5
$\begingroup$

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.

2 Answers 2

2

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.

  • 1
    @Michael: (i) Yes, exactly! That shows the *necessity* of the existence of a square root of $-1$ in $\mathbb{F}_p$ for the action to be defined. (ii) Yes, because if you have a field $\mathbf{F}$, and $\alpha\neq 0$, then the map $x\mapsto \alpha x$ is an abelian group automorphism of the additive group of $\mathbf{F}$. You can also see where the necessity breaks down for $\mathbb{F}_p^2$: for fields of prime order, *every* additive automorphism is of this form (because the additive group is cyclic), but not for fields of order $p^2$.2011-04-12
1

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 $p$.

  • 0
    @Michael: Indeed: *if* there is an element $\alpha$ whose square is $-1$, then you can turn it into a $\mathbb{Z}[i]$ module by defining $(a+bi)\cdot x = (a+b\alpha)x$ and verifying this turns it into a module. But you need to say so explicitly. For necessity, you need to show that there is no automorphism of the abelian group whose square is equal to multiplication by $-1$.2011-04-12