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Let $f: X\to \text{Spec}(R)$ be a proper and smooth morphism, with $R$ a strictly henselian dvr. Call $s = \overline{s}$ the closed point and $\eta$ the geometric point of $\text{Spec}(R)$.

Call $i_s : X_{\overline{s}}\to X$ the closed immersion and $h : X_{\overline{\eta}}\to X$ the inclusion of the geometric generic fiber.

The proper base change theorem and universal local acyclicity of smooth morphisms give an isomorphism

$$(*)\ \ \ \ H^p(X_{\overline{s}},F)\xrightarrow{\simeq} H^p(X_{\overline{\eta}},F),$$

where $F$ is a locally constant constructible $\ell$-adic sheaf.

Suppose that $f$ is not proper anymore but affine of finite type and separated. We still assume that $f$ is smooth.

Do we still have an isomorphism between $\ell$-adic cohomologies with proper supports:

$$H^p_c(X_{\overline{s}},\mathbf{Q}_{\ell})\xrightarrow{\simeq} H^p_c(X_{\overline{\eta}},\mathbf{Q}_{\ell})\ ?$$

ATTEMPTED SOLUTION:

The main idea I have in mind is as follows.

Choose a proper $g : \overline{X}\to \text{Spec}(R)$ that compactifies $f$.

Although $g$ may not be smooth, the inclusion $j : X\to \overline{X}$ is quasi-compact by the assumptions, and $j_!\mathbf{Q}_{\ell}$ is constructible on $\overline{X}$. Proper base change still applies to $j_!\mathbf{Q}_{\ell}$.

Call $j_s$ the base change of $j$ along $s\to\text{Spec}(R)$, and same for $j_{\eta}$.

Call $i_s$ and $h$ still the maps $\overline{X}_s\to \overline{X}$ and $\overline{X}_{\overline{\eta}}\to \overline{X}$ by abuse of notation.

I want to show that $i_s^*\Psi_g(j_!\mathbf{Q}_{\ell}) = 0$, for $\Psi_g$ the nearby cycles functor. If this is true, then the natural map

$$(j_s)_!\Psi_f(\mathbf{Q}_{\ell}) \to \Psi_g(j_!\mathbf{Q}_{\ell})$$ is an isomorphism, since its cofiber is $(i_s)_*i_s^*\Psi_g(j_!\mathbf{Q}_{\ell})$ (shifted). This is because $j_s^*\Psi_g = \Psi_f$ (where I use that $j$, an open immersion, is smooth).

In other words the question has a positive answer if the nearby cycles functor commutes with extension by zero.

Suppose that this is true. By properness of $g$, we have

$$R\Gamma(\overline{X}_{\overline{s}}, \Psi_g(j_!\mathbf{Q}_{\ell})) = R\Gamma(\overline{X}_{\overline{\eta}}, j_!\mathbf{Q}_{\ell}).$$

By the claim we have $R\Gamma(\overline{X}_{\overline{s}}, \Psi_g(j_!\mathbf{Q}_{\ell})) = R\Gamma(\overline{X}_{\overline{s}}, (j_s)_!\Psi_f(\mathbf{Q}_{\ell}))$. By smoothness of $f$ we have $\Psi_f(\mathbf{Q}_{\ell}) = \mathbf{Q}_{\ell}[0]$.

Putting everything together we have an isomorphism

$$R\Gamma(\overline{X}_{\overline{s}},j_!\mathbf{Q}_{\ell}) = R\Gamma(\overline{X}_{\overline{\eta}},j_!\mathbf{Q}_{\ell})$$ which gives what we want.

So the question really is:

do we have $(j_s)_!\Psi_f(\mathbf{Q}_{\ell}) =\Psi_g(j_!\mathbf{Q}_{\ell})$?

For every closed geometric point $\overline{x}\to \overline{X}_{\overline{s}}$, we call $\overline{X}_{\overline{s}}(\overline{x})$ the strict henselianization of $\overline{X}_{\overline{s}}$ at $\overline{x}$ and $\overline{X}_{\overline{s}}(\overline{x})_{\overline{\eta}}$ its generic fiber.

Call $t_x : \overline{X}_{\overline{s}}(\overline{x})\to \overline{X}_{\overline{s}}$ the obvious map.

If $\overline{x}\to \overline{X}_{\overline{s}}$ does not factor through $X_{\overline{s}}$, then $t_x^*j_!\mathbf{Q}_{\ell} = 0$. Right?

In particular the stalk $\Psi_g(j_!\mathbf{Q}_{\ell})_{\overline{x}}$ is zero and so $i_s^*\Psi_g(j_!\mathbf{Q}_{\ell}) = 0$ and we're done.

Am I correct?

SUMMARY OF THE ANSWER BELOW: the answer is no, and below there is an example of a smooth $X$ of dimension $1$ for which $H^1_c(X_{\overline{s}},\mathbf{Q}_{\ell})\to H^1_c(X_{\overline{\eta}},\mathbf{Q}_{\ell})$ is not an isomorphism.

This also gives an explicit example for which formation of the nearby cycles complex does not commute with extension by zero. My argument above breaks down when I say "$t_x^*j_!\mathbf{Q}_{\ell} = 0$".

