Automorphism of order 2 of C[x,y] and its ring of invariants Let $A=\Bbb C[x,y]$.
For an automorphism $\sigma$ of order $2$ of the algebra $A$, let $B=\{ f \in A \mid \sigma f = f \}$ be the subalgebra of $A$
consisting of all fixed points of $\sigma$,
and denote $C=\{ f \in A \mid \sigma f = -f \}$.
For the opening stage I can prove that $C$ is a $B$-module.
Here, the action of $B$ on $C$ is given by the multiplication in $A$.
Then I came up with two questions:

(1) If $\sigma$ is defined by $(\sigma f)(x,y)=f(y,x) (f \in A)$,
  then prove that $A$ is a free $B$-module. (Here, the action of $B$ on $A$ is given by the multiplication in $A$.)

My attempt: To prove that $A$ is a free $B$-module we need to show that $A \cong B^k$ for some $k \in \Bbb Z_{>0}$. So for any $f \in A$ I try to express $f$ as $f=f_1+f-f_1$, where $f_1 \in B$... It seems that it couldn't work with this approach.

(2) If $\sigma$ is defined by $(\sigma f)(x,y)=f(-x,-y) (f \in A)$,
  determine whether $C$ is a free $B$-module or not.

As before I couldn't figure out how to use the new defining operations of $\sigma$. Moreover I'm lack of knowledge on how to prove a module that is free. Any help would be much appreciated.
 A: This is the answer suggested in my hint.
First note that for $A=B\oplus C$ as $B$ modules. Indeed, you have clearly $B\cap C=\{0\}$ and every $f\in A$ can be written $\frac{1}{2}(f+\sigma f) + \frac{1}{2}(f-\sigma f)$ with the first term in $B$ and the second in $C$.
Now for (1) you can surely use user26857 beautiful argument, but to check that $1$ and $x$ span $A$, you need to do something very close to the proof I had in mind. Here is another way. If you can show that $C$ is free, then $A=B\oplus C$ will also be free as sum of two free modules.
Let us prove that $C$ is free with basis $(x-y)$. Use euclidean division to write any $f\in C$ as $f(x,y)=(x-y)Q(x,y) + R(x)$. Now because $f$ is anti-symetric, swithing $x$ and $y$ gives
$-f(x,y)=f(y,x)=(y-x)Q(y,x)+R(y)$. Summing both equalities, we get
$$(x-y)(Q(x,y)-Q(y,x)+R(x)+R(y)=0$$
if we take $x=y$ in the above equality, we get $2R(x)=0$, hence $R=0$. You also have $(x-y)(Q(x,y)-Q(y,x))=0$, so $Q$ is symetric. We have shown that any anti-symetric $f$ can be written uniquely as $(x-y)Q$ for a symetric $Q$. Hence $C$ is free.
For (2) : Assume that $C$ as a $B$-basis $(e_i)$. Write $x=\sum a_i e_i$ and $y=\sum b_i e_i$ with the $a_i,b_i\in B$. Now consider $x^2y$. You have
$$ x^2y=(xy)x=\sum (xy)a_i e_i \\
x^2y=(x^2)y =\sum (x^2)b_i  e_i $$
Note that $(xy)a_i\in B$ and $(x^2)b_i\in B$. So because $(e_i)$ is a basis, you have for any $i$, $(xy)a_i=(x^2)b_i$. Hence $ya_i=x b_i$. This implies that $a_i=x a'_i$ (using unique factorization in $A$). So that $x=\sum (xa'_i)e_i$, hence $1=\sum (a'_i)e_i$. This is a contradiction, because for example the $e_i$'s don't have constant term.
A: (1) $B$ is the subring of symmetric polynomials, that is, $B=\mathbb C[x+y,xy]$. Then $A$ is a free $B$-module, a basis being e.g. $\{1,x\}$.
(2) In this case $B=\mathbb C[x^2,xy,y^2]$. (As a side note, $A$ is not free over $B$.) Moreover, $C=Bx+By$. Suppose $C$ is a free $B$-module and let $(e_i)$ be a basis. Then $x=\sum f_ie_i$ and $y=\sum g_ie_i$ with $f_i,g_i\in B$. Since $(xy)x=(x^2)y$ we get $xyf_i=x^2g_i$, that is, $yf_i=xg_i$. It follows that $f_i=xh_i$ and $g_i=yh_i$. From $x=\sum f_ie_i$ we get $1=\sum h_ie_i$ and sending $(x,y)$ to $(0,0)$ we reach a contradiction.
