Perron's criterion for irreducibility of polynomials

spoudyal · Poincare Conjecture Offline Joined: 29 Dec 2004 Posts: 145

Is complex analysis the non-real part of calculas?

**Posted:** Sat Apr 16, 2005 5:34 pm

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ZetaX

in its roots, complex analysis is the analysis of the functions $f$ that are 'complex derivable' = analytical (meaning that the complex differential quotient $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists and is the same independently 'in which direction' $h$ tends to $0$ in the complex numbers) on all it's points of definition.
But there are other 'definitions', since you can start looking at equivalent properties of these functions.

**Posted:** Sat Apr 16, 2005 11:16 pm

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spoudyal · Poincare Conjecture Offline Joined: 29 Dec 2004 Posts: 145

Alright. I think I understand. Probably take the class in college.

**Posted:** Mon Apr 18, 2005 9:41 am

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Robert Recorde, The whetstone of witte, London, 1557
It is confessed emongste all men, that knowe what learnyng meaneth, that besides the Mathematicalle artes, there is noe vnfallible knowledge, excepte it bee borowed of them.

DLF

Do you have a source for the above proof credited to Laurenţiu Panaitopol ?

I have a student who would like to use the proof in his Senior Paper.

**Posted:** Mon Mar 13, 2006 10:27 am

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Valentin Vornicu

**Posted:** Mon Mar 13, 2006 2:18 pm

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DLF

Thank you for your help!

**Posted:** Mon Mar 13, 2006 3:04 pm

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DLF

**Posted:** Thu Mar 16, 2006 9:25 am

ZetaX

Assume there is some factorisation $(x^m + ... + a_1 x + a_0 ) (x^n + ... + b_1 x + b_0)$ . Then the roots of one factor would all have modulus $<1$ and $\neq 0$ , thus the constant coefficient, being their product, would be an integer $c$ with $|c| <1$ and $c \neq 0$ , clearly impossible.

**Posted:** Thu Mar 16, 2006 9:41 am

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DLF

I'm not sure this completely answers the question. Your comment explains why there must be at least one root whose absolute value is larger than one.

We are wondering why having EXACTLY one root larger than one is enough to conclude the polynomial is irreducible over the rationals.

Thanks for the help!

**Posted:** Thu Mar 16, 2006 10:29 am

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ZetaX

No it shows that there must be at least two roots with modulus greater than $1$ since every factor must have one.

**Posted:** Thu Mar 16, 2006 11:20 am

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DLF

Sorry if I'm missing some important idea here.

I thought that it could be proven like this:

If P(x) has exactly 1 root whose absolute value is greater than 1 and is reducible, then it can be written as P(x)=p1(x)*p2(x). But the polynomials p1 and p2 have exactly 1 root whose absolute value is greater than 1, forcing the original P(x) to have two such roots, which would obtain the contradiction.

But then, I thought that p1 and p2 would not necessarily have integer coefficients, so they wouldn't be of the form which forced them to have exactly one such root.

What am I missing?

**Posted:** Thu Mar 16, 2006 11:38 am

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ZetaX

Gauß' Lemma: when an integer polynomial is reducible as rational polynomial, then it is reducible as integer polynomial (so rational factorisation implies integral factorisation). More exactly, the integral factors are up to multiplication of some (rational) constant the same as in the rational factorisation.

**Posted:** Thu Mar 16, 2006 11:45 am

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