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Let $B=k[x,y]$, and $A$ is the subring of symmetric polynomials which means $g(x,y)=g(y,x)$. Show that $B$ is a free $A$-module with basis $\{1,x\}$.

Any hints please!

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$$f(x,y)=\dfrac{xf(y,x)-yf(x,y)}{x-y}+x\dfrac{f(x,y)-f(y,x)}{x-y}$$ and this representation is unique.

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I used the following: Note that $A$ is subring of $B$. Call $C = A + x A$. Show $x, y \in C$. Show $x^2 \in C$. If $f,g \in C$ then $f g \in C$. Finally prove: $f + x g = 0$ with $f,g \in A$ then $f=g=0$.

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