# Polynomials: gaining irreducibility by adding a constant

EDITED:

Let $f\in\mathbb{Q}[X]$. I'm interested in the factoring properties of the family $F:=\{f+c\}_{c\in\mathbb{Q}}$ of rational polynomials that differ from $f$ only by a constant. Specifically:

1) I think there must be at least one irreducible polynomial in $F$. How can we prove it?

2) If $f$ is reducible, can we effectively determine an specific $c$ such that $f+c$ is irreducible?

3) How are the asymptotic growth and the density of those $c$ such that $f+c$ is irreducible? Are there at least an infinite number of them? How does this depend on the coefficients of the polynomial?

E.g, if $f(X):=X^2$ then any $c>0$ makes $f(X)+c$ irreducible, but it looks like $X^3+3X$ can be made reducible by infinite (ever sparser) values of $c$.

4) If $f\in\mathbb{Z}[X]$, can something be said about the residue distribution of the $c\in\mathbb{Z}$ modulo any prime?

5) How do the results generalize to other fields? What happens with degree 2 polynomials in real closed fields? What happens exactly in fields of characteristic $p$?

I have received important information from Michael Filaseta, with which we can answer:

1,2) Pick a prime $p$ big enough, put $c=p-f(0)$ and consider the polynomial $F(X)=f(X)+c$, which has $F(0)=p$. Specifically, if $f(X)=a_nX^n+\ldots+a_0$, by picking $p>|a_n|+\ldots+|a_1|$ we can guarantee that $F(X)$ has all its roots out of the unit circle, due to an iterative application of the reverse triangle inequality: $$|F(z)|\geq p-|a_1||z|-\ldots-|a_n||z^n|\geq p-(|a_1|+\ldots+|a_n|)>0,$$ where we have used that $|z|\leq1$.

Suppose $F$ factors as $F(X)=g(X)h(X)$; then $g(0)h(0)=p$ is a factorization of a prime, so for example $|g(0)|=1$. Therefore the absolute value of the product of the roots of $g$ is not greater than 1 (by Vieta, taking into account the leading coefficient of $g$). This implies that there is at least one root of $g$ inside the (closed) unit circle. But the roots of $g$ come from the roots of $F$, so we have reached a contradiction.

Now, as there are infinite primes bigger than $|a_n|+\ldots+|a_1|$, we know how to find an infinite number of $c$ such that $f+c$ is irreducible.

3) Hilbert's irreducibility theorem also answers 1), and gives the asymptotic behaviour: the polynomial $f+c$ is irreducible for almost every $c$. Concretely, if we denote $S(f,x):=\sum_{|c|\leq x, f+c\text{ irreducible}}1$ then we have $$S(f,x)=2x-o(x)$$ (the $2$ in $2x$ just comes from the fact that we consider $|c|\leq x$, so the density is computed with respect to $2x$).

In fact, it may be possible that using results close to Siegel's lemma one could prove $S(f,x)=2x-O(\sqrt{x}).$

5) For polynomials of degree 2 over the reals, as has already been mentioned, we can use the sign of the discriminant to guarantee the existence of infinite $c$, which asymptotically have $$\lim \frac{S(f,x)}{2x}=1/2,$$ as (if $a>0$, say) there is a $c_0$ such that if $c<c_0$ then $f+c$ factors, while if $c>c_0$ then $f+c$ is irreducible.

Now every real closed field is elementarily equivalent to the reals, and we can encode the condition on the discriminant in first order logic, so the same applies to real closed fields.

• I wish I could up vote this more than once, especially answer to 1,2). Do you know something about the image $P(\mathbb{Q})$ of a polynomial $P$ when $P$ is of odd degree? in particular does $P(\mathbb{Q})=\mathbb{Q}$ implies $deg(P)=1$? – Clément Guérin Feb 11 '18 at 2:19
• @ClémentGuérin Sure, $P({\mathbb Q})={\mathbb Q}$ does imply ${\sf deg}(P)=1$. Indeed, suppose ${\sf deg}(P)>1$. By this answer, $P-P(0)+p$ is irreducible over $\mathbb Q$ for large enough primes $p$. In particular, $P-P(0)+p$ has no rational roots, so $p-P(0)$ is not in the image of $P$, and $P$ is not surjective. – Ewan Delanoy Feb 11 '18 at 8:44
• @EwanDelanoy, of course. Thanks. – Clément Guérin Feb 11 '18 at 8:52
• @ClémentGuérin Well, the bounty you have given does surely account for a couple of votes! ;P – Jose Brox Feb 11 '18 at 10:36
• Your questions were nice (mine was not as interesting) and the answer surely was worth the points. That makes an interesting exercise! – Clément Guérin Feb 11 '18 at 13:22

The case of degree $2$ polynomials in $\mathbb{R}$ or $\mathbb{Q}$ can be handled. Let $f(x) = ax^2 + bx$. By the quadratic formula, if $b^2 - 4ac < 0$, then we have $f(x) + c$ irreducible.

• Alright so more generally, if $f(x)=ax^2+bx+c$ then $f(x)-d$ is an irreducible polynomial if and only if $b^2-4a(c-d)$ is not a square. – Clément Guérin Feb 10 '18 at 6:12
• @ClémentGuérin Correct. – AJY Feb 10 '18 at 15:50