Suppose we have a polynomial in integer coefficients
$$p = p_0 + p_1 x + p_2 x^2 + \ldots + p_n x^n, p_k \in \mathbb{Z}$$
Now define $M(p)$ as the maximum absolute value of the coefficients of $p$, i.e.
$$M(p) = \max \{|p_k| \: |\: 0 \leq k \leq n\}$$
Is it true that for any polynomial factor $q$ of $p$, $M(q) \leq M(p)$? I'm not sure whether this is obvious and I'm just missing something, or whether my train of thought is just completely off...
This question has been studied a bit because it is actually of practical relevance to computer algebra. The tightest known bounds are surprisingly large. A good review is Bounds on factors in $\mathbb{Z}[x]$, John Abbott, which according to his CV has been accepted for publication by the Journal of Symbolic Computation. His paper also contains a number of examples and an argument for why the bounds must necessarily be larger than you might hope.
When I wrote this answer there was another one with a couple of small examples, but it's been deleted, so I'll add some small examples in different categories from Abbott's paper.
Large height reducible factor of $x^d-1$: $$\begin{eqnarray}x^{12} - 1 & = & (x-1)(x^2+1)(x^2-x+1)\; (x+1)(x^2+x+1)(x^4-x^2+1)\\ & = & (x^5 - 2x^4 + 3x^3 - 3x^2 + 2x - 1)(x^7 + 2x^6 + x^5 - x^4 - x^3 + x^2 + 2x + 1)\end{eqnarray}$$
Large height irreducible factor of polynomial of height $1$: $$x^4 + x^3 - x^2 - 1 = (x-1) (x^3 + 2x^2 + x + 1)$$
Large height reducible factor of polynomial of height $1$: $$x^3 + x^2 - x - 1 = (x-1)(x+1)^2 = (x-1)(x^2 + 2x + 1)$$
No. There is a famous example of a factor of $x^{105}-1$
\begin{align} \Phi_{105}(x) = & \; x^{48} + x^{47} + x^{46} - x^{43} - x^{42} - 2 x^{41} - x^{40} - x^{39} + x^{36} + x^{35} + x^{34} \\ & {} + x^{33} + x^{32} + x^{31} - x^{28} - x^{26} - x^{24} - x^{22} - x^{20} + x^{17} + x^{16} + x^{15} \\ & {} + x^{14} + x^{13} + x^{12} - x^9 - x^8 - 2 x^7 - x^6 - x^5 + x^2 + x + 1 \end{align}
Also $x^4+1=(x^2+1)^2-2x^2$ has the factorisation $(x^2+\sqrt 2 x + 1)(x^2-\sqrt 2 x +1)$ which isn't integral, but does suggest that lower degree counterexamples may exist.
