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Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is not finitely generated (so not contained in $\mathbb{Z}$). If it has characteristic $p$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion and so is also not contained in $\mathbb{Z}$.

Following Gyergi's comment, if we were dealing with an arbitrary finitely-generated abelian group, the last statement could have finitely many exceptions. Then an alternate argument works: for any $p$, there is some maximum $n$ such that $\mathbb{F}_{p^n}$ is a subfield, so if the field is infinite, it must be transcendental over $\mathbb{F}_p$. But $\mathbb{F}_p(x)^\times$ is not finitely generated since there are infinitely many irreducible polynomials (e.g. those such that $\mathbb{F}_{p^n} = \mathbb{F}_p[t]/(f)$).

Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is not finitely generated (so not contained in $\mathbb{Z}$). If it has characteristic $p \neq 2$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion of order greater than 2 and so is also not contained in $\mathbb{Z}$. If $p = 2$, then the field can't contain $\mathbb{F}_4$ for the same reason, so if it's not $\mathbb{F}_2$, it must be transcendental over it. But $\mathbb{F}_2(x)^\times$ is not finitely generated again, so not contained in $\mathbb{Z}$ either.

Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is not finitely generated (so not contained in $\mathbb{Z}$). If it has characteristic $p \neq 2$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion of order greater than 2 and so is also not contained in $\mathbb{Z}$. If $p = 2$, then the field can't contain $\mathbb{F}_4$ for the same reason, so if it's not $\mathbb{F}_2$, it must be transcendental over it. But $\mathbb{F}_2(x)^\times$ is not finitely generated again, so not contained in $\mathbb{Z}$ either.

Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is not finitely generated (so not contained in $\mathbb{Z}$). If it has characteristic $p$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion and so is also not contained in $\mathbb{Z}$.

Following Gyergi's comment, if we were dealing with an arbitrary finitely-generated abelian group, the last statement could have finitely many exceptions. Then an alternate argument works: for any $p$, there is some maximum $n$ such that $\mathbb{F}_{p^n}$ is a subfield, so if the field is infinite, it must be transcendental over $\mathbb{F}_p$. But $\mathbb{F}_p(x)^\times$ is not finitely generated since there are infinitely many irreducible polynomials (e.g. those such that $\mathbb{F}_{p^n} = \mathbb{F}_p[t]/(f)$).

Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is divisible (so not contained in $\mathbb{Z}$). If it has characteristic $p \neq 2$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion of order greater than 2 and so is also not contained in $\mathbb{Z}$. If $p = 2$, then the field can't contain $\mathbb{F}_4$ for the same reason, so if it's not $\mathbb{F}_2$, it must be transcendental over it. But $\mathbb{F}_2(x)^\times$ is not finitely generated, so not contained in $\mathbb{Z}$ either.

Yes; it suffices to rule out that the multiplicative group is $\mathbb{Z}$. The field can't have characteristic zero, since $\mathbb{Q}^\times$ is not finitely generated (so not contained in $\mathbb{Z}$). If it has characteristic $p \neq 2$, then it contains $\mathbb{F}_p$, where $\mathbb{F}_p^\times$ has torsion of order greater than 2 and so is also not contained in $\mathbb{Z}$. If $p = 2$, then the field can't contain $\mathbb{F}_4$ for the same reason, so if it's not $\mathbb{F}_2$, it must be transcendental over it. But $\mathbb{F}_2(x)^\times$ is not finitely generated again, so not contained in $\mathbb{Z}$ either.