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I want to prove that $f = x^5 - 6x^3 +2x^2 - 4x +5$ is irreducible in $\mathbb Q[x]$. Since it is primitive, it is irreducible over $\mathbb Q$ iff it is irreducible over $\mathbb Z$.
Eisensteins criterion obviously doesn't work and reduction to $\mathbb Z/p \mathbb Z$ didn't work for me either.

So my attempt is to assume it factorized as two polynomials $f= gh$ with $\deg f, g \geq 1$ and solve a system of equations. I'm sure that leads to a solution/contradiction (does it?) but in algebra there always is a short or more neat solution, so I would appreciate any hint. I am only allowed to use the most basic criterions.

Staki42
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4 Answers4

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COMMENT.-There is no linear factor. If $f(x)=x^5 - 6x^3 +2x^2 - 4x +5$ and $f$ is reducible then $f(x)=g(x)h(x)$ with, say, degree of $g$ and $h$ respectivement equal to $3$ and $2$. It follows $f(n)$ is prime iff $g(n)$ or $h(n)$ is equal to $\pm1$. We can expose nine prime values for $f(n)$

$$f(0)=5\\f(1)=-2\\f(-2)=37\\f(2)=-11\\f(-4)=-587\\f(4)=661\\ f(-5)=6379\\f(-8)=-29531\\f(12)=2387\\$$ What else?

Piquito
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  • I unfortunately can't see what you are after with this approach... – Staki42 Feb 02 '18 at 10:50
  • @leo42 I think they are hinting at the method stated in Theorem 5.1 here. – Viktor Vaughn Feb 02 '18 at 23:38
  • The fact is that if $f(x)=g(x)h(x)$ then the probability of $f(n)$ be prime is very little. Besides an old conjecture still unproved says that for all irreducible polynomial $f(x)$ not constantly multiple of a number, there are infinitely many $f(n)$ primes which generalizes the Dirichlet's theorem on arithmetic progression (note that $f(x)=ax+b$ is irreducible). – Piquito Feb 03 '18 at 10:25
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First note that $f$ has no rational roots by the Rational Root Test: the possible candidates are $\pm 1$ and $\pm 5$ and none of those is a root. Thus the only possible factorization is $f = gh$ where $\deg(g) = 3$ and $\deg(h) = 2$.

Now observe that mod $2$, the factorization of $\overline{f}$ into irreducibles is $$ \overline{f} = (x+1)(x^4 + x^3 + x^2 + x + 1) \, . $$ (To see that the quartic factor is irreducible: first, note that it has no roots and then check that $x^2 + x + 1$, the only degree $2$ irreducible mod $2$, is not a factor.) This shows that the factorization above into a cubic and a quadratic is impossible, since the polynomial could only factor further mod $2$.

Addendum: Since the reduction mod $p$ map is a homomorphism, then we have $\overline{f} = \overline{g} \overline{h}$, where $\overline{f}$ denotes its reduction mod $2$. An irreducible factor $\pi$ of $\overline{f}$ must be an irreducible factor of $\overline{g}$ or $\overline{h}$, and since $\overline{g}$ and $\overline{h}$ have degrees $\leq 3$, then $\pi$ must have degree $\leq 3$, too. But we have found an irreducible factor with degree $4$, which is a contradiction. Thus the factorization $f = gh$ is impossible, so $f$ is irreducible.

Viktor Vaughn
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The polynomial is irreducible modulo $19$. So this does work. But I would also use an attempt to write it as $f(x)=g(x)h(x)$ and solve it modulo $19$. By the rational root test, there are no linear factors.

Dietrich Burde
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  • How does it help me to solve it modulo $19$? All the numbers appearing aren't so big as this would help me, are they? – Staki42 Feb 01 '18 at 20:57
  • @leo42 If you are working modulo 19, then you are working over a field instead of over $\mathbb Z$. This should make some of the computations easier. I assume. – Aaron Feb 01 '18 at 20:59
  • Leo, you said "and reduction didn't work for me either". So I thought, I could help you with that. The best way still is to write it as a product of degree $2$ and degree $3$ polynomials, and to get a contradiction modulo $19$. – Dietrich Burde Feb 01 '18 at 21:22
  • Ah, I guess I see what you mean. So no neat trick for this exercise? – Staki42 Feb 01 '18 at 21:22
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    Dietrich we may have discussed this before. Lenstra and Stevenhagen emphasize that it is enough if, factored mod two different primes, we get incompatible decomposition types. Pages 10,11 http://www.math.leidenuniv.nl/~hwl/papers/cheb.pdf – Will Jagy Feb 02 '18 at 00:46
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Hint: use the rarional root test: give you the potential solutions. Verify that these (4) candidates are not solutions.

https://en.wikipedia.org/wiki/Rational_root_theorem

Tsemo Aristide
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  • No, it is not the solution. The polynomial can potentially be reducible in $\mathbb{Q}[x]$, even if it has no rational roots at all. For example, it can be reduced to the multiplication of two polynomials with power 2 and 3 which both have rational coeffeicients and both have no rational roots. – Levon Minasian Apr 30 '21 at 09:58