*Premise* 


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 $K$.


*Questions*


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 $lct\_{\mathbb{C}}(f\_{\mathbb{C}})$?**



Comparisons of log canonical thresholds
*CORRECTION ON THE PREVIOUSLY GIVEN ANSWER*


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.


