# Smooth proper variety over a number field with prescribed bad reductions

Given a number field $$K$$, and a finite set $$S$$ containing finite places of $$K$$. When can we find a smooth proper geomerically connected variety $$X$$ over $$K$$ such that $$X$$ has good reduction outside $$S$$ and has bad reduction at every place in $$S$$?

It's better to assume $$X$$ is "simple", in the sense that it's not some product of low-dimensional varieties, or in the strong sense that the $$\ell$$-adic etale cohomology $$H^i(X_{K^{alg}},\mathbb Q_l)$$ are irreducible Galois representations.

Maybe some examples can be constructed from rational varieties, abelian varieties, taking products and doing some Galois twists tricks. But I am not sure these examples work. For example, there is no elliptic curve over $$\mathbb Q$$ with only bad reduction at $$5$$.

What if we require the chow motive of $$X$$ (under numerical equivalence) does not belong to the subcategory generated by motives of abelian varieties and Artin motives?

• Yes you can do with with plane conics, at least over $\mathbb{Q}$. This follows from the description of the Brauer group of $\mathbb{Q}$ given in class field theory. – Daniel Loughran Feb 27 '19 at 20:47
• @DanielLoughran Thank you! I see, so my main interests lie in other interesting examples like abelian varieties, or more general varieties (can't be obtained from abelian varieties and Artin motives)... – sawdada Feb 27 '19 at 21:51
• The Fermat curve $x^p+y^p=z^p$ – David Lampert Mar 2 '19 at 1:04
• What about motives of higher dimensional Fermat hypersurfaces e.g. $x_1^p+x_2^p+x_3^p+x_4^p=0$? Maybe some $\Sigma x_i^{p^{n_i}}=0$ could give irreducible Galois representations. – David Lampert Mar 2 '19 at 14:25
• What is the motivation for this question? The restrictions seem arbitrary. I think it is possible to achieve everything but the strong irreducible condition using universal families over modular curves. – Will Sawin Mar 2 '19 at 17:56

There are a few key methods I know to produce varieties with bad reduction on a specified set, most of which were mentioned already in the comments:

Twists: Given a variety $$X$$ with an action of a group $$G$$, we can produce twists of $$X$$ over $$\mathbb Q$$ from elements of $$H^1(\mathbb Q, G)$$. The variety will have good reduction at every place where $$X$$ has good reduction and the Galois cohomology class is unramified, and it is often possible to prove a converse and/or to fine-tune the set of places of bad reduction of the twist.

Daniel Loughran's example of plane conics fits into this mold, taking $$X =\mathbb P^1$$ and $$G = PGL_2$$. So does Ariyan Javanpeykar's example of the elliptic curves $$y^2= x^3+1$$, taking $$X$$ to be the elliptic curve $$y^2=x^3+1$$ and $$G$$ to be the order $$6$$ group of automorphisms. The same is more obviously true for any quadratic twist family of elliptic curves.

But the fact that an automorphism of order two is sufficient to get bad reduction t an arbitrary set of new places in the case of elliptic curves suggests this is a quite general phenomenon - many varieties have automorphisms of order two.

Outside of some explicit cases like plane conics, the main way to prove that varieties constructed this way have bad reduction is the bad reduction of their cohomology. To make this work, say in the case of an order two automorphism, it is only necessary that this automorphism act nontrivially on the cohomology. You can find many examples here from sufficiently general hypersurfaces, or maybe hypersurfaces of bidegree $$(2,d)$$ in $$\mathbb P^1 \times \mathbb P^n$$, but you will always have some starting set of bad primes to work with that may be hard to control. However, the sufficient generality will easily let you handle the strongest irreducibility condition and the "far from curves and abelian varieties" condition. This was discussed by Ariyan Javanpeykar in his second comment.

Explicit equations: For a hypersurface or other variety defined by explicit equations, one can explicitly calculate the primes where the hypersurface is singular. If the number of terms in the equation is small, it is often possible to reverse-engineer to get a hypersurface singular at a proscribed set, as for instance is the case with the Fermat curve $$x^n+y^n=z^n$$, singular at primes dividing $$n$$.

However, it is not obvious that a hypersurface has bad reduction at the primes it is singular - for instance, the Fermat curve is actually nonsingular when $$n=2$$, but the related $$x^2+y^2+z^2=0$$ is not. Again, the most convenient way to rule this out is to use the explicit cohomology.

