IMC 2018 · Day 1 · P2

There exist no such field.

Suppose that $F$ is such a field and $g : F^* \to F^+$ is a group isomorphism. Then $g(1) = 0$.

Let $a = g(-1)$. Then $2a = 2 \cdot g(-1) = g\bigl( (-1)^2 \bigr) = g(1) = 0$; so either $a = 0$ or $\operatorname{char} F = 2$. If $a = 0$ then $-1 = g^{-1}(a) = g^{-1}(0) = 1$; we have $\operatorname{char} F = 2$ in any case.

For every $x \in F$, we have $g(x^2) = 2 g(x) = 0 = g(1)$, 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.