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Unit in a ring (abstract algebra)
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[QUOTE="hapefish, post: 4287120, member: 464225"] it has been a long time since I studied Ring Theory, but here is what I remember that might be relevant: A unit is an element that has an inverse. So in order for ##x^2-1## to be a unit, there would have to exist an inverse of ##x^2-1## in your field. Your suggestion of ##\frac{1}{x^2-1}## is a reasonable candidate, but I do not believe it is an element of your field. This is because I always took ##F[x]## to represent the ring of finite degree polynomials over the field, F, and the Taylor expansion of ##\frac{1}{x^2-1}## is infinite. In my opinion, the answer needs to be "no." I think proving this hinges on the fact that we're in a field (and hence an integral domain) so there is no terms that can multiply by ##x^2## to make the leading coefficient zero. Good Luck! [/QUOTE]
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Unit in a ring (abstract algebra)
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