Suppose the symmetric group $S_2$ of order 2 acts on $k^4=Spec \;k[x_1, x_2, y_1, y_2]$ by the following: for $\sigma\not=e$, $\sigma\circ(x_1, x_2, y_1, y_2)=(x_2,x_1,y_2,y_1).$
That is, the nontrivial element in $S_2$ swaps the indeterminants $x_1$ and $x_2$, and $y_1$ and $y_2$, simultaneously.
Let $k[x_1, x_2,y_1, y_2]^{\epsilon}$ denote the vector subspace of $k[x_1, x_2,y_1, y_2]$ consisting of $S_2$-alternating polynomials, and let $k[x,y]\wedge k[x,y]$ denote the $2^{nd}$ exterior power of the vector space $k[x,y]$ of polynomials in $2$ variables.
Then how is $k[x_1, x_2,y_1, y_2]^{\epsilon}$ identified with $k[x,y]\wedge k[x,y]$?
$\mathbf{Examples}:$ Some examples of polynomials in $k[x_1, x_2,y_1, y_2]^{\epsilon}$ include the following: $g_1 = x_1-x_2,\; g_2= x_1-x_2+y_1-y_2, \;\mbox{ and } \;g_3 = (x_1 + x_2)(x_1-x_2)$
since $g_i(x_2,x_1,y_2,y_1)=-g_i(x_1,x_2,y_1,y_2)$. This basically means when you swap the $x_i$'s and the $y_i$'s, the polynomial only changes by a sign.
At the moment, this identification is not obvious to me.
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