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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 $(K^*)^2$.2012-05-11
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    So your question is "for $K$ an algebraically closed field, can there be an epimorphism $K^*\to(K^*)^2$?". Right?2012-05-11
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    @ArturoMagidin I see that i was not clear. By $(K^*)^2$ i meant $K^∗\times K^∗$.2012-05-13
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    @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\}.$2012-05-13
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    @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.2012-05-13
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    @ArturoMagidin, Done.2012-05-14

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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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    Ok everything is correct. Thank you.2012-05-14
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    @Hector: Please accept the answer to show that the question has been answered to your satisfaction.2012-05-14
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    @TaraB, i am not sure how to do that. Sorry i am new here.2012-05-18
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    @Hector: You have done it. It is the green tick to the left of the answer.2012-05-18
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    @Hector: Yes, you've done it now. You hadn't yet when I wrote my comment.2012-05-18