Let $f(X,t\_1,\dots,t\_s)$ be an irreducible polynomial with coefficients in $\mathcal{O}\_K$, the ring of integers of a number field $K$. By work of S. D. Cohen (<http://plms.oxfordjournals.org/content/s3-43/2/227.full.pdf>) I know that the number of specializations $\vec t = (t\_1,\dots, t\_s)$ in $(\mathcal{O}\_K)^s$ of height at most $N$, such that the specialization does not remain irreducible (or even does not retain the original Galois group of $f$) is $O(N^{s-1/2}\log(N))$, with the implied constant depending only on $f, s,$ and $K$.


1. Is this the best result known, in terms of the order of growth in $N$?
2. Is it conjectured that the order can be improved, say to $O(N^{s-1})$? (Maybe not for $s=1?$)
3. Regarding the first two questions, is more known/conjectured in the special case that $f$ is a polynomial in $X$ where precisely $s$ of the coefficients are allowed to vary in $\mathcal{O}\_K$? (i.e. $f \in \mathcal{O}\_K[t\_1,\dots,t\_s][X]$, with each coefficient having degree 0 or 1 in $t\_1,\dots,t\_s$? In this special case, the best results I know of, also due to Cohen (<http://www.projecteuclid.org/euclid.ijm/1256048323>)
are of the same order as the more general results, except for $s=1$, where the log factor can be omitted.)
4. As above, but in the special case $K=\mathbb{Q}$?



The best possible density in Hilbert's Irreducibility Theorem
Have you stated this correctly? Assuming there are $O(N^s)$ possible specializations of height at most $N$, don't you want to say that the specialization remains irreducible *except for* $O(N^{s-1/2}\log N)$ values. 


As for question (2), I don't think you can do better than $s-\frac12$. Certainly not for $s=1$, where $X^2-t$ is clearly reducible for $O(\sqrt N)$ values of $t$ with $|t|<N$. Actually, that example works for any $s$, since $X^2-t\_1$ is irreducible in $\mathbb{Q}[t\_1,\ldots,t\_s]$. But if you really want to include all the variables, use $X^2-t\_1-t\_2-\cdots-t\_s$, which is reducible whenever $t\_1+\cdots+t\_s$ is a square.


The underlying reason for the $\frac12$ is because the exceptional values come from rational (or integral) points on non-trivial finite covers. If there happens to be a double cover with a "lot" of rational or integral points, particularly a copy of affine or projective space, then the fact it is a double cover means that it more-or-less squares the height, and then one gets a square root of the number of points in the target space. In other words, the size of the exceptional set is determined by the geometry of the covers that arise in the proof.