However, if $X$ is of dimension $d$, smooth and with geometrically connected fibers, the specialization map is indeed an isomorphism in degree $0$ and $2d$ even though $X$ is not proper.

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The statement is false without proper assumption.

Consider any "degeneration of a smooth elliptic curve to a nodal curve" and delete a singular point in a special fibre. This will give you a counterexample for the dimension reasons.

Details: Start with any proper morphism $f: \mathcal E' \to \operatorname{Spec}R$ such that its generic fibre is an elliptic curve (smooth geometrically connected curve of genus $1$ with a fixed section) and its special fibre is a nodal curve. For example, take $R=\overline{\mathbf F_p}[[t]]$ and consider a curve $\mathcal E'\subset \mathbf P^2_{R}$ given by the equation $$Y^2Z-X^3-X^2Z-t^3=0. $$

This curve has exactly one singular point in a special fibre given by $p=[0:0:1] \subset \mathbf P^2_{R}(\overline{\mathbf F_p})$. Define $$ \mathcal E:=\mathcal E' \setminus {p}. $$ By the construction $\mathcal E'$ is a smooth $R$-scheme of relative dimension one, but it is not proper. Let us compute cohomology with compact support of the geometric fibres of this curve.

Geometric generic fibre: We know $\mathcal E_{\overline{\eta}}$ is a connected proper smooth curve of genus 1, so $$ \mathrm{H}^1_{c}(\mathcal E_{\overline{\eta}}, \mathbf Q_l)=\mathrm{H}^1(\mathcal E_{\overline{\eta}}, \mathbf Q_l)=\mathbf Q_l^2. $$

Geometric special fibre: Here we see that $\mathcal E_{\overline{s}}$ is a nodal cubic $Y^2Z-X^3-X^2Z=0$ minus one point $[0:0:1]$. It is a standard computation to show that this scheme is actually isomorphic to $\mathbf G_m$. We know that $$ \mathrm{H}^1(\mathbf G_m, \mathbf Q_l)=\mathbf Q_l. $$

And finally using Poincare duality between usual cohomology and cohomology with compact support (using that $\mathbf G_m$ is smooth) we conclude that $$ \mathrm{H}^1_{c}(\mathcal E_{\overline{s}}, \mathbf Q_l) = \mathrm{H}^1_{c}(\mathbf G_m, \mathbf Q_l)=\mathrm{H}^1(\mathbf G_m, \mathbf Q_l)^{\vee}=\mathbf Q_l. $$

So, just for dimension reasons we can't have an isomorphism

$$ \mathrm{H}^1_{c}(\mathcal E_{\overline{s}}, \mathbf Q_l) \to \mathrm{H}^1_{c}(\mathcal E_{\overline{\eta}}, \mathbf Q_l). $$

Remark 1: I assume everywhere that $\ell$ is coprime with $p$.

Remark 2: This example may be a little bit misleading. Actually the failure of a specialization map to be an isomorphism has nothing to do with singularities of the "compactified family". You can produce a lot of examples by deleting points from a special fibre of smooth proper families of curves as well. But I think that this example is easier to visualize.

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    $\begingroup$ OK excellent. Thanks. I’m wondering if, however, if my $X$ is smooth as in the question, and in addition connected of pure dimension $d$, the specialization map in top degree is indeed an isomorphism. It should be, since we have trace maps $H^{2d}_c(X_{\overline{s}},\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}$ and $H^{2d}_c(X_{\overline{\eta}},\mathbf{Q}_{\ell})\to\mathbf{Q}_{\ell}$, and both trace maps are isomorphisms. Also, traces are compatible with base change so they fit into a commutative square with the specialization map on top and the identity on $\mathbf{Q}_{\ell}$ at the bottom $\endgroup$
    – Ari
    Feb 20, 2019 at 2:21
  • $\begingroup$ Am I right at least in this special case when $p = 2d$? Thanks a lot $\endgroup$
    – Ari
    Feb 20, 2019 at 2:22
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    $\begingroup$ It seems reasonable to me. But there is an issue with twists, so things should work well once you assume that $\mu_{l^{\infty}}$ lie inside $R$. Part of the Poincare duality says that Trace map $R^{2d}f_!\mathbf Q_l(d) \to \mathbf Q_l$ is an isomorphism for smooth $R$-schemes with geometrically connected fibres of dimension $d$. Since formation of $R^i f_!$ commutes with base change by the Proper Base Change Theorem you get your statement (since it holds for $\mathbf Q_l$). $\endgroup$
    – gdb
    Feb 20, 2019 at 2:34
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    $\begingroup$ Ok, sorry, my comment about twists was rather dumb. It works well for $\mathbf Q_l$ just because $R^{2d}f_! \mathbf Q_l \cong \mathbf Q_l(-d)$. So there is no need to make any assumption about roots of unity. $\endgroup$
    – gdb
    Feb 20, 2019 at 2:45
  • $\begingroup$ Yes I agree. Excellent. Great answer too. $\endgroup$
    – Ari
    Feb 20, 2019 at 3:33

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