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I'm somewhat confused by the following question:

Suppose $p$ is an odd prime. Show that the map $\phi : \mathbb{F}_p^* \rightarrow \mathbb{F}_p^*$ (where $\mathbb{F}_p$ is $\mathbb{Z}/p\mathbb{Z}$, and $\mathbb{F}_p^*$ denotes the set of units of $\mathbb{F}_p$) given by $x \mapsto x^{(p-1)/2}$ has kernel of size at most $(p-1)/2$, and hence has image $\{\pm1\}$. Deduce that if $p \equiv 1 \ \text{mod} \ 4$ then $-1$ is a square modulo $p$

Firstly, it's obvious that $\mathbb{F}_p^* = \mathbb{F}_p \backslash \{0\}$ because $p$ is prime. But I'm confused how the kernel comes into this mapping because the set of units doesn't include the $0$ element - although I'm sure I'm just being dense here.

Thanks for any help

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You look at $\mathbb{F}_p^*$ as a multiplicative group so then the idenity is $1$ and the homomorphism is a group hom not a ring hom. –  Ryan Sullivant Mar 2 at 0:26
    
@RyanSullivant Thanks Ryan, that should have been obvious. Am I right in thinking it's fairly straightforward to deduce $|\ker \phi| \leq (p-1)/2$ because only the 'even' elements of the set can generate 1 mod p and they only account for half, (p-1)/2, of the elements of the set? –  Noble. Mar 2 at 0:29
    
Glad I could help. Hmm, I think the problem with that is that you are raising the elements to a power not multiplying. What I was just thinking is this: suppose $x$ is in the kernel of this map. Then we have $x^{(p-1)/2} = 1$ or equivalently, $x^{(p-1)/2} -1 = 0$. Now, $\mathbb{F}_p$ is a field, so this equation cannot have too many solutions ... Edit: I don't think your idea would work, for example, $1$ which is odd will be sent to $1$ under this map. –  Ryan Sullivant Mar 2 at 0:36

1 Answer 1

up vote 2 down vote accepted

One way is to note that $\ker(\phi)$ is precisely the solutions to the polynomial $X^{(p-1)/2}-1\in\mathbb{F}_p[X]$, which has at most $(p-1)/2$ solutions as a polynomial of degree $(p-1)/2$ over a field. To see that there are at least $(p-1)/2$ solutions, in order to conclude $\mathrm{im}(\phi)=\{\pm 1\}$, you can use the fact that $\mathbb{F}_p^\ast$ is cyclic, and then consider the even powers of the generator.

Then from the isomorphism theorems, $$ \mathbb{F}_p^\ast/\ker(\phi)\simeq\operatorname{im}(\phi) $$ so that $$ |\operatorname{im}(\phi)|=|\mathbb{F}_p^\ast/\ker(\phi)|=(p-1)/((p-1)/2))=2. $$ But the only subgroup of order $2$ in $\mathbb{F}_p^\ast$ is necessarily $\{\pm 1\}$, as it must contain $1$ and an element of order $2$. But $-1$ is the unique element of order $2$. To see this, note any element of order $2$ is a root of $X^2-1=(X-1)(X+1)\in\mathbb{F}_p[X]$, and $\mathbb{F}_p[X]$ is an integral domain...

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Thanks for this. Sorry, I'm not seeing how showing that there are at least $(p-1)/2$ solutions to the polynomial implies that the image is $\{\pm1\}$. Thanks again –  Noble. Mar 2 at 0:44
    
@Noble. I've added a brief note on how to see that. –  000 Mar 2 at 0:51
    
Thanks, got it! –  Noble. Mar 2 at 1:27

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