International Mathematics Competition for University Students

July 22 – 28 2018, Blagoevgrad, Bulgaria

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

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

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