Non-commutative group scheme of prime square order

December 25, 2022 mayorliatmath Leave a comment

Every time I write christmas, I’ll have to double check with Google to see if I mispelled.

Lately I’ve had an interesting discussion with Oort and Petrov about constructing an example of finite flat non-commutative group scheme $G/S$ of order $p^2$ with $p \not= 0$ on $S$ .

Before going into details, let’s first recall that in our first class on abstract algebra, we probably should have learned that a finite group of prime or prime square order is automatically commutative. Next let’s also recall that in the classical paper “Group schemes of prime order” by Oort–Tate, they showed that any finite flat order $p$ group scheme is automatically commutative. In the same paper, they also gave an example of order $p^2$ which is non-commutative: it’s the semi-direct product of $\alpha_p$ and $\mu_p$ with the latter acting on the former in the ”usual” way. Concretely it can be realized as the Frobenius kernel of the Borel in $PSL_2$ .

I deliberately didn’t say what base I’m working over in the above example. One can obviously realize the above example over $\mathbb{F}_p$ , it’s unclear if that example can be deformed over a base on which $p \not= 0$ . The task of this post is to show you there exists such $G/S$ for which $S$ is irreducible and dominates $\mathrm{Spec}(\mathbb{Z}_p)$ .

The idea of Petrov’s is as follows: we simply try to find a nontrivial action of $\mathbb{Z}/p$ on an order $p$ group scheme, then form the semi-direct product. Lastly we need to recall the following key fact from the aforementioned Oort–Tate paper: in it the authors basically constructed a versal family of order $p$ group schemes on $\mathbb{Z}_p$ . It’s a group scheme denoted $G_{a, b}$ over the spectrum of $R = \mathbb{Z}_p[a, b]/(ab - p)$ with underlying algebra $R[x]/(x^p - ax)$ and the co-multiplication is given by $m(x) = x \otimes 1 + 1 \otimes x + b f(x \otimes 1, 1 \otimes x)$ where $f(x,y)$ is a homogeneous degree $p$ polynomial with coefficients in $\mathbb{Z}_p$ .

Now we are ready to make the example: let $R' = \mathbb{Z}_p[a, b, c]/(ab - p, ac, bc, c^p)$ which receives a map from $R$ , then the base changed group scheme $G_{R'}$ has an action of $\mathbb{Z}/p$ via $x \mapsto (1+c) x$ . We leave it to the reader to see this action has order $p$ and preserves both multiplication and co-multiplication on the Hopf algebra associated with $G$ . Final remark: the ideal cutting out the ”commutative locus” is given by $(c)$ . This ideal has length $(p-1)$ . This ideal also happens to cut out the closure of the characteristic $0$ fiber. Of course, by the elementary fact about finite groups I summoned in the beginning, we know the “commutative locus” has to include this closure, and in this example they agree!

Hope you enjoy this read, merry christmas!

Some families of finite flat group schemes

October 20, 2021 mayorliatmath Leave a comment

When people say finite flat group scheme, what exactly do they mean? Sometimes, they just mean finite flat group scheme, presumably over some prefixed base. But often people meant finite flat *commutative* group scheme. It’s confusing. In this blog, I shall always mean the former, and add the adjective “commutative” when I mean it. Anyway, a while back, I was wondering about: can a finite flat commutative group scheme be deformed to a finite flat non-commutative group scheme? Weird random question, I know. But let’s still discuss it, because why not.

Since the moduli of finite flat group scheme of a fixed order is of finite type over $\mathbb{Z}$ , we are basically asking: can there be a finite flat group scheme $G$ over a DVR, such that the special fibre is commutative whilst the generic fibre is not? There are 3 types of DVR: equal characteristic 0 or p, and mixed characteristic. In the equal characteristic 0 case, we can’t have such an example, as the finite flat group scheme has to be finite étale, and hence the specialization map induces an isomorphism, so the family of groups is essentially constant (appropriately interpreted).

