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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 $K^*.$?

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The squaring map from $K^*$ to $(K^*)^2$ is onto (since $K$ is algebraically closed), and is a group homomorphism since $K^*$ is abelian. So you could take $G=K^*$ and get an epimorphism from $G$ to $(K^*)^2$. –  Arturo Magidin May 11 '12 at 20:50
    
So your question is "for $K$ an algebraically closed field, can there be an epimorphism $K^*\to(K^*)^2$?". Right? –  Daan Michiels May 11 '12 at 21:02
    
@ArturoMagidin I see that i was not clear. By $(K^*)^2$ i meant $K^∗\times K^∗$. –  Hector Pinedo May 13 '12 at 17:38
    
@DaanMichiels, yeah, i want to know if there exists an epimorphism $K^*\to K^*\times K^*,$ where $K$ is an algebraically closed field and $K^*=K\setminus\{ 0\}.$ –  Hector Pinedo May 13 '12 at 17:41
    
@Hector: Then please edit the question and write it correctly. $(K^*)^2$, in the context of fields, is naturally interpreted to be $\{a^2\mid a\in K^*\}$. If you meant $K^*\times K^*$, then you need to write it as such. –  Arturo Magidin May 13 '12 at 20:10

1 Answer 1

up vote 5 down vote accepted

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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Ok everything is correct. Thank you. –  Hector Pinedo May 14 '12 at 12:25
    
@Hector: Please accept the answer to show that the question has been answered to your satisfaction. –  Tara B May 14 '12 at 13:28
    
@TaraB, i am not sure how to do that. Sorry i am new here. –  Hector Pinedo May 18 '12 at 19:21
    
@Hector: You have done it. It is the green tick to the left of the answer. –  Chris Eagle May 18 '12 at 19:23
    
@Hector: Yes, you've done it now. You hadn't yet when I wrote my comment. –  Tara B May 18 '12 at 21:26

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