Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
Aternative hint: Do you know any Galois theory? –  David Speyer Feb 25 '13 at 20:39
4  
Eisenstein's criterion + minimal knowledge of irreducible polynomial is enough to prove this. –  achille hui Feb 25 '13 at 21:01
1  
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
show 1 more comment

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.

share|improve this answer
2  
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
add comment

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.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
That said, I consider achille hui's solution to be the most elegant. –  David Speyer Feb 26 '13 at 0:23
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.