If $rk J=0$ then $R$ is a field

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If $R$ is a regular local ring with maximal ideal $J$ and $rk J=0$ why is $R$ then a field?

I have this as a statement in my notes and I feel that it should be obvious but I have no idea why it is the case.

Is it because as the ring is regular we have $dim J/J^2=0$ and hence $J=J^2$ and so $J$ must be equal to $0$ for some reason? I could use Nakayama's Lemma to show this if $J$ was f.g. but I can't see why that is the case.

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Ok, I think you meant

$$\text{rk}\,J=0\iff \dim J/J^2=0\iff J=J^2\implies J=0\;\;\text{by Nakayama's Lemma}$$

and we're done.