Let $K$ be a field of characteristic zero and $f\in K[X_1,\dots,X_m]$. By Hironaka's theorem, there exists a log resolution (over $K$) of the ideal $(f)$. Let $\{(N_i,\nu_i)\}_i$ be the numerical data of a fixed log resolution. The quantity $$ lct_K(f):=\min_{i}\frac{\nu_i}{N_i} $$ does not depend on the choice of the log resolution and it is called the log canonical threshold of $f$ over $K$.


Let $f\in \mathbb{Q}[X_1,\dots,X_m]$. By definition, we have $$ lct_{\mathbb{Q}}(f)\ge lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) \ge lct_{\mathbb{C}}(f_{\mathbb{C}}). $$ On the other hand, from Denef's formula for the motivic Igusa zeta function it follows that for all but finitely many $p$ one has $$ lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) \ge lct_{\mathbb{Q}}(f_{\mathbb{Q}}). $$ This shows that $$ lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) = lct_{\mathbb{Q}}(f_{\mathbb{Q}}) \quad \forall\forall p. $$

1. Is this equality actually true for all $p$?

In all the counterexamples I have found in the literature for the validity of Denef's formula for the "bad" primes (in the sense of Denef) one still has $lct_{\mathbb{Q_p}}(f_{\mathbb{Q}_p}) = lct_{\mathbb{Q}}(f_{\mathbb{Q}})$ also for bad primes $p$. Were this not always the case, has anybody a counterexample at hand?

2. What can we say about the comparison with $lct_{\mathbb{C}}(f_{\mathbb{C}})$?

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    $\begingroup$ Is it easy to see why $\operatorname{lct}_{\mathbb{Q}_p}(f) \geq \operatorname{lct}_{\mathbb{C}}(f)$? $\endgroup$ – Zach Teitler Aug 1 '18 at 6:02
  • $\begingroup$ My reasoning was that, since we can embed $\mathbb{Q}_p$ into $\mathbb{C}$, we can get log-resolutions over $\mathbb{C}$ by possibly adding exceptional components. However, now that I look back at it, I think I overlooked the fact that the embedding is NOT continuous, which is probably necessary. However, the inequality is still true for almost all $p$ as a consequence of Denef's formula (because the above reasoning works for the embedding of $\mathbb{Q}$ in $\mathbb{C}$). Should I edit the question? $\endgroup$ – Maurizio Moreschi Aug 1 '18 at 6:56
  • $\begingroup$ An explanation for the downvote? $\endgroup$ – Maurizio Moreschi Aug 1 '18 at 18:36


As remarked in the comments, Hironaka's construction behaves well under extension of the base field, that is a log-resolution for $(X_L,D_L)$ can be obtained via base change from a log-resolution of $(X_K,D_K)$ for any field extension $K\hookrightarrow L$. Also, an irreducible smooth divisor after base change is still smooth (although not necessarily irreducible). Since the irreducible components of a SNC divisor are smooth by definition, it follows that the numerical data of the resolution are left the same after base change (apart for possible repetitions). Therefore $lct(X_K,D_K)=lct(X_L,D_L)$.

Notation: $K$ is a field of characteristic zero, $(X_K,D_K)$ is a pair of a smooth variety over $K$ together with an effective non-zero divisor on it.

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    $\begingroup$ Hmmm, I wonder if it is perhaps simpler than this? Start with a log resolution over the smaller field $K$. It remains a log resolution over $L$, does it not? The only change is that possibly some of the exceptional divisors which were irreducible over $K$ become reducible over $L$, but in that case all the multiplicities stay the same. You just have several copies (a Galois orbit) of "the same" exceptional divisor. One must be careful, but if the original singularity was defined over $K$ then I think the $K$-resolution is still an $L$-resolution...? $\endgroup$ – Zach Teitler Aug 31 '18 at 17:32
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    $\begingroup$ I agree with Zach, lct over ${\mathbb Q}$ and over ${\mathbb C}$ are the same for the above reason. Explicitly, if I have a log resolution of $({\mathbb Q}, f)$ where $X$ and $f$ are defined over ${\mathbb Q}$, then after base change, it is still a log resolution over ${\mathbb C}$. $\endgroup$ – Karl Schwede Oct 30 '18 at 14:22
  • $\begingroup$ Yes, I agree. Hironaka's construction behaves well under extensions of the base fields. So after base change I still have a log-resolution. Also, if a divisor E is smooth and irreducible, then after base change might not be irreducible, but it is still smooth, so it cannot have multiple components. It follows that the numerical data are the same (apart possibly being repeated more times). $\endgroup$ – Maurizio Moreschi Oct 30 '18 at 15:15
  • $\begingroup$ $E$ might have multiple components, they are just pairwise disjoint. $\endgroup$ – Zach Teitler Oct 30 '18 at 16:12
  • $\begingroup$ I meant components of multiplicity bigger than 1. $\endgroup$ – Maurizio Moreschi Oct 30 '18 at 16:19

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