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Is there some algebraic construction/extension to make a reducible polynomial over $\mathbb{Q}$ irreducible?

For example: consider the polynomial $x^4-x^3-x^2+1=(x-1)(x^3-x-1)\in \mathbb{Q}[x]$. Is there some space where this polynomial becomes (or can be easily seen to be) irreducible? And possibly can this construction/extension be expressed in terms of the number fields associated to the roots of the factors?

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I am not completely sure I understand what you mean. But extending the field from Q to something larger could make an irreducible polynomial reducible, yet not the converse. (You have 'more' choices for factors.) One might consider the eplynomial over a subring of Q, but then Z is the smallest one and then by Gauss's Lemma (except for 'constants') regarding irreducibility nothing changes. So, if I understand the question correctly then, no this is not possible, and you might find Gauss's Lemma interesting. But perhaps I misunderstand this question. –  quid Mar 7 '13 at 16:31
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Except on second though, in principle, one could do somethings like this, via localising the polynomial ring appropriately so that all but one factor will become invertible; this might work always except for the polynomial being a perfect power in the ring of polynomials. But then you get something quite different with this. Yet perhaps you meant this. –  quid Mar 7 '13 at 16:37

3 Answers 3

As a more elementary variant of Joel's answer, you may also consider something like this: Let $f(X)\in\mathbb Z[X]$, and let $a_n$ be the leading coefficient. Then $f_p(X)=f(X)+\frac{1}{p}$ is irreducible for every prime $p$ not dividing $a_n$, because $pf_p(X)$ is the reciprocal of an Eisenstein polynomial.

So irreducible polynomials become arbitrarily close to the polynomial you start with.

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The given polynomial $x^4-x^3-x^2+1$ is irreducible in the ring $(x-1){\mathbb Z}[x]$ (the ring of polynomials with integer coefficients that are multiples of $(x-1)$.

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This is not a ring! :-) –  quid Mar 7 '13 at 18:37
    
the poster did not ask for a ring... –  Dima Pasechnik Mar 8 '13 at 3:49
    
This is a ring (just not one with unity). In any case, the OP asked for "an algebraic construction". –  Kevin O'Bryant Mar 12 '13 at 14:56
    
@Dima Pasechnik: in the answer it is said it is a ring. @Kevin O'Bryant: yes, of course. My comment was basically a joke. (But, as I sketched before your answer in my second comment one could do something like this getting a ring with unity.) –  quid Mar 18 '13 at 13:18

I am not sure exactly what you have in mind. But

(1) If your polynomial $P \in \mathbb Q(x)$, and $K%$ is some field containing $\mathbb{Q}$, then obviously it will stay reducible over $K$.

(2) You can assume that your polynomial is monic, and then its coefficients are in $\mathbb Z[1/N]$ for some positive integer $N$, so you can reduce your polynomial mod $p$ for all primes $p$ not dividing $N$. Unfortunately, and obviously, they polynomials you get in $\mathbb F_p[x]$ this way will still be reducible.

So there is no way to make it irreducible by reduction/extension. However, depending on what you want to do after, there is something that you can do:

(3) Put your polynomials $P(x)$ in a family of polynomials $P_t(x)$ depending on some parameter $t \in \mathbb Q$, such that $P_0(X)=P(x)$. It easy to choose the family such that $P_t(x)$ becomes irreducible on $\mathbb Q(x,t)$, and this will imply that $P_t(x)$ will be irreducible over $\mathbb Q(x)$ for all values of $t$ except a (edited:) small set (including your starting point $t=0$) by Hilbert's Irreducibility Theorem (see e.g. wikipedia). Hence, you may be able to prove things for the irreducible polynomials in your family, and deduce similar things by "continuity" for your initial $P_0(X)$.

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Why $P_t$ is irreducible for all but finitely many $t$'s? It is not true, e.g., if $P=X^2$ and $P_t=X^2-t$. Hilbert irreducibility theorem ensures that the set of $t$'s for which irreducibility fails is small, but not necessarily cofinite. Do I miss something? –  Lior Bary-Soroker Mar 18 '13 at 7:49
    
You're right, I correct. –  Joël Mar 18 '13 at 12:53

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