Automorphism of $K$ extending to $K[x_1,\dots,x_n]$. I think it is well known that if $K$ is field, and $\sigma \in Aut(K)$,
then the extended map  $\sigma: K[x,y] \rightarrow  K[x,y]$, given by 
$\sum a_{ij}x^iy^j \mapsto \sum \sigma(a_{ij})x^iy^j$ is an isomorphism.
Proving that 
$$\sigma(f(x,y)\cdot g(x,y))=\sigma(f(x,y))\cdot \sigma(g(x,y))$$
was particularly challenging for me. I had to play a bit with the monomials of $f$ and $g$, and arrange them in a suitable way. I don't even know if I can extend that procedure to 
the general ring $K[x_1,\dots,x_n]$.
I was wondering if someone know of  an  alternative (preferably neat)  proof of
$\sigma(fg)=\sigma(f)\sigma(g)$
 A: The right way to do this is to invoke the universal property of the polynomial ring. If $R$ is any commutative ring with identity and $S=R[X_1,\ldots,X_n]$, then $S$ is characterized up to isomorphism by the following property: for every commutative ring $R^\prime$ with identity and every ring homomorphism $\varphi:R\rightarrow R^\prime$ sending $1_R$ to $1_{R^\prime}$ and any $n$ elements $r_1^\prime,\ldots,r_n^\prime$, there is a unique ring homomorphism $\psi:S\rightarrow R^\prime$ with $\psi(r)=\varphi(r)$ for each $r\in R$ and $\psi(X_i)=r_i^\prime$. 
This is the way one defines homomorphisms out of polynomial rings. So in your case, you would take $R=k$ and $R^\prime=S$ and let $k\rightarrow S$ be the automorphism of $k$, $\sigma$, followed by the inclusion $k\hookrightarrow S$. This extends uniquely to a ring homomorphism $\psi:S\rightarrow S$ restricting to $\sigma$ on $k$ and sending each $X_i$ to $X_i$, by the universal property, and you can check directly that it is an automorphism because $\sigma$ is an automorphism (you can also check that is the map you want, applying $\sigma$ to each of the coefficients of a polynomial). But you don't need to check that $\psi$ is a homomorphism. It is a homomorphism by definition.
A: Hint
$$K[x_1,\cdots,x_n] = K[x_1][x_2,\cdots,x_n] = K[x_1][x_2]\cdots[x_n]$$
If you prove this, and use what you proved for one variable, then by induction you can extend $\sigma$ $n$ times.
