Proof of associates in Zp[x]

chaotixmonjuish

Show that a nonzero polynomial in Zp[x] has p-1 associates.

I don't have a proof as much as a fairly weak (in my opinion) arguement.

Suppose you had a series of functions with coefficients p. If the coeffients are p in Zp[x], all those functions go to zero. In this case you just have a bunch of zeroes, which can't happen since there is no zero associate in Zp[x]. I think I may have butchered this proof as we just went over associates and I'm not sure how to do this.

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Hurkyl

Staff Emeritus
Gold Member
I'm not familiar with the term; what's an "associate"?

chaotixmonjuish

An associate is a nonzero constant that a polynomial is multiplied by

e.q 2x+2 is 2(x+1)...2 is the associate

Hurkyl

Staff Emeritus
Gold Member
An associate is a nonzero constant that a polynomial is multiplied by

e.q 2x+2 is 2(x+1)...2 is the associate
Er... so "an associate of a polynomial" is simply "the leading coefficient of that polynomial"? But any polynomial would have only one associate then.

chaotixmonjuish

An associate is a nonzero constant multiple of a polynomial.

Hurkyl

Staff Emeritus
Gold Member
An associate is a nonzero constant multiple of a polynomial.
Ah, okay. Well... since each associate is determined by a "nonzero constant", we can simply count those, can't we? Well, we do have to make sure that the same associate isn't counted multiple times by this method. (Or make a correction if that does happen)

chaotixmonjuish

So here is what I got:

Suppose yyou have a polynomial x^n....+x+c such that p is an associate:

p*(x^n+....+x+c) => px^n+...+p*x+p*c => 0+....0+0

This contradicts the statement.

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