### International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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### Problem 2

Problem 2. Does there exist a field such that its multiplicative group is isomorphic to its additive group?

(Proposed by Alexandre Chapovalov, New York University, Abu Dhabi)

Solution. There exist no such field.

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

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

For every $\displaystyle x\in F$, we have $\displaystyle g(x^2)=2g(x)=0=g(1)$, so $\displaystyle x^2=1$. But this equation has only one or two solutions. Hence $\displaystyle F$ is the $\displaystyle 2$-element field; but its additive and multiplicative groups have different numbers of elements and are not isomorphic.