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August 1[edit]

Factoring polynomials[edit]

Given P=P(x₁, ..., x_n) and Q=Q(x₁, ..., x_n) polynomials in F[x₁, ..., x_n], where F is a field. I have P(a₁, ..., a_n)=0 iff Q(a₁, ..., a_n)=0 for any choice of values a₁, ... , a_n in F, and I want to conclude that P and Q have the same irreducible factors. This isn't valid for F = R (the reals) since a²+b² = 0 iff a=0 and b=0 iff a²+ab+b² = 0 but x²+y² and x²+xy+y² are both irreducible over R. Is it valid when F = C? --RDBury (talk) 04:31, 1 August 2019 (UTC)[reply]

I think this follows from one version of Hilbert's Nullstellensatz, but it's been a few decades since I took commutative algebra so I'm not sure of how to get the associated machinery working. It's unlikely though that that's what the author of the proof I'm trying to decipher had in mind since it was published 5 years before Hilbert's theorem. Also it was just one step in the proof and the author seemed to think it was an obvious inference not needing the authority of a theorem. Perhaps by the standards of the time it wasn't something that needed proof or perhaps it was a logical gap that Hilbert's theorem happened to fill in, or maybe there was a weaker form of the theorem that was well known at the time but is now unknown (at least by me). So slight progress but I'm just as stumped as before. --RDBury (talk) 23:42, 1 August 2019 (UTC)[reply]