I am trying to prove the following from Tadao Oda's "Convex bodies and algebraic geometry"
Let $N \cong \mathbb Z^n$ be a free $\mathbb Z$ - module of rank $n$ and $M = \text{Hom}_{\mathbb Z}(N,\mathbb Z)$ be its dual and $T_N = \text{Hom}_{\mathbb Z}(M, \mathbb C^{\times})$ where $\mathbb C^{\times}$ is the multiplicative group of non zero complex numbers.
(#) Prove that the set of $\mathbb C$ - algebra homomorphisms from $\mathbb C [M]$ to $\mathbb C$ coincides with $T_N$ where $\mathbb C[M]$ is the group ring of $M$ over $\mathbb C$.
I have seen the definition of group rings where the group is finite (Dummit Foote). Does the same definition hold for an infinite group? I think understanding the structure of $\mathbb C [M]$ is important to proving (#) because my intuition (given below) doesn't seem to be leading me in the right direction.
Now if $m_1, \cdots , m_n$ is a $\mathbb Z$ basis of $M$ then is it true that $\mathbb C [M] = \mathbb C [m_1, \cdots , m_n]$?
If 1 is true then since $m_1, \cdots , m_n$ are independent can I say that $\mathbb C[M]$ is a polynomial ring in $n$ variables?
If 2 is true then $\text{Hom}_{\mathbb C - \text{alg}}(\mathbb C [m_1, \cdots, m_n],\mathbb C) \cong \mathbb C^n$ but $T_N \cong (\mathbb C ^{\times})^n$ which goes againt (#).
So what am I missing?
Thanks!
$\mathbb{C}[M]$ will be Laurent polynomials in $x_1,\dots,x_n$. Algebra homomorphisms to $\mathbb{C}$ are thus determined by sending each $x_i$ to an element of $\mathbb{C}^\times$ and therefore the set of all such homomorphisms is associated with $(\mathbb{C}^\times)^n$
More generally, the group algebra functor is left adjoint to the functor taking an algebra to it's group of units. The problem you are given is to prove a special case of this property.