Prove a ring homomorphism is flat

Let $E$ be a field of characteristic $p$, and $\phi$ be the Frobenius map. We have the Cohen ring associated to $E$, denoted by $\mathcal{O}$, which is a absolutely unramified complete discrete valuation ring of characteristic $0$ with residue field $E$. We also have a ring homomorphism (not unique) lifting $\phi$, also denoted by $\phi$, $\phi \colon \mathcal{O} \to \mathcal{O}$. Why is this ring homomorphism flat?

This is from Lemma 3.2.3 of "CMI summer school notes on $p$-adic hodge theory". I don't have any thoughts, thank you for any help.

Solution 1:

As you say yourself, $\mathcal{O}$ is a DVR. Over any DVR (actually, any Dedekind ring), a module is flat iff it is torsion-free.

Here, the module in question $M$ is $\mathcal O$ itself but via the homomorphism $\phi$, i.e. the module structure is given via $r \cdot m := \phi(r)m$ for $r \in \mathcal O, m \in M$.

Can you take it from here? (It's particularly easy to check for torsion because $\mathcal O$ is a domain.)