Now let’s discuss the equal characteristic p situation. By the way, p is always a fixed prime for me. Already there are some weird examples in characteristic p setting: one can have a semi-direct product of $\mu_p$ by $\alpha_p$ where the action of former on the latter is what you think it is. Said differently, take the Borel in $PGL_2$ and look at the Frobenius kernel, which exactly gives the said extension. In characteristic 0, there cannot exist such a noncommutative finite group scheme of order $p^2$ . Finally, I shall leave it to the reader to check that the above group scheme can be specialized to the product of two copies of $\alpha_p$ . The idea is that one can deform $\mu_p$ to $\alpha_p$ (comultiplication reads: $x \mapsto t \cdot x \otimes x + x \otimes 1 + 1 \otimes x$ ), and we can extend the action of $\mu_p$ on $\alpha_p$ to this family (action reads: $s \mapsto (tx+1) \otimes s)$ ). This gives an order $p^2$ example in equal characteristic p.

Lastly, what if the base has mixed characteristic p? The generic point has characteristic 0. By what I said above, order $p^2$ group scheme over the generic point is necessarily commutative. The best thing to hope for is to have an example of order $p^3$ finite flat group scheme over some mixed characteristic $\mathcal{O}_K$ , such that the special fibre is commutative and the generic fibre is not. To that end, let’s start with a central extension of $(\mathbb{Z}/p)^2$ by $\mathbb{Z}/p$ . When $p \not= 2$ , there is exactly one such non-commutative group: commutator map gives rise to an alternating pairing on $(\mathbb{Z}/p)^2$ valued in $\mathbb{Z}/p$ , and we want it to be non-degenerate. When $p=2$ , there are two such groups, one is $D_4$ , the other one is $\mathcal{O}_H^{\times}$ where $H$ is the standard Hamiltonian over $\mathbb{Z}$ . In any case, such central non-commutative extension exists. Let $\mathcal{O}_K = \mathbb{Z}_p[\zeta_p]$ , over which we have a tautological character $\mathbb{Z}/p \to \mu_p$ , pushing out the above central extension along this character, gives us a central extension of $(\mathbb{Z}/p)^2$ by $\mu_p$ . The point is that now the special fibre of this extension splits, because the map $\mathbb{Z}/p \to \mu_p$ is trivial in the special fibre.

I learned the above example from reading Raynaud’s “p-torsion du Schema de Picard”. BTW, in the same paper, he explains why the last example has to be over a base with ramification index at least $(p-1)$ . For such a $G$ , if you look at its classifying stack, then its $Pic^{\tau}$ will have different size on the special and generic fibre. But Raynaud showed that the $Pic^{\tau}$ of a smooth proper scheme (e.g. a “good enough” approximation of the aforesaid classifying stack) is flat if the ramification index is less than $(p-1)$ . Isn’t he amazing??!!

Frobenius linear algebra

November 27, 2020 mayorliatmath Leave a comment

Let me report something I’ve been thinking about for the past 2 weeks. So in this post I’ll tell you some weird fact, and then ask a few questions. Choose your favorate algebraically closed characteristic $p$ field, call it $k$ .

Now let $V \text{ and } W$ be two finite dimensional $k$ -vector spaces, let $F \colon V \to W$ be a Frobenius semilinear map, and $G \colon V \to W$ be a linear map. Consider the difference $F-G \colon V \to W$ , notice that this is only linear over $\mathbb{F}_p$ . I used to think this map is somewhat random, and maybe cannot really say anything about it. But recently I found out the following:

Proposition: The cokernel of $F-G$ is finite if and only if it is $0$ .

Let me omit the proof here, I think a (quite) advanced senior undergrad perhaps should be able to do this, once they are equipped with relevant knowledges. Notice that this statement wouldn’t hold if I replaced cokernel with kernel. So this suggests that an $\mathbb{F}_p$ -linear map such as $F-G$ is not random after all, its kernel and cokernel cannot be treated at equal foot.

