International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria


Ivan is Watching You
Ivan's Office

Day 1
    Problem 1
    Problem 2
    Problem 3
    Problem 4
    Problem 5

Day 2
    Problem 6
    Problem 7
    Problem 8
    Problem 9
    Problem 10


    Day 1 questions
    Day 1 solutions
    Day 2 questions
    Day 2 solutions
    Closing Ceremony

Official IMC site

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.