So I am studying Lie Theory via Humphreys Introduction to Lie Algebras and Representation Theory and using Erdmann and Wildon's Introduction to Lie Algebras as a bit of a down-to-earth supplement, and have been having a bit of trouble understanding some of the Integrality Properties introduced by (or implied by) the only proposition made in Section 8.4 on page 39 of Humphreys, which is analogously proven in Proposition 10.10 (page 101-102) of Erdmann.
In particular, I want to know about the truth of the statement: Let $\alpha$ and $\beta$ be roots (i.e. $\alpha, \beta \in \Phi$). Then if $\beta + k\alpha$ is a root, then k is in $\mathbb{Z}$. In the discussion of root systems in Chapter 9, this seems to be something taken for granted, but I have been dissatisfied about the treatment of this proving this statement holds for the root systems associated to Lie Algebras.
To be honest, I'm not quite sure if Humphreys is even making this claim. Certainly he claims that if $\alpha$ and $\beta$ are roots then $\beta - \beta(h_\alpha)\alpha$ is a root, and $\beta(h_\alpha) \in \mathbb{Z}$, and part (e) gives conditions on when $\beta + i\alpha$ is a root for integer value of $i$, but this is still not addressing the fact that $i$ must be an integer for $\beta + i\alpha$ to be a root. In fact, in the proof of the proposition initially sets up $K = \sum_{i \in \mathbb{Z}} L_{\beta + i\alpha}$, which seems to be already assuming that the only possible roots $\beta + i\alpha$ have $i \in \mathbb{Z}$.
Erdmann definitely wants to prove this: Proposition 10.10(ii) on page 101 states that there are integers $r, q \geq 0$ such that $\beta+k\alpha$ is a root if and only if $k \in \mathbb{Z}$ and $-r \leq k \leq q$.
In Erdmann's proof, upon setting up the $\mathfrak{sl}(\alpha)$ module, $M = \bigoplus_{k} L_{\beta + k\alpha}$, she doesn't specify the index of the sum, but based on Example 10.8(3), where she indexes the same sum over $\mathbb{C}$, I am going to assume that the sum is over $\mathbb{C}$ (and obviously finite since our Lie Algebra is in finite dimensional).
From there we know $\beta(h_\alpha)$ is in $\mathbb{Z}$, which gives that $\beta(h_\alpha) + 2k = (\beta+k\alpha)(h_\alpha)$ are integers since they're the eigenvalues of $h_\alpha$. From there, the claim is made that these eigenvalues are all even or all odd, which is I can only conclude is based on the classification of irreducible $\mathfrak{sl}(\alpha)$ modules (So if we know $M$ is irreducible, we are done). However, she has not yet established that $M$ is an irreducible $\mathfrak{sl}(\alpha)$ module, and this fact that the eigenvalues are all even or all odd is what she uses to conclude that $M$ is irreducible. Now, certainly if $k \in \mathbb{Z}$ then these eigenvalues are all even or odd, but if the sum was meant to be over $\mathbb{Z}$ instead of $\mathbb{C}$, then this only establishes one direction of the proof, and not the direction I'm interested in.
However, Erdmann's treatment establishes that $\beta(h_\alpha) + 2k$ are integers, which convinces me that the $k$ must be half-integers since it was previously established that $\beta(h_\alpha)$ is an integer.
So, ultimately, I am asking for a proof (or counterexample, if it is not true) that if $\alpha$ and $\beta$ are roots, and $\alpha + k\beta$ is a root, then $k \in \mathbb{Z}$.
If I am missing something in my understanding of the proofs of Humphreys and Erdmann,and they do actually prove this, then I would greatly appreciate any explanation of where I am going wrong!
Here we go, in four sections. I use Bourbaki's book on Lie Groups and Algebras as reference.
A. The roots $\Phi$ form a root system in the sense of Bourbaki (VI 1), where we define the linear form (coroot) $\check{\alpha}$ as evaluating at $h_\alpha$, i.e. $\check{\alpha} (\beta) := \beta (h_\alpha)$ for all roots $\alpha, \beta$.
