Character group of diagonalizable group scheme 
Suppose $k$ is an integral domain, $\Lambda$ is a commutative group, $k[\Lambda]$ is the corresponding group ring and $\text{Diag}(\Lambda) = \text{Spec}(k[\Lambda])$ is the diagonalizable group scheme. Prove that the character group $X(\text{Diag}(\Lambda))$ of $\text{Diag}(k[\Lambda])$ is isomorphic to $\Lambda$. Here $X(G) := \text{Hom}(G, G_m)$.

I got stuck on this problem, so I would really appreciate if anyone can give me any hint.
What I tried so far:

$X(\text{Spec}(k[\Lambda])) = \text{Hom}(\text{Spec}(k[\Lambda]), G_m) \cong \text{Hom}(k[X, 1/X], k[\Lambda])$.

Any $\varphi \in \text{Hom}(k[X, 1/X], k[\Lambda])$ can be identified with $\varphi(X) \in k[\Lambda]$. So it would be done if we can prove that $\varphi(X) \in \Lambda$. I guess the fact that $\varphi(X)$ must be invertible and $k$ is a domain should be used here, but I still can't prove it.
 A: Morphisms of affine group schemes are the same thing as morphisms of their corresponding Hopf algebras (a morphism of Hopf algebras is a $k$-linear map such that it is both a morphism of algebras and of coalgebras), so the set $X(G)$ is the set $\{ \varphi \colon k[\mathbb{G}_m] \to k[\Lambda] \}$ of Hopf algebra morphisms.
The coalgebra structure on $k[\mathbb{G}_m] = k[x^{\pm 1}]$ is $\Delta(x) = x \otimes x$, and $\epsilon(x) = 1$: in other words, the element $x$ is group-like, and therefore must be sent to another group-like element by any homomorphism $\varphi$.
The coalgebra structure on $k[\Lambda] = \bigoplus_{g \in \Lambda} kg$ is $\Delta(g) = g \otimes g$ and $\epsilon(g) = 1$ for each basis element $g \in \Lambda$. So clearly the basis elements are group-like, and setting $\varphi(x) = g$ for any $g \in \Lambda$ will yield a bialgebra homomorphism. The question is: are there any other group-like elements in $k[\Lambda]$?
The answer is no: the basis of original group elements are precisely the set of group-like elements in the Hopf algebra $k[\Lambda]$. (This is fairly easy to prove, and is where you can use the fact that $k$ is a domain).
