = 0$, since $\Omega_{B/k}$ is the free module generated by the symbol $dx$. I am brand new to modules of differentials and so am very clearly missing some basic point here. 
 A: Even if $B$ is an integral domain, the kernel of the multiplication map $\mu: B \otimes B \to B$ is not in general $0$, since for any $b \in B$, we have $1 \otimes b - b \otimes 1 \in \text{ker}(\mu)=I$. In fact, the map $B \to I/I^2$ given by $b \mapsto 1 \otimes b - b \otimes 1$ satisfies the univeral property of the module of Kahler differentials so that $\Omega_{B/A} \cong I/I^2$. 
A: The tensor product is over $k$, not $k[x]$. An example of an element
in the kernel $I$ is $x^r\otimes x^s-x^s\otimes x^r$ with $r<s$.
This is a nonzero element of $k[x]\otimes_k k[x]$ as the $x^r\otimes x^s$
form a vector space basis for this tensor product. These
$x^r\otimes x^s-x^s\otimes x^r$ form a vector space basis for $I$
and the $\phi_s=1\otimes x^s-x^s\otimes 1$ for $s\ge1$ form a free generating
set for $I$ as a $k[x]$-module.
Now
$$\phi_s\phi_t=1\otimes x^{s+t}-x^s\otimes x^t-x^t\otimes x^s
+ x^{s+t}\otimes1=\phi_{s+t}-x^s\phi_t-x^t\phi_s$$
so that
$$\phi_{s+t}\equiv x^s\phi_t+x^t\phi_s\pmod{I^2}.$$
By induction
$$\phi_s\equiv sx^{s-1}\phi_1$$
so that $I/I^2$ is generated by the image of $\phi_1$ in $I/I^2$
which we call $dx$. The existence of the standard derivation on
$k[x]$ shows that $dx\ne0$ and indeed $I/I^2$ is free of rank one
on $dx$.
