# admissible subcategories over non algebraically closed fields

Let $X$ be a smooth projective variety over a field $k$ and $D^b(X)$ its bounded derived category. Let $\bar{X}$ the base change to $\bar{k}$. Let $A$ be a triangulated subcategory of $D^b(X)$ that $\bar{A}$ is admissible inside $D^b(\bar{X})$ over $\bar{k}$. Is $A$ admissible also inside $D^b(X)$ over $k$? (admissible means that the embedding functor admits both left and right adjoint functors)

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Notice that your definition of admissible doesn't depend on a ground field. – Fernando Muro May 27 '13 at 20:25
Yes, in fact I edited the question a little bit in order to make it clearer what I meant. Thank you. – IMeasy May 28 '13 at 6:19
What's $\bar A$? – Fernando Muro May 28 '13 at 9:08
It is the subcategory of $D^b(\bar{X})$ obtained as a scalar extension of $A$. – IMeasy May 28 '13 at 11:08
@IMeasy, beware that $D^b(\bar X)$ is not any scalar extension of $D^b(X)$. If you take a triangulated category $T$ defined over a field $k$ (i.e. hom sets are $k$-vector spaces) then $T\otime_k\bar k$ is just a $\bar k$-linear additive category, it doesn't carry any induced triangulated structure. – Fernando Muro May 28 '13 at 11:21

I think the answer should be yes, but the statement should be more accurate. First, in the definition of $\bar{A}$ you first extend scalars and then take the triangulated hull and add all direct summands. consequently, if you want the statement to be true you should add all direct summands to $A$ as well (or assume that $A$ is Karoubian complete from the start).
The proof should go as follows. Assume that $D(\bar{X}) = \langle \bar{A}, \bar{B} \rangle$ be a semiorthogonal decomposition. First one should check that it is invariant under the Galois action. Then one should check that an object in $D(\bar{X})$ is Galois invariant if and only if it is in $D(X)$. Then one should restrict the semiorthogonal decomposition of $D(X)$ by intersecting the above decomposition with $D(X)$.