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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 $\mathbb{Z}_{2}$?

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

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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.

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