Given a (split connected reductive) root datum $\Psi$, there is another associated root datum $\Psi'$ and a morphism of root data $\Psi \to \Psi'$ called the canonical isogeny. My question is, is the canonical isogeny ever not an isomorphism?
I ask this question because I believe the answer is yes—simply from the name—but I don't see how it's possible. Specifically, it seems to me that $X \cong X/X_0 \oplus X/\tilde{Q}$, implying $\Psi \cong \Psi'$. What am I missing?
In the first chapter of the Corvallis proceedings (by Springer), a root datum is defined as a quadruple $\Psi = (X, \Phi, X^{\vee}, \Phi^{\vee})$, satisfying some axioms. Most relevant here are that $X$ and $X^{\vee}$ are dual free abelian groups of finite rank and $\Phi \subset X$ is a finite subset in bijection with $\Phi^{\vee} \subset X^{\vee}$. Given a root datum $\Psi$, we define $Q = \mathbb{Z} \Phi \subset X$ and $X_0 = (\Phi^{\vee})^{\perp} \subset X$. More generally, for any subgroup of a free abelian group $L \subset M$ we have $\tilde{L}$ is the maximal subgroup $L \subset \tilde{L} \subset M$ such that $\tilde{L}/L$ has finite order. Springer notes that $L = \tilde{L}$ if and only if $L$ is a direct summand of $M$, but note also that every element $\ell \in \tilde{L}$ has a multiple $m \ell \in L$, as well as $\tilde{\tilde{L}} = \tilde{L}$.
We have (Lemma 1.2) that $X_0 \cap Q = 0$, and $X_0 + Q$ is of finite index in $X$. It is easy to check that $X_0 = \tilde{X}_0$, and so both $X_0$ and $\tilde{Q}$ are direct summands of $X$, they span $X$, and they must have trivial intersection. Therefore we have both that $X \cong X_0 \oplus \tilde{Q}$ and $X \cong X/X_0 \oplus X/\tilde{Q}$. (Note: I believe this paragraph is correct as written, but it is not as strong as it appears. I was trying to claim that $X = X_0 \oplus \tilde{Q}$, which is not necessarily true—only that $X \supset X_0 \oplus \tilde{Q}$.)
How is it not the case that the canonical isogeny $f: X \to X/X_0 \oplus X/\tilde{Q}$ induces an isomorphism of root data? Is it some slightly more subtle possible difference in $\Phi$ "$\subset$" $X/X_0$ vs $\Phi \subset \tilde{Q}$ (and dual)? Or is the canonical isogeny actually necessarily an isomorphism of root data?
I think I figured this out with the simplest example imaginable, $GL_2$. We have $X = \mathbb{Z}^2$, with basis $\{\lambda_1, \lambda_2\}$, where $\lambda_i \begin{pmatrix} t_1 & 0 \\ 0 & t_2 \end{pmatrix} = t_i$, and $\Phi = \{\pm(\lambda_1 - \lambda_2)\}$. Then $X_0 = \mathbb{Z}(\lambda_1 + \lambda_2)$ and $\tilde{Q} = Q = \mathbb{Z}(\lambda_1 - \lambda_2)$. Each is a direct summand and the two subgroups do have trivial intersection, but they are not complementary. Most immediately, $X_0 \oplus \tilde{Q} \subsetneq X$.
Basically I was thinking of lattices too much like vector spaces. In particular, the word "span" that I used in the question does not apply—despite being linearly independent and full rank, the subgroups $\tilde{Q}$ and $X_0$ do not necessarily generate $X$.