Character of $G_m$

Question

I am reading a text about algebraic group, which claims that character of $G_m$ take the forms :$x\to x^r$. Then I am confused. I guess this question is very stupid, if it is please forgive me. I appreciate any help!

Answer 1

By definition, $\Bbb G_m := \operatorname{Spec}\Bbb Z[x,x^{-1}]$ (if you only want to think of $\Bbb G_m$ over a field, replace $\Bbb Z$ with your favorite field $k$; the details will essentially remain the same). A character is a morphism (of group schemes) $\Bbb G_m\to\Bbb G_m$, which is equivalent to a morphism of $\Bbb Z$-algebras $\Bbb Z[x,x^{-1}]\to\Bbb Z[x,x^{-1}]$ which is compatible with the group structure of $\Bbb G_m$ (equivalently in our dual point of view, the Hopf algebra structure on $\Bbb Z[x,x^{-1}]$). Because the image of $\Bbb Z$ must be $\Bbb Z$ (isomorphically), the only choice we can make is where to send $x$, and we are restricted in that we must map $x$ to a unit in (the codomain) $\Bbb Z[x,x^{-1}]$, as $x$ is invertible in (the domain) $\Bbb Z[x,x^{-1}]$. $\Bbb Z[x,x^{-1}]^\times = \{\delta x^n\mid \delta = \pm 1, n\in\Bbb Z\}$ So a character must be a homomorphism of $\Bbb Z$-algebras of the form \begin{align*} \chi_{n,\delta} : \Bbb Z[x,x^{-1}]&\to\Bbb Z[x,x^{-1}]\\ x&\mapsto \delta x^n \end{align*} for $\delta = \pm 1$ and $n\in\Bbb Z$. However, we need to check which of these maps $\chi_{n,\delta}$ satisfy the relations $\mu\circ \chi_{n,\delta} = (\chi_{n,\delta}\otimes\chi_{n,\delta})\circ\mu$, $\epsilon\circ\chi_{n,\delta} = \epsilon$, and $\iota\circ\chi_{n,\delta} = \chi_{n,\delta}\circ\iota$ (the requirements for a map of algebras to be a map of Hopf algebras), where \begin{align*} \mu : \Bbb Z[x,x^{-1}]&\to\Bbb Z[x,x^{-1}]\otimes_{\Bbb Z}\Bbb Z[x,x^{-1}]\\ x&\mapsto x\otimes x\\ \epsilon : \Bbb Z[x,x^{-1}]&\to\Bbb Z\\ x&\mapsto 1\\ \iota : \Bbb Z[x,x^{-1}]&\to\Bbb Z[x,x^{-1}]\\ x&\mapsto x^{-1}. \end{align*} As all of these are maps of $\Bbb Z$-algebras, it suffices to check that they satisfy the conditions on the element $x$. Looking at the second requirement, we have \begin{align*} \epsilon\circ\chi_{n,\delta}(x)&= \epsilon(\delta x^n)\\ &=\delta\epsilon(x^n)\\ &=\delta\epsilon(x)^n\\ &=\delta\\ \epsilon(x) &= 1, \end{align*} which forces $\delta = 1$. I will leave it to you to check that the $\chi_n = \chi_{n,1}$ all satisfy the rest of the axioms for maps of Hopf algebras, from which it follows that all characters of $\Bbb G_m$ are induced by maps of the form $\chi_n : \Bbb Z[x,x^{-1}]\to\Bbb Z[x,x^{-1}]$ (with $n\in\Bbb Z$) as above, which when looking at points over your favorite field $k$ means the induced maps are of the form \begin{align*} \chi_n(k) : \Bbb G_m(k)\cong k^\times&\to \Bbb G_m(k)\cong k^\times\\ \alpha &\mapsto \alpha^n, \end{align*} if you trace through the identification of $\Bbb G_m(k)$ with $k^\times$ and how a map $\chi_n : \Bbb G_m\to\Bbb G_m$ induces a map $\chi_n(k) : \Bbb G_m(k)\to\Bbb G_m(k)$.