# Isomorphic group to the multiplicative group of a field.

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

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 $\mathbb{C}^* \times \mathbb{C}^*$.