Next, I want to tell you another related fact. More notations: let $R$ be the ring of integers in the completed algebraic closure of $k(\!(t)\!)$ , like your favorite $\mathcal{O}_C^{\flat}$ you know. (Here obviously I’m following notation in earlier papers of Fontaine’s.)

Again let $M \text{ and } W$ be two finitely generated $R$ -modules, let $F \colon M \to N$ be a Frobenius semilinear map, and $G \colon M \to N$ be a linear map. Consider the difference $F-G \colon M \to N$ , notice that this is also only linear over $\mathbb{F}_p$ . Similar to the above Proposition, we have:

Proposition: The cokernel of $F-G$ is finite if and only if it is $0$ .

Any quick way to see this? The proof I came up with is slightly weird, and not really comeupbyundergrad-able (surely I’ve gotta make up new words in every post, right?). OK, I just think these are some random weird facts, and somehow it comes up in a joint work that I’m about to finish. It would be quite interesting if one can prove some similar random facts of the kernel, of course that can’t be the exact same statement (which is trivially false). Here are questions that I’d love to know answers.

Suppose now we have two finitely generated *torsion* $k[\![t]\!]$ -modules $M \text{ and } N$ , and let $F$ and $G$ be similar (semilinear and linear) as the setting before. Now after base changing along the natural map $k[\![t]\!] \hookrightarrow R$ , we can get naturally induced maps $\widetilde{F} \colon M_R \to N_R$ and $\widetilde{G} \colon M_R \to N_R$ . Here’s my question:

If now I tell you that $\widetilde{F} - \widetilde{G} \colon M_R \to N_R$ is a bijection, what can we say about the quadruple $(M, N, F, G)$ ?

Here is a thought. If there is only semilinear part, then obviously we will have $p \cdot length(M) = length(N)$ (notice that by torsion assumption, these two have finite length); if there is only linear part, then also obviously we will have $length(M) = length(N)$ . So now I’m wondering if we will always have an inequality: $length(M) \leq length(N) \leq p \cdot length(M)$ ? And presumably we can say a lot more when one of the equality holds? Is there a situation which is NOT successive extensions of the two cases considered above? Is there any hope to classify these data? An interesting case (which more or less already comes up in the aforementioned joint work) is when $N$ is a finite direct sum of $k[t]/t^p$ , answering these questions in this particular case seems really doable (maybe I should make it an REU project?) and will lead to some statement concerning $u$ -power torsions inside Breuil–Kisin-prismatic cohomology and Nygaard filtration within some range (talking about throwing big words huh).

Oh almost forgot to say, happy thanksgiving!

Entertaining with PD envelope

April 1, 2020 mayorliatmath Leave a comment

“An expert is a man (or woman) who has made all the mistakes which can be made in a very narrow field.”–Niels Bohr

So let’s make some mistakes ~~hence the title~~. No, I don’t mean entertaining with PD envelope is a mistake. In this blog post, I want to report 2 mistakes I’ve made when I was trying to understand the following:

Question: what is the PD envelope of $\mathbb{Z}_p$ along the ideal $(p)$ ? And what does the filtration of the PD ideals $(p)^{\{[r]\}}$ look like, for instance, what is the associated graded ring?

Let us denote this PD envelope by $D$ , with the induced PD ideal $I$ . A result of Bhargav Bhatt (applied to our situation here) says that the associated graded ring is $\Gamma_{\mathbb{F}_p} ((p)/(p)^2) = \mathbb{F}_p\{x\}$ , the free divided algebra over $\mathbb{F}_p$ in one variable. So in particular we see that $I^{\{[r]\}}/I^{\{[r+1]\}}$ is $D/I$ as modules over $D$ . Let’s try to see this directly, shall we? Here comes my first mistake.

