Why this is an exact sequence? Let k be a field. Consider the following sequence, where all tensors are over k: $$0\rightarrow k[x]\otimes k[x]\xrightarrow{x\otimes 1-1\otimes x} k[x]\otimes k[x]\xrightarrow{\pi}k[x]\rightarrow 0$$ The last morphism is multiplication, so its surjective and its a complex due to commutativity in k[x]. What I don't see is that it is in fact an exact sequence...
 A: Let $R:=k[x]$. A much better way to see these objects is as $k[x]\otimes k[x]=k[x,y]=R[y]$, i.e. the first $x$ remains $x$ and the second one becomes a new variable $y$. One can prove this by considering the bilinear map
\begin{align*}
k[x]\times k[x] &\longrightarrow k[x,y] \\
(f,g) &\longmapsto f(x)\cdot g(y)
\end{align*}
which factors as $k[x]\otimes k[x]\to k[x,y]$ by the universal property. It also maps the basis $\{ x^n\otimes x^m \mid n,m\in\Bbb N\}$ to the basis $\{ x^ny^m \mid n,m\in\Bbb N\}$, so it is an isomorphism.
From this perspective, the sequence becomes
$$
0 \to R[y] \xrightarrow{\quad\iota\quad} R[y] \xrightarrow{\quad\pi\quad} R
$$
where $\pi(f)=f(x)$ and $\iota(f)=(y-x)\cdot f$. Note that $f(x)$ now means that I plug $x$ into the polynomial $f\in R[y]$, which is a polynomial in $y$. For example, if $f=y^2-x$, I get $f(x)=x^2-x$.
Now all you need to do is check that the kernel of $\pi$ is actually the ideal generated by $y-x$. One inclusion is clear. Now assume that $f(x)=0$ and perform polynomial long division $f=(y-x)\cdot q + r$ with $r\in R$ because $y-x\in R[y]$ is a monic polynomial of degree one. Now the usual trick shows that 
$$
r = r(x) = (x-x)\cdot q + r(x) = f(x) = \pi(f) = 0
$$
and we may conclude that $f=(y-x)q$ is in said ideal.
