No induction. . Since is an integral domain it means or .
8.4 For any field F, F[x],
c) P(x)Q(x) = P(x)R(x) [P(x) d.n.e. 0] implies that Q(x)=R(x).
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My professor says that this is an induction problem. So I'm assuming that this is going to involve stating a particular term in LHS is equal to the same term in the RHS, up until a certain term. When multiplying two polynomials, let and represent the sum of all coefficients of the term when the two polynomials are multiplied. Basically just grouping all like powers and having one coefficient per power. Let's say E(k) is a propositional statement that " ", where is the coefficient of the respective term in the product of P(x)Q(x) and is the coefficient in the product of P(x)R(x).
So now assume that P(k) is true for all k<i. We must now show that P(i) -> P(i+1), then all terms of the two products are equal and Q(x)=R(x).
Look like a good plan?
If only I could do that! The thing is we have only defined a very few things about polynomials. We have no theorem to justify that polynomials may be factored as such. When multiplying polynomials, the only information known about the product (from this text) is:
, where for each n, .
So, this proof has to be done term by term, so to speak as I cannot justify any operations on the product of two polynomials.
Ok well apparently I missed the theorem right above this one which says that F[x], the set of all polynomials over F, satisfies all axioms for the real numbers except for the multiplicative inverse. So I can use your proof which is so much easier than doing it by induction. I still think she wants us to solve it that way, I'll be curious to see how this turns out.