Mistake 1: everyone knows that the pair $(\mathbb{Z}_p, (p))$ has a PD structure. Hence shouldn’t the PD envelope just be $\mathbb{Z}_p$ itself with the PD ideals given by $I^{\{[r]\}} = (p^n/n!; n \geq r)$ ? But then the graded pieces don’t match with the prediction above. In the extremal case with $p=2$ , the PD filtration stays constant! What’s going on?

Okay, the PD structure on the pair $(\mathbb{Z}_p, (p))$ only tells us that there is a splitting $D \twoheadrightarrow \mathbb{Z}_p$ . But the kernel is kind of mysterious. Formally $D$ is obtained by adjoining symbols $\gamma_i(p)$ for $i \geq 1$ and quotient out the ideal generated by the relations dictated by a PD structure. So for instance, we only know that $\gamma_p(p) - p^{p-1}/(p-1)!$ is $p$ -torsion, but not necessarily zero.

Alright, now let’s follow the above logic. Since the ideal $(p)$ is a free module over $\mathbb{Z}_p$ , we know that $D = \mathbb{Z}_p\{x\}/(x - p)$ . But this seems so nasty, no? Bhargav saved me by reminding me that the above ring can be alternatively written as $\mathbb{Z}_p\{y\}/(y)$ via change of variable $y = x - p$ . This is because adjoining PD variable $x$ and $x - p$ makes no difference, thanks to the PD structure on $(p) \subset \mathbb{Z}_p$ (from now on I will write $\gamma_i(p)$ to mean the actual $p$ -adic integers).

With this simple presentation $\mathbb{Z}_p\{y\}/(y)$ , everything seems so easy and promising! And here comes my second mistake.

Mistake 2: surely the PD filtrations are generated by combinations of divided powers of $p$ and $y$ , right? Oh but wait, still in characteristic $2$ , it seems that (after some computations) with this filtration one has $gr^3 = 0$ . Where’s the bug??

Bhargav saved me again by pointing out that in the presentation $D = \mathbb{Z}_p\{x\}/(x - p)$ the PD filtrations should be generated by the image of divided powers of $x$ . Hence after changing variable in the presentation $\mathbb{Z}_p\{y\}/(y)$ the PD filtrations should be generated by the image of divided powers of $y + p$ .

There is still an advantage of having the presentation $D = \mathbb{Z}_p\{x\}/(x - p)$ , as from this it is clear that the $i$ -th graded piece of the PD filtration is generated by single element $\gamma_i(x) = \gamma_i(y + p)$ and it’s killed by all the divided powers of $x$ and $p$ (since $x = p$ ). So what we are left is to show that the generator $\gamma_i(y + p)$ itself is not zero, namely we need to show it cannot be expressed by combinations of $\gamma_{>i}(y + p)$ .

Now we’re ready to go (unless I make another mistake?!). By the presentation $\mathbb{Z}_p\{y\}/(y)$ , we get that as a $\mathbb{Z}_p$ module, $D$ is given by $\mathbb{Z}_p \oplus \bigoplus_{n \geq 1} \mathbb{Z}_p/(n) \gamma_n(y)$ . In this big direct sum, all the terms with $p$ not dividing $n$ is automatically gone. So if we denote $\gamma_{ip}(y) = y_i$ , we see that $D = \mathbb{Z}_p \oplus \bigoplus_{i \geq 1} \mathbb{Z}_p/(ip) \cdot y_i$ . And what is the algebra structure? Well, one can directly verify that $y_i \cdot y_j = ((i,j)) u_{i,j} y_{i+j}$ where $((i,j))$ is $(i+j)$ choose $i$ and $u_{i,j}$ are some units which one can make explicit if one wants to.

