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To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound?

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, is false without additional assumptions. Since we may take $c = M(f)$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

An example of a rather different kind (in particular, having unbounded Mahler measure) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further examples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.

To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound?

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, is false without additional assumptions. Since we may take $c = M(f)^2$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

A counterexample of a rather different kind (in particular, having unbounded Mahler measure under all integer translations) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further counterexamples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.

To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound?

It seems hard enough to construct non-trivial non-trivial integer polynomials having high degree $d$ and discriminant bounded by $d^d$ (the cyclotomic polynomials have this property, but we may consider them to be trivial). There could be a finite set $T \subset \mathbb{Z}[x]$ such that, up to a translation and a reflection, all integer polynomials having non-zero discriminant of absolute value $\leq d^d$ would factor into products cyclotomic ones and polynomials from $T$.

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, does not hold without additional assumptions. Since we may take $c = M(f)$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

An example of a rather different kind (in particular, having unbounded Mahler measure) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further examples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.

To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound?

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, is false without additional assumptions. Since we may take $c = M(f)$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

An example of a rather different kind (in particular, having unbounded Mahler measure) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further examples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.

To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound? It seems hard enough to construct non-trivial examples of polynomials having degree $d$ and discriminant bounded by $d^d$ (the cyclotomic polynomials have this property, but we may consider them to be trivial); it seems as if there might be a finite set $T \subset \mathbb{Z}[x]$ such that, up to translation and a reflection, all integer polynomials having non-zero discriminant with absolute value $\leq d^d$ would factor into products cyclotomic ones and polynomials from $T$.

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, cannot be true. Since we may take $c = M(f)$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by an explicit formula for the discriminant of the general trinomial, whose statement and a simple derivation is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

An example of a rather different kind (in particular, having unbounded Mahler measure) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further examples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could very well be true. It certainly holds for all trinomials, by the formula in Swan's article. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, an those have only cyclotomic factors.

To reiterate Serre's question (Open Problem 2.5 in Odlyzko's survey): Must there be only finitely many polynomials having root discriminant below a given bound? 

It seems hard enough to construct non-trivial non-trivial integer polynomials having high degree $d$ and discriminant bounded by $d^d$ (the cyclotomic polynomials have this property, but we may consider them to be trivial). There could be a finite set $T \subset \mathbb{Z}[x]$ such that, up to a translation and a reflection, all integer polynomials having non-zero discriminant of absolute value $\leq d^d$ would factor into products cyclotomic ones and polynomials from $T$.

With this answer I just want to note that the much stronger statement formulated in the last paragraph of user631's answer, which concerned the existence of a $c > 1$ such that all irreducible polynomials of discriminant $\leq (cd)^d$ are essentially cyclotomic, does not hold without additional assumptions. Since we may take $c = M(f)$ as Mahler proved, this is a natural statement to consider: its truth would have strengthened Lehmer's conjecture.

A counterexample is given by $x^d-x-1$, whose discriminant has absolute value $d^d + (-1)^{d}(d-1)^{d-1}$. We may take more generally any sequence of irreducible trinomials with $\pm 1$ coefficients and degree going to infinity. This follows by the explicit calculation of the discriminant of the general trinomial; the statement and simple derivation of the formula is given as Theorem 2 in Swan's 1962 paper Factorization of polynomials over finite fields in the Pacific Journal of Mathematics. (Another reference is Prasolov's book Polynomials, which reproduces the same calculation).

An example of a rather different kind (in particular, having unbounded Mahler measure) is provided by the minimum polynomial of the generator $\zeta_n+\zeta_n^{-1} = 2\cos(2\pi/n)$ of the integer ring of the maximal totally real subfield of $\mathbb{Q}(\zeta_n)$. Further examples would include the minimum polynomials of the generators over $\mathbb{Z}$ of the integer rings of other bounded index monogenic subfields, if such exist, of either $\mathbb{Q}(\zeta_n)$ or the splitting fields of the $\pm 1$ trinomials considered in the previous paragraph.

However, one may wish to restrict to reciprocal polynomials; for Lehmer's problem this would be sufficient. For these the statement seems very interesting, and could well be true; by the formula in Swan's article, it certainly holds for all trinomials. The above examples are of no use here: the only reciprocal $\pm 1$ trinomials are $x^{2n} \pm x^n +1$, and those have only cyclotomic factors.