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I am asked this question:

Prove that $x^5 - 5x + 1$ has no double roots in $\mathbb C[x]$.

Now here's what I said:

$p(x) \in \mathbb C[x]$ has no double roots if and only if $gcd(p,p') = 1$. Now we need to prove that. So: $p(x) = x^5 - 5x + 1$ and $p'(x) = 5x^4 - 5$ and I start doing the Euclidean algorithm and I got to a point where I have to do $\frac{-4x+1}{ \frac{5}{4}x^3-5 }$, which we obviously can not do.

What do we do in such a situation? What does that mean that while doing the Euclidean algorithm we can not continue dividing?

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Hint: what does it mean when you do the Euclidean algorithm with integer rather than polynomial division and you get to the point where, as you say, "we can not continue dividing"? –  symplectomorphic May 4 '13 at 10:53
    
That they do not have a common divisor but 1? –  TheNotMe May 4 '13 at 10:56
    
At each step you should be dividing some polynomial $a$ by another one $b$ of smaller degree than $a$. Just as in the integer Euclidean algorithm at esch step you're dividing a number by some smaller number at each step. And then you update the two terms at each step, if p div q gives remainder r then next step working with q div r. –  coffeemath May 4 '13 at 11:07
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2 Answers 2

up vote 2 down vote accepted

You are almost finished. $\gcd(-4x+1, \frac{5}{4} x^3-5)=1$

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self facepalm I forgot that the Euclidean algorithm is recursive! Thank you so much. –  TheNotMe May 4 '13 at 11:02
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In the Euclidean algorithm, you divide the larger by the smaller. A cubic polynomial is larger than a linear polynomial.

Computing gcds by finding the prime factorization is often a feasible alternative. In this setting, prime factorization means "factor", and is almost synonymous with finding its roots.

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So what you are saying is that I should ignore $\frac{-4x+1}{\frac{5}{4}x^3 -5}$ and do $\frac{\frac{5}{4}x^3 - 5}{-4x+1}$ instead? –  TheNotMe May 4 '13 at 11:09
    
@TheNotMe: Yep. (assuming I correctly infer what you intend by "doing" a fraction) –  Hurkyl May 4 '13 at 11:11
    
Haha, to divide I mean. –  TheNotMe May 4 '13 at 11:11
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