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I'm working on an exercise that asks to compute galois group of polynomial $f(X)=X^6+22X^5-9X^4+12X^3-37X^2-29X-15$ over rationals. To compute the galois group, I'm trying to compute the galois group of the polynomial reduced modulo 2, 3, 5. Computing galois group of irreducible polynomial over finite field is straightforward, so actually the problem is mostly about finding the irreducible factors of the reduced polynomial.

So that is where I am having trouble with since my method to find the irreducible factors seems to be having too long and messy calculations. One way I'm using is explicitly finding its root via substitution of the elements of the $\mathbb{F}_{p^n} $. The other way I'm using is to assume $f(X)=f_1(X)f_2(X) $ where $f_1$ is in some form such as $X^2+aX+b$ and the coefficients are undetermined. If there is such factorzation, then comparing coefficients of both sides would yield appropriate values for the undetermined coefficients.

Since the degree is not so high, continuing with those elementary methods would eventually lead to galois group of the polynomial. However, I wonder if I could have used some simpler method to deal with reduced polynomial.

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    $\begingroup$ Modulo $p$ you can find the product of all the (distinct) linear factors by calculating the gcd with $x^p-x$, the product of all the (distinct) quadratic factors by calculating gcd with $x^{p^2}-x$, the cubic factors by calculating the gcd with $x^{p^3}-x$. Your polynomial has "obvious" linear factors modulo $p=3$ and $p=5$ that you could also find by testing for zeros in the prime field. The remaining factors turn out to be irreducible, and you can verify that with the test described above. $\endgroup$ – Jyrki Lahtonen Feb 24 '18 at 7:29

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