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Show $p(x) = x^6 + 1.5x^5 + 3x - 4.5$ is irreducible in $\mathbb Q[x]$.

By Gauss' Lemma, a primitive polynomial in $\mathbb Z[x]$ is irreducible in $\mathbb Q[x]$ if and only if it is irreducible in $\mathbb Z[x]$. We can look at $2x^6 + 3x^5 + 6x - 9 \in \mathbb Z[x]$.

Eisenstein's Criterion fails since $3^2 \mid (-9)$. I also tried replacing $x$ with $x-1$ and $x+1$ to see if I could use Eisenstein, but they didn't work. I tried reducing it mod $p$. You cant to it mod $2$ since the leading coefficient divides 2, so I tried mod 3, but it immediately factors there. I tried mod 5 and the linear terms don't have roots, but I still need to check quadratic and cubic factors. But that just seems extremely long and if it doesn't work mod 5 I'll have to keep trying mod $p$ until I reach some prime where $p(x)$ is irreducible.

I don't know where to go from here. What is the correct way to approach this?

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up vote 8 down vote accepted

Another method is to use Newton Polygon at the prime $3$. This gadget can tell you about factorizations over $\mathbb Q_3$. The vertices are $(0,2)$, $(1,1)$, and $(6,0)$, so that over $\mathbb Q_3$, there is a linear factor with root a $3$-adic unit times $3$, and an irreducible Eisenstein factor of degree five. Thus if $p$ factors at all over $\mathbb Q$, the same factorization will hold over $\mathbb Q_3$, and so there will be a $\mathbb Q$-root. But the only possibilities are $\pm3/2$ and $\pm3$, combining the above facts with the Rational Root Theorem. None of these is a root, so $p$ is irreducible.

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If you factor $2p$ modulo 5, then you get a product of two irreducible cubics. If you factor it modulo 7, then you get an irreducible quadratic times an irreducible quartic. This means that the polynomial is irreducible over $\mathbb{Z}$, since any factorisation over $\mathbb{Z}$ would induce a factorisation over $\mathbb{Z}/l\mathbb{Z}$ for any $l$. But the factorisations over 5 and 7 don't have a common coarser factorisation, other than the trivial one.

P.S.: No variation of the Eisenstein trick is going to work, since no prime is totally ramified in the number field $\mathbb{Q}[x]/(p)$.

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In the last line of your answer, you're using the fact that a prime $q$ is totally ramified in an extension $\Bbb{Q}(\alpha)$ if the minimal polynomial of $\alpha$ over $\Bbb{Q}$ is Eisenstein at $q$ yeah? In addition, how do you know that no prime is totally ramified in $\Bbb{Q}[x]/(p)$? Thanks. – user38268 Feb 20 '13 at 13:21
@BenjaLim I am in fact using the reverse implication. That's an "if and only if" in your first sentence. I asked MAGMA for the ramification in $\mathbb{Q}[x]/(p)$. This was not intended as a strategy for how to solve the problem, I just put it in to save time to those who will desperately try to prove irreducibility using their favourite technique. – Alex B. Feb 20 '13 at 15:03
I think we can also use the following criterion. Suppose that $K = \Bbb{Q}(\alpha)$ is a finite extension with the minimal polynomial of $\alpha$ being of degree $n$ and Eisenstein at $q$. Then $q^{n-1}| disc \mathcal{O}_K$, if $q \nmid n$ and $q^{n} | disc \mathcal{O}_K$, if $q |n$. I calculate using Macaulay2 that the discriminant is $- 3^6 \times 37^2 \times 733$. Thus we see that the only primes $q$ that we can use Eisenstein's for $p(x)$ is $q = 3$. I am still trying to show how we can discount the case $q = 3$ though. – user38268 Feb 20 '13 at 20:08
To tell the truth, i like this answer better than my own. – Lubin Mar 13 '13 at 16:15
@Lubin: Thank you! – Alex B. Mar 14 '13 at 21:46

By Gauss' Lemma $f(x)$ is irreducible in $\mathbb Q[x]$ if and only if it is irreducible in $\mathbb Z[x]$. So let us consider $2x^6 + 3x^5 + 6x - 9 \in \mathbb Z[x]$. By the Rational Root Theorem we find that there are no linear factors, which implies that there are no quintic factors. Suppose to the contrary that $f(x) = g(x)h(x)$ where $g(x) = 2x^k + a_{k-1}x^{k-1} + \cdots + a_1x + a_0$ and $h(x) = x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0$ where $k, m \geq 1$ and $k + m = 6$. We choose $g(x)$ to have the leading coefficient as 2 without loss of generality. Consider $\overline {f(x)} = f(x) \pmod 3$. So $\overline{f(x)} = 2x^6 = \overline{g(x)}\overline{h(x)}$ so $\overline{g(x)} = 2x^k$ and $\overline{h(x)} = x^m$. This implies that $3 \mid a_0, a_1, \ldots, a_{k-1}$ and $3 \mid b_0, b_1, \ldots, b_{m-1}$. If $k, m \geq 2$ then $9 \mid a_1b_0 + a_0b_1$ but observe that $6 = a_1b_0 + a_0b_1$ a contradiction. Hence we can conclude that one of $k, m$ is $1$. But this is a contradiction since we have already shown that $f(x)$ has no linear factors by the Rational Root Theorem. Hence we conclude that $f(x)$ is irreducible.

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The last resort is to try a hypothetical factorization $$(2x^2 + ax + b)(x^4 + cx^3 + dx^2 + ex + f)$$ or $$(x^2 + ax + b)(2x^4 + cx^3 + dx^2 + ex + f)$$ or $$(2x^3 + ax^2 + bx + c)(x^3 + dx^2 + ex + f).$$ You expand the products, and by comparing the coefficients with $2p$ you get $6$ equations. Then show that there is no solution $a,b,c,d,e,f\in \mathbb{Z}$.

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That's what I was thinking. Do I do this in $(\mathbb Z/5\mathbb Z)[x]$? – Robert Cardona Feb 20 '13 at 10:59
Try doing this in $\mathbb{Z}$. In $\mathbb{Z}/5\mathbb{Z}$, the polynomial $2p$ factors as $2(x^3 + x - 1)(x^3 - x^2 - x + 2)$, so this is not the way to go. – azimut Feb 20 '13 at 11:14
BTW, your polynomial $p$ seems to be quite special. I've just checked that the smallest prime $q$ such that $2p$ is irreducible in $\mathbb{Z}/q\mathbb{Z}$ is $q = 107$. – azimut Feb 20 '13 at 11:19
If we're looking at it in $\mathbb Z[x]$ then we cannot assume the factors are monic correct? Since our leading coefficient is 2? – Robert Cardona Feb 20 '13 at 21:04
@Robert Thanks, I didn't realize that. I will edit the post accordingly. – azimut Feb 20 '13 at 21:52

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