Proof check - $L(\mathbb{F})\to L_{V}(\mathbb{F})$ for $L=\mathfrak{sl}_{2}$ is an isomorphism.

Question

Let $L=\mathfrak{sl}_{2}$ with basis $(x,y,h)$ and maximal toral subalgebra (or CSA) $H$, $char\mathbb{F}>2$ and $V=V(m)$ an irreducible $L$-module with highest weight $m\in\mathbb{Z}^{+}$.
Let also $L(\mathbb{F})=L(\mathbb{Z})\otimes_{\mathbb{Z}}\mathbb{F}$, $L(\mathbb{Z})$ the $\mathbb{Z}$-span of a Chevalley basis, be the Chevalley Algebra and $L_{V}(\mathbb{F})=L_{V}\otimes_{\mathbb{Z}} \mathbb{F}$, where $L_{V}$ is the stabilizer of any admissible lattice in $V$. The inclusion $L(\mathbb{Z})\to L_{V}$ induces a homomorphism of Lie Algebras $L(\mathbb{F})\to L_{V}(\mathbb{F})$.
I was trying to prove, that this homomorphism is an isomorphism for any choice of $m$. But I am not used to work in fields with characteristics other than $0$ and I fear that I overlooked something. I thought of the following:
First I took a look at $L_{V}$ for any $m$. I know it is $L_{V}=H_{V}+\mathbb{Z}x+\mathbb{Z}y$ so I only have to check what happens with $h$. It is $h.v_{i}=(m+2i)v_{i}$, $(v_{0}, \dots,v_{m})$ the usual basis of $V(m)$ with $v_{i}=\frac{1}{i!}y^{i}v_{0}$, that does span an admissible lattice.
If $m$ is odd, there is nothing one can change about the operation of $h$, so it is $L_{V}=L(\mathbb{Z})$ and so the inclusion as well as the homomorphism is of course an isomorphism (even for $char\mathbb{F}=2$).
If $m$ is even, it`s $h.v_{i}=2(\frac{m}{2}+i)v_{i}$, so $L_{V}=\mathbb{Z}\frac{h}{2}+\mathbb{Z}x+\mathbb{Z}y$. In the homomorphism above $h\otimes 1$ is send to $2(\frac{h}{2}\otimes 1)$ what is $0$ for $char\mathbb{F}=2$, which is why it is no isomorphism in this case, but not $0$ for $char\mathbb{F}>2$. As $x\otimes 1$ is send to $x\otimes 1$ and $y\otimes 1$ to $y\otimes1$ this is an isomorphism.
Thank you for helping me.