Factorization into irreducible factors

Question

I am trying to find a factorization of the following polynomial $f=X^{4}Y^{2}+X^{3}+X^{2}Y^{4}+XY^{2}$ into irreducible factors over $\mathbb{Z}_{2}[X,Y]$.

I know that $\mathbb{Z}_{2}[X,Y]$ can be written as $\mathbb{Z}_{2}[X][Y]$, so we have to bring first $f$ in the form of $\sum_{i=1}^{n}f_{i}Y^{i}$, where every $f_{i}$ is an irreducible polynomial of the ring $\mathbb{Z}_{2}[X]$, but i have problems with this and in general i don't understand how to show the factorization in cases of $\mathbb{Z}[X,Y]$, $\mathbb{Q}[X,Y]$ etc. Can anybody help me or give me a hint how to find this factorization? Thank you in advance!

Answer 1

$X^{4}Y^{2}+X^{3}+X^{2}Y^{4}+XY^{2}=X^2Y^2(X^2+Y^2)+X(X^2+Y^2)=X(X+Y)^2(XY^2+1)$.

Answer 2

Hint: Group the terms for which the sums of exponents is the same, then factor the appropriate power of the form $X^aY^b$ within each group, then notice other obvious factors.

Answer 3

Thanks to everybody for the help!

After the hint of Lucian, i figured out that i can write $f$ as $X^{2}Y^{2}(X^{2}+Y^{2})+X(X^{2}+Y^{2})=X(X^{2}+Y^{2})(XY^{2}+1)=X(X+Y)^{2}(XY^{2}+1)$.

With some more details:

• $X$ is already irreducible;
• Since we have $\mathbb{Z}_{2}$, a finite field with characteristic $2$, we have that $(X^{2}+Y^{2})=(X+Y)^{2}$, because every linear factor of the form $X+a$ in a prime power is irreducible;
• Consider the last term $(XY^{2}+1)$ i look at $X$ as a constant, my variable in this case is $Y$, clearly this polynomial has no roots in the given polynomial ring.

If you have any other comments, i will be glad to read them. Thank you one more time.