Problem 1716

There exist no such field. Suppose that $F$ is such a field and $g: F^{*} \rightarrow F^{+}$ is a group isomorphism. Then $g(1)=0$. Let $a=g(-1)$. Then $2 a=2 \cdot g(-1)=g\left((-1)^{2}\right)=g(1)=0$; so either $a=0$ or char $F=2$. If $a=0$ then $-1=g^{-1}(a)=g^{-1}(0)=1$; we have char $F=2$ in any case. For every $x \in F$, we have $g\left(x^{2}\right)=2 g(x)=0=g(1)$, so $x^{2}=1$. But this equation has only one or two solutions. Hence $F$ is the 2-element field; but its additive and multiplicative groups have different numbers of elements and are not isomorphic.