Smoothness of the moduli space of Drinfeld modules

Question

I'm studying the proof of Thm 1.5.1. in Laumon's "Cohomology of Drinfeld Modular Varieties". Notation: $\mathfrak{m}$ is a square zero ideal of $\mathcal{O}$ and $k=\mathcal{O}/\mathfrak{m}$. Laumon shows that the obstruction to the existence of a lift of a Drinfeld module $\phi: A \rightarrow k[\tau]$ to a Drinfeld module $\phi':A \rightarrow \mathcal{O}[\tau]$ lies in the second Hochschild cohomology $HH^2(A,\mathfrak{m}[\tau])$.

I get that this is an obstruction to the existence of a lift of $\phi$ to a ring morphism $\phi':A \rightarrow \mathcal{O}[\tau]$. However, I don't see how the degree condition is controlled by the Hochschild cohomology. I mean just because there exists a lift as a ring morphism $\phi'$, this does not need to be of the same rank as $\phi$.

Laumon does not explain this, and neither do Blum and Stuhler in "Drinfeld Modules and Elliptic Sheaves". Can some clarify this for me? Thanks.

Answer 1

At this point of the proof, Laumon assumes that $\phi$ is standard—I'll do so as well. Let $a\in A$ be nonzero. Then

$$ \phi_a = \sum_i \phi_{a,i} \tau^i$$

with $\phi_{a,-d\deg(\infty)\infty(a)} \in k^{\times}$ and $\phi_{a,i} = 0$ in higher degrees. Note that an element $x$ of $\mathcal{O}$ is a unit if and only if is a unit mod $\mathfrak{m}$, because $\mathfrak{m}$ consists of nilpotents. Hence

$$ \phi'_a = \sum_i \phi'_{a,i} \tau^i$$

with $\phi'_{a,-d\deg(\infty)\infty(a)} \in \mathcal{O}^{\times}$ and $\phi'_{a,i} \in \mathfrak{m}$ in higher degrees. Compare this with Definition 1.2.1 to see that $\phi'$ is a Drinfeld module of rank $d$ over $\mathcal{O}$.