1. ## Induction proof about polynomials

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 $c_n$ and $d_n$ represent the sum of all coefficients of the $x^n$ 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 " $c_n x^n = d_n x^n$", where $c_n$ is the coefficient of the respective term in the product of P(x)Q(x) and $d_n$ 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?

2. No induction. $p(x)q(x) = p(x)r(x) \implies p(x)[q(x) - r(x)] = 0$. Since $F[x]$ is an integral domain it means $p(x)= 0$ or $q(x) - r(x)\equiv 0 \implies q(x) = r(x)$.

3. Originally Posted by ThePerfectHacker
No induction. $p(x)q(x) = p(x)r(x) \implies p(x)[q(x) - r(x)] = 0$. Since $F[x]$ is an integral domain it means $p(x)= 0$ or $q(x) - r(x)\equiv 0 \implies q(x) = r(x)$.
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:

$P(x)Q(x) = c_0 + c_1 x + c_2 x^2 ... + c_n x^n + ...$, where for each n, $c_n = \sum_{i+j=n} a_i j_i$.

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.

4. Originally Posted by Jameson
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:

$P(x)Q(x) = c_0 + c_1 x + c_2 x^2 ... + c_n x^n + ...$, where for each n, $c_n = \sum_{i+j=n} a_i j_i$.

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.
In your proof invent a lemma. And the lemma would be the distributive law. Maybe you can prove the distributive law using induction (rather than just expanding everything out).

5. Yes that certainly would work. I think that lemma might be more difficult than the original proof though. I'll see how induction goes.

6. Originally Posted by Jameson
Yes that certainly would work. I think that lemma might be more difficult than the original proof though. I'll see how induction goes.
You do not need to use induction on this lemma. You can do it by definition of polynomial multiplication.

7. 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.