$X^3Y+Y^3+X\in \Bbb Z_2[X,Y]$ is irreducible in $\Bbb Z_2[X,Y]$

Question

Exercise. We would like to prove that the polynomial $f(X,Y):=X^3Y+Y^3+X\in \Bbb Z_2[X,Y]$ is irreducible in $\Bbb Z_2[X,Y]$.

My attempt. Consider $f(X,Y):=X^3Y+Y^3+X\in (\Bbb Z_2[X])[Y]$ and we are looking for a factorisation in $(\Bbb Z_2[X])[Y]$. So, say $$Y^3+X^3Y+X=(p_1(X)Y+p_2(X))(p_3(X)Y^2+p_4(X)Y+p_5(Y)).$$ Then, $$p_1(X)p_3(X)=1 \implies p_1(X)=p_3(X)=1,\\ p_1(X)p_5(X)+p_2(X)p_4(X)=X^3\implies p_5(X)=X^3+p_2(X)p_4(X),\\ p_2(X)p_5(X)=X \implies X^3p_2(X)+p_2^2(X)p_4(X)=X.$$ So, $p_2(X)(X^3+p_2(X))p_4(X))=X \implies \deg_X p_2(X)+\deg_X(X^3+p_2(X)p_4(X))=1$.

Questions.

1. Is this methogology correct? And if yes, how can we reach a contradiction?
2. Can we work with the total degree for a contradiction, like this example?
3. Any other ways to prove this?

Thank you.

Answer 1

You have another approach using Eisenstein criterion for the ring of polynomials $\mathbf F_2[X][Y]$ with coefficients in the P.I.D. $\mathbf F_2[X]$:

$X$ divides all coefficients of the polynomial $Y^3+X^3Y+X$, but the leading coefficient, and $X^2$ does not divide its constant term.

Answer 2

Here is some Macaulay2 code that proves your polynomial is irreducible:

(You can run this here: http://www2.macaulay2.com/Macaulay2/TryItOut/ )

R = ZZ/2[X,Y]

I = ideal(X^3 * Y + Y^3 + X)

isPrime I

If you run this you get:

i1 : R = ZZ/2[X,Y]

o1 = R

o1 : PolynomialRing

i2 : R

o2 = R

o2 : PolynomialRing

i3 : I = ideal(X^3 * Y + Y^3 + X)

            3     3
o3 = ideal(X Y + Y  + X)

o3 : Ideal of R

i4 : isPrime I

o4 = true