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Assume $a$, $b$ are distinct roots of the following equation: $$1+x+\frac{x^2}{2!}+\cdots+\frac{x^p}{p!}=0,$$ where $p$ is a prime number and $p \gt 2$. How to prove that $ab$, $a+b$, $a-b$ are not rational numbers.

Thanks in advance

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However, we do know that the sum and product of all the roots are rational because the coefficients of the polynomial are rational. –  marty cohen Feb 25 '13 at 19:47
Do you know anything about $p$-adic numbers? –  David Speyer Feb 25 '13 at 20:18
Aternative hint: Do you know any Galois theory? –  David Speyer Feb 25 '13 at 20:39
Eisenstein's criterion + minimal knowledge of irreducible polynomial is enough to prove this. –  achille hui Feb 25 '13 at 21:01
More Hint: Let $f_p(x) = \sum_{k=0}^p\frac{x^k}{k!}$. If $a + b = r \in \mathbb{Q}$, what can you say between $f_p(x)$ and $f_p(r - x)$? –  achille hui Feb 25 '13 at 21:14

3 Answers 3

up vote 7 down vote accepted

This is an expansion of @achille hui's hint in the comments. I just show that $a+b$ cannot be rational; the other cases are simple modifications to this approach. I must thank @achille hui again for a helpful correction to this solution.

Multiply the polynomial through by $p!$ to get a monic polynomial. Note that $p$ divides each term after the leading $x^n$ term and that $p^2$ does not divide the constant term. This means we can apply Eisenstein's criterion to show the polynomial (call it $f(x)$) is irreducible.

Suppose $a$ and $b$ are roots of the polynomial $a+b=r\in\mathbb Q$. Note that $f(r-x)$ is irreducible, because $x\rightarrow r-x$ is an isomorphism of $\mathbb Q[x]$. Because $f$ is irreducible and has $a$ as a root, it is the minimal polynomial for $a$. Because $f(r-a)=f(b)=0$, $a$ is a root of $f(r-x)$, so $f(x)$ must divide $f(r-x)$. In a similar fashion, we see that $f(r-x)$ must divide $f$, so $f(x)=f(r-x)$ up to a unit. By comparing leading coefficients, we see $f(x)=-f(x-r)$. Since the degree of $f(x)$ is odd and all roots are distinct, there must be a root $c$ of $f(x)$ with $c=r-c$. Hence $c=\frac{r}{2} \in \mathbb Q$, a contradiction, as the irreducibility of $f$ over $\mathbb Q$ implies in particular that it has no rational roots.

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Nope, $a|b$ and $b|a$ only allows one to conclude $a = b$ up to a unit. By compare leading coefficients, you can conclude $f(x) = -f(r-x)$. Since the degree of $f(x)$ is odd and all roots of $f(x)$ are distinct, there must be a root of $c$ of $f(x)$ such that $c = r - c$. This leads to $c = \frac{r}{2} \in \mathbb{Q}$, a contradiction! –  achille hui Feb 25 '13 at 21:43
@achillehui Of course! Thank you for the correction. I have included it. –  Potato Feb 25 '13 at 21:47

Not a proof of the statement, but some background information. It was proved by Schur that the polynomial made from the first $n$ terms of the exponential series has full Galois group, $S_n$, for $n$ not divisible by $4$. That says there are no polynomial relations (with rational coefficients) between roots except those that are derived from symmetric function identities.

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Where can one find a proof of this (preferably a modern treatment)? –  Potato Feb 25 '13 at 21:47
The original paper appears to be in German. Do you know of an English reference? –  Potato Feb 25 '13 at 21:54
None offhand but see here for a proof the group is at least A_n : mathoverflow.net/questions/44844/… –  zyx Feb 25 '13 at 21:56

The two proofs that I was thinking of:

$p$-adic: The Newton polygon of this polynomial is a line segment connecting $(0,0)$ and $(p, -1)$. So the roots, in the algebraic closure of $\mathbb{Q}_p$, have valuation $-1/p$. That means any product of them have valuation $-2/p$. Since $2/p$ is not an integer, the product of the roots is not in $\mathbb{Q}$.

A slightly harder proof works for the sum. Let $K$ be the ramified extension $\mathbb{Q}_p(x)/(1+x+x^2/2+\cdots+x^p/p!)$ of $\mathbb{Q}_p$, and let $\pi$ be a uniformizer in $K$. I get that all of the roots of the polynomial are of the form $\pi^{-1} + \cdots$ where the $\cdots$ is in $\mathcal{O}$. So any $2$ of those roots add up to something of the form $2/\pi + \cdots$, with valuation $-1/p$, and thus isn't rational.

I didn't completely think through the case where we consider the difference of the roots.

Galois proof: Let $r_1$, $r_2$, ..., $r_p$ be the roots of this polynomial. Since the polynomial is irreducible, the Galois action on the $r_i$ is transitive. So we have $G \subseteq S_p$ and $p | \#(G)$. By the first Sylow theorem, $G$ contains a $p$-cycle. Renumber the $r_i$ so that $G$ contains $r_1 \to r_2 \to \cdots \to r_p \to r_1$. Suppose that $r_i+r_{i+a}$ is rational. Then $r_i+r_{i+a} = r_{i+a} + r_{i+2a}$ so $r_i = r_{i+2a}$. Then the Galois action, again, gives $r_i = r_{i+2a} = r_{i+4a} = \cdots $ so all the $r_i$ are equal, which is obviously wrong.

Same proof works in the product case. In the difference case, if $r_i = r_{i+a} + b$ for some rational $b$, then the Galois symmetry gives $\sum r_i = \sum r_i + pb$, so $b=0$ and, again, all the roots are equal.

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That said, I consider achille hui's solution to be the most elegant. –  David Speyer Feb 26 '13 at 0:23

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