For which prime $p$ are the additive groups $\mathbb{F}_p$ and $\mathbb{F}_p^2$ $\mathbb{Z}[i]$-modules? Homework for my algebra class. Chapter 14, Exercise 7.8 in Artin's Algebra, Second Edition:

Let $F = \mathbb{F}_p$. For which
  prime integers $p$ does the additive
  group $F^1$ have a structure of a
  $\mathbb{Z}[i]$-module? How about
  $F^2$?

I am submitting my possible solution as an answer; but I'm not sure if it's correct. Can I get a second opinion? Thank you.
P.S. If one of the regulars could tell me if questions like these are allowed (checking solutions), or how best to ask these kinds of questions, I'd be much obliged.
 A: You are close, as discussed in the comments.
Claim. An abelian group $A$ has a structure as a $\mathbb{Z}[i]$ module if and only if there exists $\phi\in\mathrm{Aut}(A)$ such that $\phi^2(a) = -a$ for all $a\in A$.
Proof. Suppose that $A$ has a $\mathbb{Z}[i]$-module structure. Define $\phi\colon A\to A$ by $\phi(a) = i\cdot a$. Then $\phi(a+b) = i\cdot(a+b) = (i\cdot a)+(i\cdot b) = \phi(a)+\phi(b)$, so $\phi$ is an endomorphism. Since $\phi^4(a) = a$ for all $a\in A$, it follows that $\phi$ is invertible as well, hence an automorphism. Also, $\phi^2(a) = i\cdot(i\cdot a) = (ii)\cdot a = (-1)\cdot a = -a$. Thus, if $A$ has a structure as a $\mathbb{Z}[i]$-module, then there exists $\phi\in\mathrm{Aut}(A)$ such that $\phi^2(a)=-a$.
Conversely, suppose there exists $\phi\in\mathrm{Aut}(A)$ such that $\phi^2(a)=-a$ for all $a\in A$. Define the action by $(r+si)\cdot a = (r+s\phi)(a) = ra+s\phi(a)$. which is easily verified to be a module structure. QED
From this it easily follows that if $-1$ is a square in $\mathbb{F}_p$, then we can make $\mathbb{F}_p$ into a $\mathbb{Z}[i]$ module by considering the automorphism of $\mathbb{F}_p$ (as an abelian group) given by multiplication by $\alpha$, where $\alpha^2 = -1$. It also shows, via your argument, that $\mathbb{F}_p^2$ is a $\mathbb{Z}[i]$ module, by considering $\phi(a,b) = (-b,a)$. 
You still have to show the necessity for $\mathbb{F}_p$, though. 
Hint: See what $i\cdot 1$ is.
A: For $F = \mathbb{F}_p$ to be a $\mathbb{Z}[i]$-module, the scalar multiplication by elements of the form $a + 0i$ is well defined. This leaves the element $i$. For $i$, note that there must be some action such that for all $a \in F$, $-a = i^2 \cdot a = i \cdot ( i \cdot a)$. Therefore $F$ must have an element whose square is $-1$. This is true precisely where $x^2+1$ is reducible, i.e. when $\mathbb{Z}/(p,x^2+1) = \mathbb{Z}[i]/p$ is not a field. So $F$ is a $\mathbb{Z}[i]$-module precisely when $p$ is reducible in $\mathbb{Z}[i]$ (equivalent to when $p$ is not a Gaussian prime, and also equivalent to when $p \equiv 1$ modulo 4).
For $F^2 = {\mathbb{F}_p}^2$, use the notation for elements $(a,b)$, where $a$ and $b$ are in $F$. Again, only need to find an action of $i$ such that $i^2 \cdot (a,b) = (-a,-b)$. Let $i \cdot (a,b) = (-b,a)$. This fulfills the necessary condition. So $F^2$ is a $\mathbb{Z}[i]$-module for all $p$.
