Finite free graded modules and the grading of their duals

Question

Let $S$ be a $\mathbb{Z}$-graded ring and $F$ a $\mathbb{Z}$-graded module that is free of finite rank $n$. Then we can write $F = \oplus_{i=1}^n S(\nu_i)$, where $S(\nu_i)$ is a graded ring isomorphic to $S$ given by $S(\nu_i)_k = S_{k+\nu_i}$.

Question 1: the ring isomorphism $S(\nu_i) \rightarrow S$ given by $x \mapsto x$ is not graded, right? This is because the element $1$ has degree $-\nu_i$ in $S(\nu_i)$, yet it is mapped to an element of degree $0$. Is my argument correct?

Now let $\phi: F \rightarrow G$ be a morphism of graded, free, finite-rank $S$-modules. This induces a morphism of the duals $\operatorname{Hom}(G,S) \rightarrow \operatorname{Hom}(F,S)$.

Question 2: How exactly is the grading of the dual module $\operatorname{Hom}(G,S)$ defined? From what i have seen in Eisenbud, it seems that the degrees all change sign upon dualization, but i don't understand if indeed this is the case and if so, why is it that way.

Answer 1

As YACP has already said you are correct on $1$. Here's some elaboration about $2$.

Let's let $\hom(A, B)$ be ungraded homomorphisms between left $S$-modules $A$ and $B$. For any $i$ let's let $\hom^i(A, B)$ be homogeneous maps of degree $i$ from $A$ to $B$. So left $S$-module homomorphisms $\phi\colon A \to B$ satisfying $\phi(A_n) \subseteq B_{i + n}$ (where $A_n$ is the $n^\text{th}$ graded piece).

Now let $\hom^\bullet(A, B) = \bigoplus_{i \in \mathbb Z}\hom^i(A, B)$. If we declare $(\phi\cdot s)(a) = s\phi(a)$ then $\hom^\bullet(A, B)$ is a right $S$-module. Moreover, if we declare $\hom^i(A, B)$ to be the $i^\text{th}$ graded piece of $\hom^\bullet(A, B)$ then $\hom^\bullet(A, B)$ is a graded right $S$-module.

Now we naturally have $\hom^\bullet(A, B) \subseteq \hom(A, B)$ but in general these need not be equal. For example let $S = \mathbb Z$ with everything in degree $0$. Let $A = B = \mathbb Z[x]$, but we will let $x^i$ have degree $0$ in $A$ and have degree $i$ in $B$. Now the identity map of $\mathbb Z[x]$ is an element of $\hom(A, B)$, but if you try and write it as a sum of homogeneous maps you'll find you need a map in every positive degree, so it is not an element of $\hom^\bullet(A, B)$.

So $\hom^\bullet(A, B)$ is a graded right $S$-module which in general is not equal to $\hom(A, B)$. But, when $A$ is finitely generated they are equal. I'll let you figure out the proof, it's not to hard. I'll just give you the hint that the finitely many generators can only land in finitely many different degrees.

Finally, let me note that as $S$-modules we have $\hom(S, S) \simeq S$. One can check that if we choose the gradings on these $S$'s to be $\hom(S(v_i), S) \simeq S(-v_i)$ then this map is a graded map (the element $1$ on the right has degree $v_i$ and corresponds to the identity map on the left. This identity map sends $1 \in S(v_i)$ which is of degree $-v_i$ to $1 \in S$ which is of degree $0$, so it is homogeneous of degree $v_i$ as well.)