If we expand $\gamma_i(y + p)$ we get $\sum_{0 \leq i \leq n/p} y_i \cdot \gamma_{n - ip}(p)$ . Now let’s see the first few of the elements $\gamma_i(y + p)$ (up to unit):

$p , p^2, \ldots, p^{p-1}, y_1 + \text{unit} \cdot p^{p-1}$ ;

$p^p , p^{p+1}, \ldots, p^{2p-2}, y_2 + \text{unit} \cdot p^{2p-2}$ ;

$\ldots, p^{(p-1)^2}, y_{p-1} + \text{unit} \cdot p^{(p-1)^2}$ ;

$p^{(p-1)^2+1}, p^{(p-1)^2 + 2}, \ldots, p^{(p-1)p}$ ,

$y_p + \text{unit} \cdot p^{(p-1)p - 1}, py_p + \text{unit} \cdot p^{(p-1)p}, p^{(p-1)p + 1}$ and so on.

So the general pattern is that when the $p$ -adic valuation of the constant part being non-increasing, one sees these $y_i$ -terms showing up in the expression. To prove in general the $i$ -th graded piece of the PD filtration is nonzero, one perhaps needs to introduce some kind of degree function that keeps track of both $p$ -adic valuation and the degree of $y_i$ ‘s.

If you are Bhargav (or any intelligent reader for that matter) who reads up to here, do you see a way to make a direct argument in this specific case? Anyways, sorry I have to leave with such a loose end, but I really need to eat this congee before it becomes disgusting.

Okay, Bhargav suggests an argument which doesn’t need this presentation involving $y$ at all. Here is how it goes. Let $\mathbb{Z}_p\{x\}$ be PD algebra in one variable over $\mathbb{Z}_p$ with the PD filtration given by divided powers of $x$ . It is clear that the graded pieces are given by $Gr^i = \mathbb{Z}_p[x]/(x) \cdot \gamma_i(x)$ . Now because $\mathbb{Z}_p\{x\}$ and these $Gr^i$ ‘s have no $(x-p)$ -torsion (so Tor independent over $\mathbb{Z}_p[x]$ with $\mathbb{Z}_p[x]/(x-p) = \mathbb{Z}_p$ ), one sees that the PD filtrations on $D = \mathbb{Z}_p\{x\}/(x-p) = \mathbb{Z}_p\{x\} \otimes_{\mathbb{Z}_p[x]} \mathbb{Z}_p[x]/(x-p)$ has graded pieces given by directly tensoring $Gr^i(D) = Gr^i \otimes_{\mathbb{Z}_p[x]} \mathbb{Z}_p[x]/(x-p) = \mathbb{F}_p$ . More concretely the Tor-independence thing says that any element $f \in \mathbb{Z}_p\{x\}$ has the property that $f \cdot (x-p) \in Fil^i$ if and only if $f \in Fil^i$ .

It’s sad that in the end, one doesn’t need this $y$ -presentation to understand the graded pieces, but since it took me quite long time to type them, let me not deleting them.

Yet another example

June 9, 2019 mayorliatmath Leave a comment

Some while ago, I learned that one can have a pair of unramified liftings of a finite group representation in positive characteristic with the property that their geometric generic fibres are different.

I started recording this example, but got constantly interrupted by all these graduation -and-travel-and-moving nonsenseseseses. Finally, I finished the writing, which is only 2 pages… And here you are.

My original motivation of finding this kind of example was to hope that in the end I can produce a geometric example of similar flavor, i.e. different unramified liftings. Anyways, I couldn’t quite make it to work. Hopefully I will come back on this.

an example I learned yestoday

March 7, 2019 mayorliatmath Leave a comment

Okay, yestoday is officially a word that stands for yesterday and today.

Anyways, I learned this cool example at this year’s AWS. It is the work of a project group associated with Matthew Morrow.

First off, some background. We all know that a ring homomorphism $A \to B$ is called formally étale if the infinitesimal lifting criterion is satisfied (see https://stacks.math.columbia.edu/tag/00UQ). Note that in the criterion, you are testing liftability against *first order* thickenings. You may wonder about the following: what if we replace the condition of $I$ being square zero by asking $I$ being locally nilpotent, that is we only require every element of $I$ being nilpotent without having a bound of the nilpotency. For an obvious reason, let’s call the corresponding morphism *being ln-formally étale*.

