What are elements of the field $\frac{\mathbb{F}_2[x,y]}{x^3-y^2+x+1}$? Is that polynomial irreducible?

Question

On a previous problem, I had the field $A=\mathbb{F}_3[x]$ and the polynomial $p(x)=x^3+x+1$, where that polynomial is reducible in $\mathbb{F}_3$, then I had $\frac{A}{p(x)}$ and I had to find its elements, which is easy knowing that $\frac{A}{p(x)}$ is isomorphic to $\frac{\mathbb{F}_3[x]}{x-1} \times \frac{\mathbb{F}_3[x]}{x^2 + x + 2}$. So now I've got this field $\frac{\mathbb{F}_2[x,y]}{x^3-y^2+x+1}$. And I need to conclude how would the previous problem change if now I worked with this new field, but I've got no clue on how to work with it since it works on two variables. Is the polynomial irreducible? How are its elements defined? Any help? I don't really need to do much, I just need to come to some conclusion.

Answer 1

Since $1=-1$ in $\mathbb{F}_2$, we can replace all the minus sings ($-$) by plus signs ($+$).

First we prove that the polynomial $p(x,y)=y^2+x^3+x+1\in\mathbb{F}_2[x,y]$ is irreducible. To do so, note that $\mathbb{F}_2[x,y]=(\mathbb{F}_2[x])[y]$, that is, we can see $p(x,y)$ as a polynomial in the indeterminate $y$ with coefficients in the ring $\mathbb{F}_2[x]$. As such, $p(x,y)$ is a polynomial of degree $2$ in $y$ so, if its is reducible, it must be either a product of a degree two and a degree zero polynomial in $y$ or a product of two polynomials of degree one in $y$. The former case is easy to rule out. In the latter we have $$ p(x,y)=(c(x)y+a(x))(d(x)y+b(x)), \quad a(x),b(x),c(x),d(x)\in\mathbb{F}_2[x]. $$ By comparing the coefficients of $y^2$ in both sides, you can conclude that $c(x)$ and $d(x)$ are units, and we can assume that both equal $1$. So we have $$ p(x,y)=(y+a(x))(y+b(x)). $$ Expanding this product, you obtain that $a(x)+b(x)=0$ and $a(x)b(x)=x^3+x+1$. But then $a(x)=b(x)$ (recall that $1=-1$ in $\mathbb{F}_2$), that is, $(a(x))^2 = x^3+x+1$. But this is impossible, since $(a(x))^2$ must have even degree, and $x^3+x+1$ has odd degree. This proves that $p(x,y)$ is irreducible.

Now, to describe the elements of $R=\mathbb{F}_2[x,y]/(p(x,y))$, you have already the answer in the comments. I just complement it (you must verify the details):

Let $\varphi:\mathbb{F}_2[x]\oplus\mathbb{F}_2[x]\to R$ be the $\mathbb{F}_2$-linear map given by $$ \varphi(a(x),b(x)) = a(x)+b(x)y + (p(x,y)). $$ Let $\psi':\mathbb{F}_2[x,y]\to \mathbb{F}_2[x]\oplus\mathbb{F}_2[x]$ defined by $$ \sum a_i(x)y^{i}\mapsto \left( \sum a_{2i}(x)(x^3+x+1)^{i},\sum a_{2i+1}(x)(x^3+x+1)^{i} \right). $$ Note that $(p(x,y))$ is contained in the kernel of $\psi'$ so this induces a $\mathbb{F}_2$-linear map $\psi:R\to \mathbb{F}_2[x]\oplus\mathbb{F}_2[x]$. After a simple calculation, you can show that $\varphi$ and $\psi$ are mutually inverse linear maps, so that $R$ is isomorphic, as a vector space, to $\mathbb{F}_2[x]\oplus\mathbb{F}_2[x]$. Under this isomorphism, the pair $(a(x),b(x))\in \mathbb{F}_2[x]\oplus\mathbb{F}_2[x]$ corresponds to $a(x)+b(x)y+(p(x,y))$. Thus, if we write $\overline{x}$ and $\overline{y}$ to denote the image of $x$ and $y$ in $R$, respectively, we see that the elements of $R$ are all of the form $a(\overline{x})+b(\overline{x})\overline{y}$ for $a(x),b(x)\in \mathbb{F}_2[x]$.