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Plan of attack for showing irreducibility? 
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#1
Mar2405, 11:17 AM

P: 2

I'm trying to show that a polynomial of two variables is irreducible over a unique factorization domain, namely the complex numbers, but I don't know where to begin. Any help is appreciated, thanks



#2
Mar2405, 05:18 PM

P: 1,572

What's the polynomial?
My guess is that if the polynomial is of degree greater than 1, then it is not irreducible. 


#3
Mar2405, 05:20 PM

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PF Gold
P: 16,091

That's not true. For example, x^2+y.
Anyways, one thing you can try is proving C[x, y] / (f) is not an integral domain, which means (f) is not prime, and f has factors. Er, I misread  you want to prove it is irreducible. So, you just want to prove (f) is prime, which is true iff C[x, y]/(f) is an integral domain! 


#4
Mar2405, 05:28 PM

P: 1,572

Plan of attack for showing irreducibility?
Shoot.
Now is showing that C[x,y]/(f) is not an integral domain would involve having two equivalence classes of polynomials [g] and [h] whose product is the zero class which is [f]. So does that mean find g and h such that their product is a multiple of f? Oh and [g] and [h] can't be [0]=[f], so neither g nor h can be a multiple of f. 


#5
Mar2405, 05:29 PM

P: 2

the polynomial



#6
Mar2405, 05:36 PM

Emeritus
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PF Gold
P: 16,091

Note that if g and h
(i) aren't both functions of x alone and (ii) aren't both functions of y alone Then gh has x^m y^n terms with m, n > 0. 


#7
Mar2405, 06:01 PM

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P: 9,470

remark: an open dense set of all polynomials of given degree consists of irreducible ones.
a nice criterion for irreducibility of polynomials of two variables is "non singularity". i.e. if a polynomial f(x,y) has no roots in common with its two partials fx, and fy, and ifn this is also true for both alternative dehomogenizatrions of f, i.e. if it is also true at "infinity", then f is irreduible. to find another dehomogenization of f, write f(x,y) as f(u/w,v/w) and multiply by w^d where d is the degree of f. this gives the homogeneous version of f, as as homogeneoius polynomial in three variables. then to dehomogenize f again, just set either u, v, or w equal to 1. for example note that setting w equal to one gives back the original f. it would also suffice to check that the three partials of the homogenous version have no common zeros except for (0,0,0). the principle behind this criterion is "bezout's theorem". i.e. if a polynomial is reducible, then its set of zeroes has two components which must meet in the extended plane, at a point where the partials must vanish. of course this criterion is not necessary and hence often fails. since many singular polynomials are also irreducible. in general irreducibility is hard to check. example: to prove that x^3 + y^3 + 1 is irreducible, we homogeneize it as [(u/w)^3 + (v/w)^3 + 1] w^3 = u^3 + v^3 + w^3. The three partials are 3u^2, 3v^2, and 3w^2, which have only the common zero (0,0,0). 


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