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I have the following problem:

Factor the integer polynomial $x^5+2x^4+3x^3+5$ modulo 2, and over $\mathbb{Q}$.

For $2$, the polynomial becomes $f(x)=x^5+x^3+1$. It is easy to check if it has roots in $\mathbb{Z}_2$, namely $f(0)=1$ and $f(1)=1$, so, it doesn't have roots. Wouldn't this mean the polynomial is irreducible? Or is there something I am missing?

As for $\mathbb{Q}$, I'd think I can also try to find the roots, but it's a polynomial of degree five, so finding the roots is not trivial. So, there probably is a different method. I'd appreciate any help because I really don't know.

user26857
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Schach21
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    You still need to exclude the possibility of factoring $f(x)$ as a product of a degree $2$ and a degree $3$ polynomial. The rational roots you can exclude using the rational root theorem. – plop May 11 '21 at 02:41
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    Just to follow up, the irreducible polynomials of degree 2 and 3 in $\mathbb{Z}_2$ are $x^2+x+1$, and $x^3+x^2+1$, $x^3+x+1$, respectively. I can simply try both combinations and find that they are not factor of $x^5+x^3+1$. So, would that be enough to show that that polynomial is irreducible in $\mathbb{Z}_2$? – Schach21 May 11 '21 at 03:04
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    Correct, @Schach21. Actually checking divisibility by that quadratic factor is enough. For if a degree $n$ polynomial factors then one factor has degree at most $n/2$. Just like when you test an integer $q$ for primality you only ever need to est potential factors up to $\sqrt{q}$. With polynomials we use the degree as a substitute for size. Only this time the size of the product is the SUM of the sizes of the factors (rather than their product). Here $n/2=5/2$ so one factor would have to be at most quadratic. – Jyrki Lahtonen May 11 '21 at 03:27
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    Irreducibility over $\Bbb{Q}$ follows from irreducibility over $\Bbb{Z}_2$ and Gauss's lemma. I try to explain it here. The theme has been touched by many other users (in other threads) also. – Jyrki Lahtonen May 11 '21 at 03:31
  • @JyrkiLahtonen So, the polynomial is irreducible in $\mathbb{Z}_2$, but it's reducible in $\mathbb{Z}_3$, so wouldn't that contradict what you said in your answer (and also here), namely that the polynomial in question is irreducible in $\mathbb{Q}$. – Schach21 May 12 '21 at 01:34
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    No. The flow of logic is a one-way street. If a monic polynomial with integer coefficients is irreducible over $\Bbb{Z}_p$ for some prime $p$, then it is irreducible over $\Bbb{Q}$. But the other direction fails. Such a polynomial can even be reducible over $\Bbb{Z}_p$ for ALL PRIMES $p$, and still be irreducible over $\Bbb{Q}$. See here. – Jyrki Lahtonen May 12 '21 at 03:52
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    Similarly, if such a polynomial is reducible over $\Bbb{Q}$, by Gauss's lemma it is reducible over $\Bbb{Z}$. Hence automatically reducible over $\Bbb{Z}_p$ for all primes $p$. The difficult (sometimes impossible) direction is to "lift" a factorization modulo $p$ to a factorization over $\Bbb{Z}$. – Jyrki Lahtonen May 12 '21 at 03:53

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