Let $K$ be a field and $G:=SL\_2(K)$, then $G$ is a $K-$split reductive group (to use some big words). These groups are classified by a based root datum $(X,D,X',D')$. Let $G'$ be group associated to $(X',D',X,D)$, the so called dual group.
Is it correct that $G'=PGL\_2(K)$? 


I"m wondering how $PSL\_2(K)$ fits into this picture.
I'm aware of the fact that if C is algebraically closed, then $PSL\_2(C) \cong PGL\_2(C)$ as abstract groups; can this be made into an isomorphism of algebraic groups, i.e. is $PSL$ a $K-$form of $PGL$?



What is the difference between PSL_2 and PGL_2?
Yes, the dual of $SL\_2$ is $PGL\_2$.


But you're not going down the right track with $PSL\_2$. The problem with $PSL\_2$ is that it's not a variety at all! You can quotient out the variety $SL\_2$ by the subgroup $\pm1$ but the quotient is the variety $PGL\_2$ (recall that quotients in the category of sheaves (for these are really fppf sheaves) don't have to be surjective on global sections, so the statement that there's a surjection $SL\_2\to PGL\_2$ does not imply that the induced map $SL\_2(\mathbf{Q})\to PGL\_2(\mathbf{Q})$ is a surjection). 


The problem with $PSL\_2$ is that it is a functor from, say, rings to groups, but it's not a representable one, so in particular it's not an algebraic group. If you like, you can imagine $PSL\_2$ as a presheaf quotient, and $PGL\_2$ as the associated (representable) sheaf.




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Edit: I tried to think of a way to make this observation more enlightening. Let's for example try and build an affine variety over $\mathbf{Q}$ representing the $PSL\_2$ functor. Well we can certainly build an affine variety over $\mathbf{Q}$ representing the $SL\_2$ functor: it's just $A:=\mathbf{Q}[a,b,c,d]/(ad-bc-1)$. Now let's see what happens if we try and quotient out by the group $\pm1$. The quotient is affine, and is represented by the invariants of the action, that is, the subring of $A$ consisting of polynomials in $a$, $b$, $c$, $d$ with the property that every monomial mentioned in the polynomial has total degree even. Now here's the problem: I can see a $\mathbf{Q}$-point of this subring (that is, a map from this subring to $\mathbf{Q}$) that corresponds to the matrix $(s,0;0,1/s)$ for $s=\sqrt{p}$, $p$ a prime number! It's the point that sends $a^2$ to $p$, $ab$ to $0$, and so on and so on, and finally $d^2$ goes to $1/p$ and $ad$ goes to $1$, $bc$ goes to $0$, so $ad-bc=1$. The map from the subring to $\mathbf{Q}$ doesn't extend to a map from the whole ring to $\mathbf{Q}$, so our putative construction has failed because the $\mathbf{Q}$-points of this ring are a group that canonically but strictly contains $PSL\_2(\mathbf{Q})$ (this subgroup being the $\mathbf{Q}$-points that extend to $\mathbf{Q}$-points of $A$).


