For simplicity, I will be talking only about connected groups over an algebraically closed field of characteristic zero.
The basic theorem of affine algebraic groups is that they all admit faithful, finite-dimensional representations. The fundamental theorem for semisimple groups is that these representations are all completely reducible, but unfortunately there is no reason that any irreducible summand of a faithful representation should be faithful, only that the kernels of all these representations intersect trivially.
My question is whether such a representation does, in fact, exist.
(Answered: iff the center is cyclic.)
This does not hold of general reductive groups for the following reason: if $T$ is any torus of rank $r > 1$, then its irreducible representations are all characters $\chi \colon T \cong \mathbb{G}_m^r \to \mathbb{G}_m$, which therefore have nontrivial kernels. More generally, any reductive group $G$ has connected center a torus of some rank $r$, so by Schur's lemma this center acts by a character $\chi$ in any irreducible representation of $G$ and if $r > 1$, therefore does not act faithfully.
The exceptional case $r = 1$ does have an example, namely $\operatorname{GL}_n$, whose standard representation is faithful and irreducible and whose center has rank 1. A more general version of this question might be, then:
Does any reductive group whose center has rank at most 1 have a faithful irreducible representation?
(Answered: when not semisimple, iff the center is connected.)
Another special case is that if $G$ is simple and of adjoint type, then its adjoint representation is irreducible and faithful by definition (or, depending on your definition, because the center is trivial). A constructive version of this question for any $G$ (semisimple or reductive of central rank 1) is then:
Can we give a construction of a faithful, irreducible representation of $G$ from its adjoint representation?
(Not yet answered!)
This is deliberately a little vague since I don't want to restrict the possible form of such a construction, only that it not start out with "Throw away the adjoint representation and take another one such that..."
Finally, suppose the answer is "no".
What is the obstruction to such a representation existing?
(Answered: for $Z$ the center, it is the existence of a generator for $X^*(Z)$.)
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