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Prove that $\mathbb{Q}_p$ always contains $p$ solutions $a_0,a_1,...,a_{p-1}$ to the equation $x^p-x=0$ satisfying $a_j\equiv j$ (mod p).

This is an intense problem as far as I'm concerned. I'm not even sure how I'd go about proving that $x^2-1=0$ has a solution in the reals, without appealing to some sort of geometric argument such as the intermediate value theorem.

Besides the trivial solutions: $0,1,-1$, the remaining solutions would be the roots of unity in the Euclidean completion and algebraic closure, and thus in $\mathbb{Q}_p$, the remaining solutions are going to have infinite non-repeating p-adic decimal expansions.

I decided to take a look at $\mathbb{Q}_5$ for some intuition and thus in this case the remaining solutions are the roots of the equation $x^2+1$ and in the 5-adic numbers $-1$ is representable by $.444...$ and so I'm essentially looking for some infinite non-repeating 5-adic expansion whose square is $-1$. I really have no idea how one would do this.

In conclusion I feel like this problem probably needs to be attacked at a higher level, using some piece of the theory I apparently don't understand.

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Try this: first prove it has $p$ solutions modulo $p$, then apply Hensel's Lemma. – Gerry Myerson Mar 2 '13 at 5:55
@Gerry Myerson: mod p would be by Fermat's little theorem correct? – esproff Mar 2 '13 at 6:01
Yes. ${}{}{}{}$ – Gerry Myerson Mar 2 '13 at 6:56
Well, $x^2 - 1 = (x - 1)(x + 1)$, so it's quite easy to show that it has a solution in the reals, or for that matter in any ring... – Qiaochu Yuan Mar 31 '15 at 6:02
up vote 5 down vote accepted

For proving things like this, the stronger form of Hensel’s Lemma works very well. It says this:

Let $F(x)$ be a $\mathbb Z_p$-polynomial that, when looked at modulo $p$, factors into two relatively prime polynomials: $\overline F = gh$, where $g,h\in\mathbb F_p[x]$ and $\gcd(g,h)=1$. Then there are polynomials $G,H\in\mathbb Z_p[x]$ such that $F=GH$, $\overline G=g$, $\overline H = h$, and $\deg(G)=\deg(g)$.

In your case, $x^p-x=\prod_m(x-m)$, where $m$ ranges through all $p$ elements of $\mathbb F_p$. The individual factors are pairwise relatively prime, so you can wave the magic Hensel wand and get your $p$ roots of $x^p-x$ in $\mathbb Z_p$.

In fact Hensel applies to any complete noetherian local ring, not just $\mathbb Z_p$, but that’s for a later chapter of your book.

For computational purposes, if you actually want to get a numerical root of the polynomial $x^p-x$, you can use Newton-Raphson. For this you should have a good notation for describing $p$-adic numbers. I like $p$-ary expansions extending (potentially) infinitely to the left, but there are others. You start with an approximate root, say $r_1$, then evaluate $f(r_1)/f'(r_1) = c_1$, and set $r_2=r_1-c_1$. In the case $p=5$ and $r_1=2$, you get $r_2=2-\frac{30}{79}$, whose square is already congruent to $-1$ modulo $5^3$. Like most Newton approximations, this method converges very rapidly to a square root of $-1$ in $\mathbb Z_5$.

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The way to think about such questions is not in terms of $p$-adic expansions (although that can help you get a feel for the concrete solutions) but in terms of modular arithmetic.

Having a solution over $\mathbb{Z}_p$ is the same as having a solution mod $p^k$ for any positive integer $k$. Now there is a brilliant result called Hensel's lemma which says that typically (under certain conditions) you only have to consider solving mod $p$ and then mod higher powers of $p$ you get unique "lifts" of solutions, in other words you get unique "lifts" of solutions to $\mathbb{Z}_p$.

If we can show that $x^p - x$ has exactly $p$ solutions mod $p$ then after lifting there must be exactly $p$ solutions in $\mathbb{Z}_p\subset\mathbb{Q}_p$.

Showing the above is not difficult, it is really Fermat's little theorem.

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This is a really enlightening answer, thanks for this. – esproff Mar 3 '13 at 4:40

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