Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Problem

Show that $$X^3-2008X^2+2010X-2009$$ is irreducible in $\mathbb{Q}[X]$.

Progress

I considered applying Eisenstein's Theorem, but there are no primes $p$ such that $p|2008$, $p|2009$ and $p|2010$. (This is quite clear as $\nexists p$ prime such that p divides consecutive integers for any choice of p.)

I think this may require an application of Gauss's Lemma, but I'm yet to successfully show it. Any help would be appreciated.

Regards.

share|improve this question

2 Answers 2

up vote 11 down vote accepted

HINT $\ $ By Gauss's lemma if $\rm\:f(x)\:$ splits over $\rm\:\mathbb Q\:$ then it splits over $\rm\:\mathbb Z\:,\:$ hence it splits over $\rm\:\mathbb Z/3\:,\:$ contra $\rm\ f(x)\ \equiv\ x^3-x^2+1\ $ has no roots $\rm\ mod\ 3\:.$

REMARK $\ $ Gauss's Lemma is a bit overkill here. But it's trivial to prove the simple monic case:

LEMMA $\rm\ \ f\ g\in \mathbb Z[x]\ \Rightarrow\ g \in \mathbb Z[x]\ \ \ if\ \ \ g\in \mathbb Q[x],\ \ monic\ f\in \mathbb Z[x]$

Proof $\ $ Write $\rm\ g = c\ x^n + g'\:,\ \ deg\ g' < deg\ g = n\:.\ $ Since $\rm\:f\:g \in \mathbb Z[x]\:$ its lead coeff $\rm\:c\in\mathbb Z\:.\:$ Thus $\rm\:f\ g' =\: f\ g - c\ x^n\:f\: \in\: \mathbb Z[x]\:,\:$ so $\rm\:g'\in \mathbb Z[x]\:$ by induction on deg $\rm\:g\:,\:$ so $\rm\ g = c\ x^n + g'\in \mathbb Z[x]\:.\: $ QED

share|improve this answer
    
Hadn't thought about applying splitting fields in this way; thanks! –  Mathmo Dec 10 '11 at 10:59
    
@TJO Above "f splits" means f $\ne 0$ is reducible, i.e. it "splits" into a product of two nonunit factors. No knowledge of splitting fields is employed. –  Bill Dubuque Dec 10 '11 at 14:41

Since the polynomial is a cubic, if it can be nontrivially factored, then one of the factors must be linear. So it suffices to show that the polynomial does not have rational roots. From the rational root theorem, any rational root must also be an integer. Moreover, it must be a divisor of $2009$, i.e., one of $\{ 1, 7, 41, 49, 287, 2009 \}$. [We can rule out the negative numbers because the polynomial is strictly negative for $X < 0$.] Therefore, to prove that the polynomial is irreducible, it suffices to check none of these $6$ numbers is a root of the polynomial.

share|improve this answer
    
+1: That's exactly what I wanted to write! –  M Turgeon Dec 9 '11 at 23:17
    
Does the proposed method require factoring $2009$ or did you have something simpler in mind? I cannot tell from the hint. No factorization is needed if you work mod $3$, as I hinted. –  Bill Dubuque Dec 9 '11 at 23:32
    
@Bill, Yes, you're right. I had factoring $2009$ in mind, nothing simpler. –  Srivatsan Dec 9 '11 at 23:41

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.