Two people in Morrow’s project group produced the following example: for any algebra $A$ , there is an $A$ algebra $B$ with a splitting surjection $f \colon B \to A$ such that $f$ is ln-formally étale, but NOT flat. In fact, later on we will also see that the cotangent complex of $f$ vanishes as well.

Now let me get to the construction. Consider the ordered group $\Lambda := \mathbb{Z}[1/2]^{\mathbb{N}}$ , where we put the lexicographical order. Just to make sure that we are on the same page: $(1,0,0,\ldots) > (0,1,0,\ldots)$ . Then let us consider the group algebra $A[\Lambda]$ . There is a natural valuation on this algebra sending $[\lambda] \to \lambda$ . Take the “ring of integers” of this valuation, which is our $B$ . This is a valuation $A$ -algebra, with a splitting map back to $A$ (quotienting all the $[\lambda]$ ‘s). Okay, the claim is that this morphism is ln-formally étale and non-flat. To see the former, just note that $(1,0,0,\ldots) = 2^n \cdot (1/2^n,-1,0,\ldots) + 2^n \cdot (0,1,0,\ldots)$ and “shift this equality to the right” and stare at the diagram long enough. To see the later, we need to show the kernel of $f$ is not a pure ideal (see https://stacks.math.columbia.edu/tag/04PQ). And it is again an easy exercise to show that the ideal $([(1,0,0,\ldots)])$ is NOT equal to the kernel times $([(1,0,0,\ldots)])$ by using the valuation we start with.

We can also see that the cotangent complex $\mathbb{L}_{A/B} \simeq 0$ . To see this, use the triangle for $A \to B \to A$ , along with the following facts:

(0) $\mathbb{L}_{A/A} \simeq 0$ ;

(1) $B$ is a filtered colimit of smooth $A$ -algebras, hence $\mathbb{L}_{B/A}$ is qis to a flat $B$ -module sitting in degree 0 and;

(2) $B \to A$ is a surjection with the kernel squaring to itself, hence $\mathbb{L}_{A/B}$ is supported in degree at most $-2$ .

Cool example, huh?

Here is one thing that I’ve been thinking about (with other students in our team) during this winter school (which is another student’s project given by Morrow at this winter school). Say $R$ is a local Artinian $\mathbb{Q}$ -algebra $R$ with residue field $\mathbb{Q}$ . Take any nonzero nilpotent $f \in R$ . Is it true that $df \not= 0 \in \Omega^1_{R/\mathbb{Q}}$ ? More generally, is it true that the de Rham complex of $R$ over $\mathbb{Q}$ is quasi-isomorphic to $\mathbb{Q}[0]$ ?

new domain name

January 25, 2019 mayorliatmath Leave a comment

I’m so excited about this new domain name of my website. Please check it out!

a brain teaser

December 21, 2018 mayorliatmath 7 Comments

Consider the group $G = PSL_2(\mathbb{F}_5)$ , by which I mean quotient of $SL_2(\mathbb{F}_5)$ by its center. So this group has order $60$ . If an interested reader looks at the classification of groups of order $60$ :”apparently”, says this reader, “this group is isomorphic to $A_5$ !”

Now $A_5$ naturally acts on a set of 5 elements. On the other hand, $G$ naturally acts on a set of 6 elements, namely the $\mathbb{F}_5$ points of $\mathbb{P}^1$ .

The brain teaser is: how to find a natural action of $A_5$ on a set of 6 elements, and how to find a natural action of $G$ on a set of 5 elements? I couldn’t find these in a nice way. For instance, we can let $G$ act on (the projectivization of) degree 4 homogeneous polynomials (with coefficients in $\overline{\mathbb{F}_5}$ ). The orbit of $x^4 - 2\sqrt{2} x^2 y^2 + y^4$ consists of $5$ elements (okay, let me not prove this here). So this must be it. But this is not satisfying. So the brain teaser is really: how can you realize these actions in a nice way?

