# Working with polynomials in $\mathbb{Z}_{2}[X]$, finding roots and splitting fields

I'm a beginner in the area and yet can't see how to work with a polynomial in other fields diferents from $$\mathbb{Q}$$.

I have the polynomial $$f(X)=X^5 -X^2+1 \in \mathbb{Z}_{2}[X]$$ and must prove that if $$\mathbb{K}$$ is his splitting field then $$[\mathbb{K},\mathbb{Z}_{2}]=5$$

$$\mathbb{Z}_{2}$$ have just two elements, $$0$$ and $$1$$ mod $$2$$, i just have no idea of how to use this field information to handle this polynomial

If it was $$f(X)=X^5 -X^2+1 \in \mathbb{Q}[X]$$ then i would use some criteria (must search something) to show that $$f$$ is irreducible over this field and then the result would follow, but with $$\mathbb{Z}_{2}$$ things just dont flow.

Anyone can give a hint to look at this polynomial as an object of $$\mathbb{Z}_{2}$$?

• Actually factorizing polynomials over finite fields is easier computationally than doing it over ${\mathbb Q}$. – Derek Holt Oct 29 '18 at 7:44

Note that $$X^5-X^2+1=X^5+X^2+1$$. As it has no root in $$\mathbf F_2$$, the only possible factorisation is as the product of irreducible polynomials of degree $$2$$ and $$3$$.
Now the only irreducible polynomial of degree $$2$$ in $$\mathbf F_2[X]$$ is $$X^2+X+1$$, so the factorisation should be \begin{align} X^5+X^2+1&=(X^2+X+1)(X^3+aX^2+bX+1)\\ &=X^5+(a+1)X^4+(b+a+1)X^3+(1+b+a)X^2+(1+b)X+1. \end{align} By identification of the coefficients, we obtain the equations $$\begin{cases} a=1,\quad b=1,\\b+a+1=0,\end{cases}$$ which are incompatible.
Obviously $$X^5+X^2+1$$ has no root, so if it's not irreducible then it's $$(X^3+aX^2+bX+1)(X^2+cX+1)$$ Note that $$c=1$$ because the quadratic must be irreducible. So multiplying it out we get $$X^5+aX^4+bX^3+X^2+X^4+aX^3+bX^2+X+X^3+aX^2+bX+1$$ Try to find values to satisfy this. We need $$a=1$$, $$b=0$$, but, uh oh, then the $$X^2$$ term cancels. So it's irreducible.