Proof: Bourbaki's first criterion: We certainly assume that $\Phi$ is non-empty, and know it is finite and does not contain $0$.
Bourbaki's second criterion: For all $\alpha, \beta \in \Phi$, we have $\check{\alpha}(\alpha) = 2$ (clear), and the following is a root: $$s_\alpha(\beta) := \beta -\check{\alpha}(\beta) \alpha = \beta - \beta(h_\alpha) \alpha$$ To see this last claim, note that $-\beta(h_\alpha) =q-r$ and this is an integer between $-r$ and $q$, so by the direction of Erdmann's proof that you accept, indeed $s_\alpha(\beta) \in \Phi$.
Bourbaki's third criterion: $\check\alpha(\Phi) \in \mathbb{Z}$ for all $\alpha \in \Phi$. You imply this as proven in your sixth paragraph.
B. Let $R$ be a root system and $\alpha, \beta \in R$. Let $k \in \frac{1}{2}\mathbb{Z} \setminus \mathbb{Z}$ such that $\beta +k\alpha$ is a root. Then $R$ is not reduced, i.e. there is a root $\gamma \in R$ such that $\frac{1}{2}\gamma \in R$.
Proof: We choose a $W(R)$-invariant scalar product $(\cdot,\cdot)$ on $V := \mathbb{R}R$. Note that $(\beta +k\alpha, \beta +k\alpha) = (\beta, \beta) + k^2 (\alpha, \alpha) + 2k(\alpha, \beta)$. Note further that for all roots $\delta$, we have $\check{\delta}(\cdot) = \displaystyle \frac{2(\delta, \cdot)}{(\delta, \delta)}$. What we do in the following is basically comparing root lengths, although the formulas look nasty sometimes. It helps a lot to draw pictures!
By the third criterion, $\check\beta(\beta +k\alpha) = \check\beta(\beta) +k\check{\beta}(\alpha) = 2 + k\check{\beta}(\alpha)\in \mathbb{Z}$, hence $k\check{\beta}(\alpha) \in \mathbb{Z}$, hence $\check{\beta}(\alpha) \in 2\mathbb{Z}$. By Bourbaki VI 1 no. 3, the only possibilities are $\check{\beta}(\alpha) \in \{0, \pm 2, \pm 4\}$.
Case 1: $\check{\beta}(\alpha) = 0 \Leftrightarrow \check{\alpha}(\beta) = 0 \Leftrightarrow \alpha \perp \beta$. Since $\beta +k\alpha$ is neither perpendicular to nor a multiple of either $\beta$ or $\alpha$, we must have $$\{\frac{1}{3}, \frac{1}{2},1,2,3\} \ni \displaystyle \frac{(\beta +k\alpha, \beta +k\alpha)}{(\beta, \beta)} = \displaystyle \frac{(\beta, \beta) + k^2 (\alpha, \alpha)}{(\beta, \beta)} = 1 + k^2 \displaystyle \frac{(\alpha, \alpha)}{(\beta, \beta)}$$ as well as $$\{\frac{1}{3}, \frac{1}{2},1,2,3\} \ni \displaystyle \frac{(\beta +k\alpha, \beta +k\alpha)}{(\alpha, \alpha)} = \displaystyle \frac{(\beta, \beta) + k^2 (\alpha, \alpha)}{(\alpha, \alpha)} = \displaystyle \frac{(\beta, \beta)}{(\alpha, \alpha)} + k^2 .$$ One checks that this necessitates $k = \pm\frac{1}{2}$ and $(\alpha, \alpha) = 4 (\beta, \beta) = 2(\beta +k\alpha, \beta+k\alpha)$. But then $$s_{\beta +k\alpha}(\beta) = \beta - \displaystyle \frac{2(\beta +k\alpha, \beta)}{(\beta +k\alpha, \beta +k\alpha)}(\beta +k\alpha) = -k\alpha$$ is a root, so the claim is proven with $\gamma =\alpha$. (And indeed this case occurs in all root systems $BC_{n\ge 2}$.)
