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Let $\phi$ be a Drinfeld module $\phi:A\to \ell$ and let $k = Fr(A)$. Then recall that the smallest field of definition is defined as the field constructed from $k$ and the coefficients of $\phi$, and the minimal field is defined as the extension of $k$ which is contained in every field of definition of $\phi$ (recall that a field of definition of $\phi$ is an extension of $k$ by the coefficients of some $\psi$ isomorphic to $\phi$).

When we construct the normalization field for an $\epsilon$-normalized Drinfeld module (i.e., the smallest field over which $\phi$ is defined), we show in the general theory that this is independent of our choice of $\phi$, i.e., we can construct it out of any $\epsilon$-normalized Drinfeld module (and later show that this is the extension by the narrow class group). Then why isn't this field the minimal field of definition of $\phi$? (It isn't--the minimal field will correspond to an extension by the class group, which differs from this extension by a cyclic group as long as the degree of our chosen point at $\infty$ is $>1$).

I think clearing this up will also help in understanding more generally why the theory seems to require that we construct the narrow Hilbert class field first and not directly the Hilbert class field as in the case of complex multiplication elliptic curves :)

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This field is dependent on our choice of sign function $\epsilon$, although not dependent on $\phi$. To have an $\epsilon$-normalized Drinfeld module, we must at least extend by $\mathbb{F}_{r^{\deg\infty}}/\mathbb{F}_r$. – David Tweedle Dec 17 '12 at 15:29

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