International Mathematics Competition
for University Students
2018

IMC2018: Day 1, 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.