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Let $f\in\mathbb{Z}[X]$ be an irreducible polynomial. Is there an integer $a\neq 0$ such that $f(X)+a$ is also irreducible in $\mathbb{Z}[X]$?

Can this be also extended to $\mathbb{Q}[X]$?

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Yes, and you don't need $f$ irreducible. The following irreducibility criterion suffices and shows that infinitely many $a$ work.

Lemma: Let $g(x) = a_n x^n + ... + a_0 \in \mathbb{Z}[x]$ be such that $a_0$ is prime and

$$|a_0| > |a_1| + ... + |a_n|.$$

Then $g(x)$ is irreducible.

Proof. The condition on the coefficients ensures that all complex roots of $g$ have absolute value greater than $1$. But if $g$ is reducible then at least one of its irreducible factors has constant term $\pm 1$, hence at least one of its roots has absolute value less than or equal to $1$; contradiction. $\Box$

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Let $t$ be another variable. Then $f(X)+t$ is irreducible over $\mathbb Q(t)$. By Hilbert's Irreducibility Theorem there are infinitely many $a\in\mathbb Z$ such that $f(X)+a$ is irreducible over $\mathbb Q$. Often that is only formulated for $a\in\mathbb Q$, but the proofs which I know actually yield integers $a$.

As to irreducibility over $\mathbb Z$ versus over $\mathbb Q$, it seems to me that you should look up the Gauss Lemma.

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  • $\begingroup$ Thank you. Indeed, I forgot about Gauss Lemma, so I feel silly asking about $\mathbb{Q}[X]$ now. However, Hilbert's Theorem is too advanced for me, unless there is a simple proof for this case. I was thinking to look for big primes, but I haven't found anything useful yet. $\endgroup$ – l.t Dec 2 '13 at 15:34
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    $\begingroup$ O.K., Qiaochu Yuan's elegant and elementary answer does what you want. $\endgroup$ – Peter Mueller Dec 2 '13 at 17:04

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