Suppose that I have an abelian group $G$ and an epimorphism from $G$ to $(K^*)\times (K^*), $ where $K$ is an algebraically closed field. Is it true that $G$ cannot be isomorpic to $K^*.$?
Isomorphic group to the multiplicative group of a field.
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abstract-algebra
group-theory
field-theory
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0@ArturoMagidin, Done. – 2012-05-14
1 Answers
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The multiplicative group $\mathbb{C}^*$ is isomorphic to the product of additive groups $\mathbb{R}/\mathbb{Z} \times \mathbb{R}$, the first component being the argument over $2 \pi$ and the second being the log of the modulus. As additive groups, $\mathbb{R}$ is isomorphic to $\mathbb{R}^4$, since they're both continuum-dimensional rational vector spaces. There's clearly an epimorphism from $\mathbb{R}^4$ onto $(\mathbb{R}/\mathbb{Z} \times \mathbb{R})\times(\mathbb{R}/\mathbb{Z} \times \mathbb{R})$, and hence an epimorphism from $\mathbb{C}^*$ to $\mathbb{C}^* \times \mathbb{C}^*$.
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0@Hector: Yes, you've done it now. You hadn't yet when I wrote my comment. – 2012-05-18