$6x^5+14x^3-21x+35$ is irreducible in $\mathbb{Q}[x]$

Question

I am trying to show that $p(x)=6x^5+14x^3-21x+35$ is irreducible in $\mathbb{Q}[x]$.

I would like to be able to use Eisenstein's Criterion which states:

Let $P$ be a prime ideal of the integral domain $R$ and let $f(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x+a_0$ be a polynomial in $R[x]$, where $n \geq 1$. Suppose that $a_{n-1}, \ldots, a_1 \in P$ and suppose that $a_0$ is not an element of $P^2$. Then $f(x)$ is irreducible in $R[x]$.

The problem is that the polynomial $p(x)$ that I am given does not have a leading coefficient of 1. Is there another approach I can take? I would like to be able to write a self-contained proof, if possible.

You can still use Eisenstein's criterion provided that the prime in question doesn't divide the leading coefficient. — carmichael561, Apr 18 '17 at 01:46
I'm sure you see it yourself. The polynomial is Eisenstien with prime 7. Therefore irreducible. It's a one-line proof! — Kaynex, Apr 18 '17 at 01:56
Your approach might be to divide the whole thing by $6$ and use Eisenstein, but with the Principal Ideal Domain $\Bbb Z[1/6]$ instead of $\Bbb Z$. — Lubin, Apr 18 '17 at 02:49
It is probably a typo, but you need that $a_0\in P$ as well. — Servaes, Apr 19 '17 at 18:19
$13$ is the smallest prime $p$ for which $6x^5+14x^3-21x+35$ is irreducible in mod $p$ but it'll probably be a lot of work to prove that without a computer. Nevertheless, see https://math.stackexchange.com/a/175253/589. — lhf, Apr 19 '17 at 18:50
Reducing mod $5$ already shows that $p(x)$ must have two nonlinear factors, maybe that helps? — Servaes, Apr 20 '17 at 00:21
@Ana There is an extended Eisenstein's criterion that fits to your case. (The accepted answer is simply a proof of this criterion for your particular example.) — user26857, Apr 21 '17 at 15:40

score 3 · Answer 1 · edited Apr 19 '17 at 22:34

3

You can get rid of the coefficient $6$ by multiplying your equation by the quantity $6^4$, then performing the substitution $y = 6x$; this will give you a MONIC polynomial in y with integer coefficients. Eisenstein now applies.

edited Apr 19 '17 at 22:34

Smylic

6,715

answered Apr 19 '17 at 20:48

PMar

31
1

1

If I show that this new polynomial is irreducible over $\mathbb{Z}$ using Eisentein's Criterion then how would this imply that the original polynomial is irreducible? – Ana Apr 19 '17 at 23:14

score 3 · Accepted Answer · answered Apr 19 '17 at 23:11

Because $p(x)\in\Bbb{Z}[x]$ is primitive, by Gauss' lemma it suffices to show $p(x)$ is irreducible in $\Bbb{Z}[x]$. Suppose toward a contradiction that $p(x)$ is reducible, say $p(x)=f(x)g(x)$ with $\deg f,\deg g>0$. Reducing modulo $7$ yields two polynomials $\overline{f}(x),\overline{g}(x)\in\Bbb{F}_7[x]$ such that $\overline{f}(x)\overline{g}(x)=\overline{p}(x)=\overline{6}x^5$. This means $\overline{f}(x)=ax^m$ and $\overline{g}(x)=bx^n$ for some $a,b\in\Bbb{F}_7$ and $m,n\in\Bbb{N}$. Because the product of the leading coefficients of $f$ and $g$ is $6$, the leading coefficients of $f$ and $g$ are not divisible by $7$, so $$m=\deg\overline{f}=\deg f>0\qquad\text{ and }\qquad n=\deg\overline{g}=\deg g>0.$$ This means the constant coefficients of $f(x)$ and $g(x)$ are divisible by $7$, and so the constant coefficient of $p(x)=f(x)g(x)$ is divisible by $7^2$. But $35$ is not divisible by $7^2$, a contradiction. Hence $p(x)$ is irreducible in $\Bbb{Z}[x]$, and by Gauss' lemma it is irreducible in $\Bbb{Q}[x]$.

What do you mean by $p(x)$ being primitive? – Ana Apr 20 '17 at 02:23 — Ana, Apr 20 '17 at 02:23
That the $\gcd$ of its coefficients is $1$. – Servaes Apr 20 '17 at 10:07 — Servaes, Apr 20 '17 at 10:07

Servaes · Answer 3 · 2017-04-19T18:22:58.147

Take $R=\Bbb{Z}[\frac16]$, $P=7R$, and $a_4=0$, $a_3=2\cdot\frac{1}{6}\cdot7$, $a_2=0$, $a_1=-3\cdot\frac{1}{6}\cdot7$ and $a_0=5\cdot\frac{1}{6}\cdot7$.

Verify that $5\cdot\frac16\cdot7\notin7^2R$ to find that $$f(x)=x^5+\frac{14}{6}x^3-\frac{21}{6}x+\frac{35}{6},$$ is irreducible in $R[x]$, hence so is your polynomial $6f(x)$ because $6$ is a unit in $R$. Then certainly your polynomial is also irreducible in $\Bbb{Z}[X]$ because $\Bbb{Z}\subset R$, and so by Gauss' lemma it is irreducible in $\Bbb{Q}[X]$.

Needless to say this argument holds for any polynomial whose leading coefficient is not in $P$, and it is worth verifying that Eisenstein's criterion works not only for monic polynomials, but for all polynomials satisfying $a_n\notin P$, where $a_n$ is the leading coefficient.

What exactly is the set $7R$? Is it ${ 7p(1/6): p(1/6)\in R}$? — Ana, Apr 19 '17 at 14:55
Yes; it is the ideal of $R$ generated by $7$, so all $R$-multiples of $7$. — Servaes, Apr 19 '17 at 18:16

score 1 · Answer 4 · answered Apr 19 '17 at 18:24

Use p=7. 7 does not divide 6 it divides 14,21,and 35 but 7^2=49 does not divide 35. Hence, the polynomial is Eisenstein at p=7 and thus is irreducible in $\mathbb{Q}[x]. Now, this is an equivalent formulation of the Eisenstein Criterion you stated above.

$6x^5+14x^3-21x+35$ is irreducible in $\mathbb{Q}[x]$

4 Answers4