Good reduction or not, that is the question

December 6, 2018 mayorliatmath Leave a comment

Ok, admittedly this title is not the best one that I can come up with. I just read (small part of) this paper. The authors constructed a family of degenerating Calabi–Yau 3-folds (over a punctured disc), with trivial monodromy action and have a (potentially) semistable reduction which is not a smooth filling.

If I understand it correctly, the authors believe that this family has no (potentially) good reduction. Namely, however we pull this family back along maps from another punctured disc to the current one given by “ $z^n$ “, it won’t admit a smooth filling.

Now the (philosophical) question is, how can we tell if a smooth variety over the fraction field of a DVR does NOT have a smooth integral model over that DVR?

I know this (kind of fake) example: if we take a non-archimedean Hopf surface, then its étale cohomology groups are unramified (if you use l-adic coefficients) or Crystalline (if you use p-adic coefficients). But it doesn’t have a smooth reduction. Why? Because the weight is wrong. Ok, so this example is lame.

So I think there is a task I can give myself: to find a smooth projective variety over a non-archimedean field such that (1) its étale cohomology groups are “good reduction” meaning unramified/Crystalline of the correct weight and, (2) it does not admit any potential good reduction. I assume it would be easy to find a potential example, like the example discussed in the above link, but it perhaps is quite difficult to prove that such a thing is an actual example.

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Update on Dec. 8th: Bhargav reminded me that a curve of compact type is one such example. Meanwhile Daniel said that one can detect (this example) by looking at the monodromy action on the (algebraic) fundamental group of this curve and decided it doesn’t have (potential) good reduction. On the other hand, Dingxin pointed the existence of this paper of Robert Friedman. This paper exhibits an example of degenerating family of quintic surfaces over a punctured disk, which doesn’t have a (potential) good reduction, with the property that a finite power of the monodromy action is homotopic to identity!

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In memory of Michael Zhao.

Explicit surfaces in characteristic p with non-reduced Picard.

November 10, 2018 mayorliatmath Leave a comment

In this notes of Jesse Kass, an example due to Igusa is discussed. One takes an ordinary elliptic curve $E$ over characteristic $2$ with an actual $2$ -torsion point $P$ , and form the quotient of $E \times E'$ by a (fixed point free) action of $\mathbb{Z}/2 = \langle \sigma \rangle$ where the automorphism $\sigma$ acts by $\sigma(x,y) = (x+P,-y)$ and here $E'$ is another elliptic curve (could be the same as $E$ ). The point is that this quotient is then a smooth projective surface in characteristic $2$ with non-reduced Picard scheme.

One can simply make similar examples in other characteristics too. For instance, in characteristic $3$ , take ordinary $E$ with an actual $3$ -torsion point. Then take $E'$ to be the supersingular elliptic curve with an automorphism by “ $\omega$ “. Do the similar construction above, and run the same argument, you can prove that the Picard scheme has tangent space of dimension $2$ whereas the Albanese has dimension $1$ (use $\omega^2 + \omega + 1 = 0$ ).

As for characteristic $p \geq 5$ , take ordinary $E$ with an actual $p$ -torsion as usual. Then let $C$ be the hyperelliptic curve given by $y^2 = x^p - x$ , which admits the automorphism which fixes $y$ and sends $x$ to $x+1$ . This is an automorphism of order $p$ . So do the similar construction, by Hochschild–Serre spectral sequence, you will find that the tangent space of the Picard scheme is at least $2$ . However, its Albanese has dimension $1$ again, this is because that our automorphism acts on $H^1_{crys}(C)[\frac{1}{p}]$ via those non-trivial $p-1$ characters of order $p$ , hence there is no invariants.