Case 2: $\check{\beta}(\alpha) = -2 \Leftrightarrow \check{\alpha}(\beta) = -1$. In this case, $(\alpha, \alpha) = 2(\beta, \beta) = -2(\beta, \alpha)$. ($\alpha$ and $\beta$ are a basis of $B_2$, $\alpha$ is the longer one, the angle between them is $3\pi/4$.)
If $k = \frac{1}{2}$, we have $s_\beta(\beta+\frac{1}{2}\alpha) = \frac{1}{2}\alpha$, so again we are done with $\gamma = \alpha$ (and again, this case occurs in $BC_{n\ge2}$).
If $k \neq \frac{1}{2}$, we have that $\alpha$ is neither perpendicular to nor a multiple of $\beta +k\alpha$, hence $$\{\frac{1}{3}, \frac{1}{2},1,2,3\} \ni \displaystyle \frac{(\beta +k\alpha, \beta +k\alpha)}{(\alpha, \alpha)} = k^2 - k + \frac{1}{2}.$$ which is impossible for $k \in \frac{1}{2}\mathbb{Z} \setminus \mathbb{Z}$.
Case 3: $\check{\beta}(\alpha) = 2 \Leftrightarrow \check{\alpha}(\beta) = 1$. In this case, $(\alpha, \alpha) = 2(\beta, \beta) = 2(\beta, \alpha)$. ($s_\beta(\alpha)$ and $\beta$ are a basis of $B_2$, $\alpha$ is the longer one, the angle between $\alpha$ and $\beta$ is $\pi/4$.)
If $k = -\frac{1}{2}$, we have $s_\beta(\beta-\frac{1}{2}\alpha) = -\frac{1}{2}\alpha$, so again we are done with $\gamma = \alpha$ (and again, this case occurs in $BC_{n\ge2}$).
If $k \neq -\frac{1}{2}$, we have that $\alpha$ is neither perpendicular to nor a multiple of $\beta +k\alpha$, hence $$\{\frac{1}{3}, \frac{1}{2},1,2,3\} \ni \displaystyle \frac{(\beta +k\alpha, \beta +k\alpha)}{(\alpha, \alpha)} = k^2 + k + \frac{1}{2}.$$ which is impossible for $k \in \frac{1}{2}\mathbb{Z} \setminus \mathbb{Z}$.
Case 4: $\check{\beta}(\alpha) = \pm 4$. Then necessarily $\alpha = \pm 2\beta$ and hence $\frac{1}{2}\alpha$ is a root. This case occurs in every non-reduced root system.
Q.E.D. Notice that in all cases, $k =\pm\frac{1}{2}$, and $\alpha$ is actually one of the roots whose half is a root too.
C. The root system $\Phi$ of a splitting Cartan subalgebra $\mathfrak{h}$ in a semisimple Lie algebra $\mathfrak{g}$ is reduced.
Proof: Assume there were a root $\alpha \in \Phi$ with $2\alpha \in \Phi$. Choose an $\mathfrak{sl}(\alpha)$-triple. Via this, we view the space $E:= \bigoplus_{i \in \mathbb{Z}} \mathfrak{g}_{i\alpha}$ as $\mathfrak{sl}_2$-representation. We need the following fact from the theory of $\mathfrak{sl}_2$-representations: If $\lambda > 0$ is an $h_\alpha$-eigenvalue in $E$, then $x_\alpha$ (the element of the triple from $\mathfrak{g}_\alpha$) maps the eigenspace $E_{\lambda-2}$ surjectively onto $E_\lambda$ (see Bourbaki VIII 1 no. 2, corollary(ii) to prop. 2). Applying this to our situation, we get surjective linear maps $$[x_\alpha, \cdot] : E_0 = \mathfrak{g}_0 = \mathfrak{h} \twoheadrightarrow \mathfrak{g}_{\alpha} = E_2$$ and $$[x_\alpha, \cdot] : E_2 = \mathfrak{g}_\alpha \twoheadrightarrow \mathfrak{g}_{2\alpha} = E_4.$$ The first map shows that $\mathfrak{g}_\alpha = [\mathfrak{h}, x_\alpha] = K x_\alpha$ is one-dimensional and generated by $x_\alpha$. But then the second map is surjective and identically 0 (because $[x_\alpha, x_\alpha] = 0$), hence $\mathfrak{g}_{2\alpha} = \{0\}$, contradiction to $2\alpha$ being a root.
