Let $\mathbb{F}_2$ be the field with $2$ elements. Can we write the following quotient of the ring $\mathbb{F}_2[x,y]$ simpler? $$\frac{\mathbb{F}_2[x,y]}{\langle x^{2^a}-1,y^{2^b}-1\rangle}$$ Actually, I want to find a ring isomorphic to the above ring which working with its elements can be simpler.
Thanks for your help.
Note that in characteristic $2$, $(x-1)^{2^{a}}=x^{2^{a}}-1$. Since $\mathbb{F}_2[X,Y]=\mathbb{F}_2[X-1,Y-1]$ we have that the ring in question is isomorphic to $\mathbb{F}_2[X,Y]/(X^{2^{a}},Y^{2^{b}})$, which is a bit easier to work with. This is a local ring with unique maximal ideal $(X,Y)$, because the units are exactly the polynomials with a constant term $1$. In fact $(X,Y)$ is the unique nonzero prime ideal, because it is nilpotent.