Doubt in the correctness of the proof by induction of the corollary to the Fundamental Theorem of Algebra

Question

I came across the following proof of the corollary of the Fundamental Algebra Theorem, which I shorten as follows:

"Every polynomial $p(z) = a_nz^n + a_{n-1}z^{n-1}+...+a_1z+a_0$ has a factorisation $p(z) = a_n(z-\alpha_1)(z-\alpha_2)...(z-\alpha_n)$, where the complex numbers $\alpha_1$, $\alpha_2$, ..., $\alpha_n$ are the roots of $p(z)$."

The proof follows the method of mathematical induction and takes $P(n)$ to be the statement above.

After proving that $P(1)$ is true, the proof continues with the proof that $P(k) \rightarrow P(k+1)$ is true, which is the step where my doubts arose. Note that I am not going to reproduce the proof rigorously, but only to summarise the main passages.

The statement $P(k+1)$ is considered:

$p(z) = a_{k+1}z^{k+1} + a_{k}z^{k}+...+a_1z+a_0$.

Then, by the Fundamental Theorem of Algebra $p(z)$ must have at least one complex root, say, $\alpha_{k+1}$, so that we can write:

$p(z) = (z-\alpha_{k+1})q(z)$, where $q(z)$ is a polynomial of degree $k$.

By polynomial long division of $p(z) = a_{k+1}z^{k+1} + a_{k}z^{k}+...+a_1z+a_0$ by $z-\alpha_{k+1}$ we find that the coefficient of $z^k$ in $q(z)$ is $a_{k+1}$.

Hence, by $P(k)$ the polynomial $q(z)$ must have factorisation $q(z) = a_{k+1}(z-\alpha_1)(z-\alpha_2)...(z-\alpha_k)$, but (and this is my first question)

(1) this has been claimed by replacing the coefficient $a_k$ of $z^k$ in $P(k)$ with $a_{k+1}$. Is this legitimate? If so, why? Is this because these variables simply represents coefficients and therefore it doesn't matter how we name them?

The proof continues by deducing from what above that $p(z) = (z-\alpha_{k+1})a_{k+1}(z-\alpha_1)(z-\alpha_2)...(z-\alpha_k) = a_{k+1}(z-\alpha_1)(z-\alpha_2)...(z-\alpha_k)(z-\alpha_{k+1})$.

This should show that $P(k) \rightarrow P(k+1)$ is true. But (and here is my second question):

(2) the proof started by reasoning on $P(k+1)$ (what we need to deduce) and deduced that it is true as it is a product (and therefore an algebraic manipulation) of the factorisation of $q(z)$, which is true as it is a re-statement of $P(k)$ (which is what was initially assumed to be true!). Isn't this against the deductive step, in that we should start from what we assumed to be true ($P(k)$) to arrive to what we need to deduce ($P(k+1)$)?

Hence my third and last question:

(3) If I want to prove the implication $P(k) \rightarrow P(k+1)$ starting by manipulating $P(k)$ (which in my opinion is more rigorous), I suppose I should first state $P(k)$ concisely as

$a_kz^k + a_{k-1}z^{k-1}+...+a_1z+a_0 = a_k(z-\alpha_1)(z-\alpha_2)...(z-\alpha_k)$

and then multiplying both sides by $z-\alpha_{k+1}$. The prove is true if what I get on both sides is $P(k+1)$. This seems to work as we can see by starting with the multiplication by $z$ of the left-hand side:

$a_kz^{k+1} + a_{k-1}z^{k}+...= a_k(z-\alpha_1)(z-\alpha_2)...(z-\alpha_k)(z-\alpha_{k+1})$.

Here the left-hand side resembles $P(k+1)$, but not in the coefficients. How can I fix this problem to succeed in the proof?

I hope to have been able to make you clearly understand what my queries exactly are.

Many thanks in advance

Answer 1

1. This is legitimate because we have $p(z)$ having a leading coefficient of $a_{k+1}$ and $(z-\alpha_{k+1})$ having a leading coefficient of $1$ and thus since $p(z)=(z-\alpha_{k+1})q(x)$, we find that the leading coefficient of $q(z)$ is $\frac{a_{k+1}}{1}=a_{k+1}$. Thus, by $P(k)$, the coefficient on the outside of the factorization is the leading coefficient of $q(z)$, whatever that leading coefficient is. In this case, the leading coefficient is $a_{k+1}$, so we substitute in $a_{k+1}$ where there was originally $a_k$ because that is the new leading coefficient.
2. No, this is not against the inductive step. We started with a polynomial $p(z)$, which has degree $k+1$. We then did some manipulations to get this down to a polynomial of degree $k$, which is $q(z)$. Then, we use our assumption of $P(k)$ to prove $P(k+1)$. Nowhere in our proof did we assume $P(k+1)$. We did some manipulations first in order to apply $P(k)$, but we never assumed our conclusion, so this is completely valid.
3. This won't work. You need to prove that every polynomial of degree $k+1$ can be completely factorized into linear factors. However, the polynomial in your equation is a polynomial of degree $k$ times a polynomial of degree $1$. You first have to prove that this can be done for all polynomials of $k+1$, which is why we had to show $p(z)=(z-\alpha_{k+1})q(z)$: This is us breaking the polynomial of degree $k+1$ into a polynomial of degree $k$ and a polynomial of degree $1$, which justifies your step. Thus, we need to do this initial manipulation before we can apply $P(k)$.