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I'm having troubles of proving that if $T: {(\mathbb{C}^{*})}^{n} \rightarrow {(\mathbb{C}^{*})}^{s}$, is a homomorphism between algebraic tori, then the homomorphic image $\thinspace$$T({\mathbb{C}^{*}}^{n})$$\thinspace$ is again a torus.

Thank you!

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Since $T\colon (\Bbb C^*)^n \to (\Bbb C^*)^s$ is a morphism of algebraic tori, there are unique vectors $a_1,\ldots,a_n\in \Bbb Z^s$ such that $$ T(t) = t_1^{a_1}\cdots t_n^{a_n},$$ where for $z \in \Bbb C$ and $u=(u_1,\ldots,u_d) \in \Bbb Z^d$, we set $z^u = (z^{u_1}, \ldots, z^{u_d})$. Let $N$ be the $\Bbb Z$-span of the $a_i$'s. Since $\Bbb Z^s$ is a finitely generated torsion free abelian group, so is $N$. Hence, $N\cong \Bbb Z^d$ for some d. Let $v_1,\ldots,v_d$ be a free $\Bbb Z$-basis of $N$. Then (prove this!) $$ \operatorname{im}T = \{t_1^{v_1}\cdots t_d^{v_d} \mid t_i \in \Bbb C^*\}.$$ Define a function $F\colon (\Bbb C^*)^d \to \operatorname{im} T$ by $$ F(t) = t_1^{v_1}\cdots t_d^{v_d}.$$ This is certainly a morphism of algebraic tori, and it's onto by the above. This, all that remains to show is that it's one-one. I'll leave that to you.

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    Avi thank you for your response! I have one question though, I can understand that we can "decompose" $T$ as a product of one-parameter subgroups, that is homomorphisms of the form $\phi : \mathbb{C} \rightarrow {\mathbb{C}^{*}}^{n}$, are these $t_{i}$ the same thing? – 2017-01-17
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    @mayer_vietoris yup. – 2017-01-17
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    ok then! Think, I got this! Thank you again! – 2017-01-17
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    One more question, where does this $d$ come from? – 2017-01-17
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    @mayer_vietoris It's the rank of $N$. – 2017-01-17