However, hypersurfaces that are simple enoguh that their singular primes and cohomology can be easily calculated will not usually have motive independent from curves and abelian varieties, and will almost never satisfy your strong irreducibility condition. To avoid the curves and abelian varieties thing, a good idea might be to use the Dwork family or similar hypersurfaces which have hypergeometric Galois representations appearing in their cohomology, and use irreducibility properties of these hypergeometric representations, but this will certainly not make the whole Galois representation irreducible. Still, these will certainly be irreducible in the sense of not being a product of lower-dimensional varieties.

Shimura varieties and automorphic forms: Shimura varieties (for instance the powers of the universal families of elliptic curves with level $$N$$ structure) usually have good reduction away from primes dividing their level, and have cohomology which can be calculated in terms of automorphic forms, which will usually have bad reduction at primes dividing their level. This makes it easy to verify bad reduction, again using the homological criterion.

A subtlety is ensuring that the varieties have a nonsingular compactification. The $$k$$-fold fiber product of the universal family of elliptic curves over the moduli space of elliptic curves with full level $$N$$ structure has a natural compactification as a ramified cover of the Deligne-Mumford moduli space $$\overline{\mathcal M}_{1,k+1}$$, but this is singular. However, the singularities are toric and should be easy to remove away from the primes dividing $$N$$. Then as soon as there is a cusp form of weight at most $$k+2$$ and level dividing $$N$$, whose level is a multiple of $$p$$, it will show up in the cohomology (by work of Deligne) and hence this space will have bad reduction at $$p$$. This is easy to ensure by taking $$k$$ or $$N$$ sufficiently large and using formulas for the number of cusp forms.

Because the Hodge structures of Galois representations of modular forms are known, it should be easy to check your motive condition as long as one of the relevant modular forms has weight $$>2$$ and is non-CM. But because the Galois representation splits into many two-dimensional pieces, your strong irreducibility criterion is essentially never satisfied here.

• I think twisting elliptic curves is a bit more subtle, even over $\mathbb{Q}$. For example, starting with $y^2=x^3+1$ and twisting is fine provided your set $S$ contains 2 and 3. In general, if you let me add one prime to $S$, I'll add a prime $p$ such that there is an elliptic curve of conductor $p$, and then twisting will get the other primes in $S$ and no other primes. But if, say $S=\{p\}$, I don't think there is a good characterization of which $S$ will work. There may also be an issue over a field $K$ whose class number is not 1, since twisting by one prime may force other bad reduction. – Joe Silverman Mar 3 '19 at 23:22
• @JoeSilverman I forgot that this question was over number fields when writing my answer. Thanks for pointing it out! But with regards to your first issue, what I suggested was possible is that starting with a fixed elliptic curve, you can add an essentially arbitrary set of bad reduction primes without taking any away (which is true over $\mathbb Q$ using quadratic twisting), not that you can obtain an arbitrary set of bad reductions. So that wasn't intended to be a complete answer to this question. – Will Sawin Mar 4 '19 at 0:28
• If $E$ is the universal elliptic curve over the modular curve $Y(N)$ with $N \geq 3$, then Deligne has constructed a canonical smooth compactification $\overline{E^k}$ of $E^k$ (with a single blowup!). The motive $H^{k+1}(\overline{E^k})$ consists of the cusp forms of weight $k+2$ and level $N$, so one can take the submotive associated to a newform of weight $k+2$ and level $N$, which exists if $k$ is large enough. – François Brunault Mar 5 '19 at 8:56
• @FrançoisBrunault Deligne finds weight $k+2$ inside $H^1 ( X(N), \operatorname{Sym}^k V)$ where $V$ is the sheaf of Tate modules. The Leray spectral sequence writes $H^{k+1}$ of this moduli space in terms of $H^1(X(N), V^{\otimes k} + \sum_{i=1}^{k/2} \frac{k!}{ i! i! (k-2i)!} V^{\otimes (k-2i)} )$ which can be further broken up into a sum of $\operatorname{Sym}^{ k-2j} V$. Sorry that was brief, I can add more details... – Will Sawin Mar 5 '19 at 18:55
• @FrançoisBrunault I guess motivically, the idea is that there is an action of $GL_{k}(\mathbb Z)$ on the interesting part of the cohomology by its action on the abelian variety $E^k$. If we look at the eigenspace of $H^{2a+b+1}$ with respect to any matrix coefficient of the representation with highest weight $(2^{a}, 1^b, 0^{k-a-b})$ we get $\operatorname{Sym}^b$ and thus modular forms of weight $b+2$. – Will Sawin Mar 5 '19 at 19:10