D. There are situations close to this where we do get a non-reduced root system, and hence a root of the form $\beta + \frac{1}{2}\alpha$ for roots $\alpha, \beta$. (In any non-reduced root system, if $\frac{1}{2}\alpha$ and $\alpha$ are roots, just set $\beta = \frac{1}{2}\alpha$.)
Namely, the following 8-dimensional Lie algebra over $\mathbb{R}$ is a matrix representation of the quasi-split form of type $A_2$: $$\mathfrak{g} = \lbrace \begin{pmatrix} a+bi & c+di & ei\\ f+gi & -2bi & -c+di\\ hi & -f+gi & -a+bi \end{pmatrix} : a, ..., h \in \mathbb{R} \rbrace.$$ The subalgebra $$\mathfrak{a} := \lbrace \begin{pmatrix} a & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -a \end{pmatrix} : a \in \mathbb{R}\rbrace$$ is abelian, consists of ad-diagonalisable elements, and is maximal with this property. Hence $\mathfrak{g}$ allows a weight decomposition $$\mathfrak{g} = \mathfrak{g}_0 \oplus \bigoplus_{\gamma \in \Phi} \mathfrak{g}_\gamma.$$ One checks that all arguments work "as usual" to show that $\Phi$ is a root system: it is just non-reduced. In fact, it consists of $\pm \beta, \pm 2 \beta$, where $\beta\left(\begin{pmatrix} a & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -a \end{pmatrix}\right) = a$. We can set $h_\beta = \begin{pmatrix} 2 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -2 \end{pmatrix}$, and we can also extend this to an $\mathfrak{sl}_2$-triple with $x_\beta = \begin{pmatrix} 0 & 1+i & 0\\ 0 & 0 & -1+i\\ 0 & 0 & 0 \end{pmatrix}$ and $x_{-\beta} = \begin{pmatrix} 0 & 0 & 0\\ 1-i & 0 & 0\\ 0 & -1-i & 0 \end{pmatrix}$.
The first place where all above reasoning fails is in part C, because $\mathfrak{a}$ is not a Cartan subalgebra, it is not self-normalising or self-centralising, specifically $\mathfrak{a} \subsetneq \mathfrak{g}_0$. In fact, $$\mathfrak{g}_0 = \lbrace \begin{pmatrix} a+bi & 0 & 0\\ 0 & -2bi & 0\\ 0 & 0 & -a+bi \end{pmatrix} : a,b \in \mathbb{R}\rbrace$$ and $$\mathfrak{g}_\beta = \lbrace \begin{pmatrix} 0 & c+di & 0\\ 0 & 0 & -c+di\\ 0 & 0 & 0 \end{pmatrix} : c,d \in \mathbb{R}\rbrace$$ are both two-dimensional, $[x_\beta, \cdot]$ defines an isomorphism between them, and $$\mathfrak{g}_{2\beta} = \lbrace \begin{pmatrix} 0 & 0 & ei\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} : e \in \mathbb{R}\rbrace$$ is one-dimensional but non-zero.
The upshot of this example is: Without somehow using the fact that our Cartan subalgebra $\mathfrak{h}$ is self-centralising, we cannot show the fact you are looking for.