Verification of a proof about when a polynomial is zero

Question

I was doing an exercise about algebraic geometry and, the final step is to prove that if the polynomial in $k[t]$ ($k$ a field) $\sum_{i=0}^na_i(1-t^2)^i(1+t^2)^{n-i}=0$ then, the coefficients $a_i$ must all be zero.

It's true for $n=0,1,2$, I tested it while trying to write a proof.

Let's assume that it's true for $n-1$. That is, if the polynomial of degree $n-1$ is zero, then all its coefficients are of degree zero.

One can construct the polynomial of degree $n$ like this: $p_{n-1}\cdot(1+t^2)+a_n(1+t^2)^n$.

If that is zero, then we can write $p_{n-1}\cdot(1+t^2)=-a_n(1+t^2)^n$.

Now the confusing part. I have to deduce that all coefficients are zero. If I put $t=0$, then $p_{n-1}=-a_n$. But $p_{n-1}$ was zero, so $a_n$ is zero, and therefore all of them are zero.

Is this correct? It's the final step of a problem and induction is not among my most effective proof tricks...

What were the previous questions? This might help to solve this one, which you say is the last one of a problem. — mathcounterexamples.net, Oct 12 '22 at 15:03
If $\operatorname{char}k=2$, $1-t^2=1+t^2$ and there exist a lot of easy counterexamples. — Acrobatic, Oct 12 '22 at 16:12
Your argument is not correct. Because we were given $p_n=0$ and we need to use if $p_{n-1}=0$ all $a_i, i\geq n-1$ are zeros. If you want to use induction first you need to show if $p_n=0$ then $a_n=0$ so we can use our induction statement — IrbidMath, Oct 12 '22 at 19:04

Jean Marie · Answer 1 · 2022-10-12T18:15:57.677

In the case $k=\mathbb{R}$, we can set $t=\tan x$.

Using classical trigonometric identities:

$$1+\tan^2 x = \frac{1}{\cos^2 x} \ \ \text{and} \ \ 1-\tan^2 x = \frac{\cos (2x)}{\cos^2 x},$$

relationship:

$$\forall t, \ \ \ \sum_{i=0}^n a_i(1-t^2)^i(1+t^2)^{n-i}=0$$

is equivalent to

$$\forall x, \ \ \ \frac{1}{(\cos^2 x)^n} \sum_{i=0}^na_i(\cos 2x)^{i}=0$$

which amounts, up to an evident change of variables, to:

$$\forall x, \ \ \ \sum_{i=0}^na_i(\cos x)^{i}=0$$

which is possible if and only if the all the $a_i$ are zero because functions defined by $f_i(x)=(\cos x)^{i}$ are linearly independent (see for example here).

Verification of a proof about when a polynomial is zero

1 Answers1