Polynomial P equals to 0

I have simple questions about polynomials (when I say "polynomials", I mean "formal polynomials", not polynomial mappings).

It might be a little bit strange, but I don't really understand why if we have a polynomial $$P = a_{0} + a_{1}X + a_{2}X^{2} + ... + a_{n}X^{n} \in \mathbb{R}[X]$$ (where $$n \in \mathbb{N}^{\ast}$$), then : $$P = 0_{\mathbb{R}[X]} \Rightarrow \forall i \in \{1, ..., n\}, a_{i} = 0$$

Also, if I write $$P = a_{0} + a_{1}X + a_{2}'X^{2} + a_{2}X^{2} + ... + a_{n}X^{n}$$, can I conclude that : $$P = 0_{\mathbb{R}[X]} \Rightarrow \forall i \in \{1, ..., n\}, a_{i} = 0 \wedge a_{2}' = 0$$ or just that : $$P = 0_{\mathbb{R}[X]} \Rightarrow \forall i \in \{1, ..., n\} - \{2\}, a_{i} = 0 \wedge a_{2} + a_{2}' = 0?$$

Thank you for your answers.

• A "formal polynomial" still has only one coefficient for each $x^i.$ So the second case is true if $a_2+a_2'=0$ and $a_i=0$ for $i\neq 2.$ – Thomas Andrews Mar 19 at 15:45
• If you write your $P$ as $a_{0} + a_{1}X +( a_{2}' + a_{2})X^{2} + ... + a_{n}X^{n}$, the question is settled. – Yves Daoust Mar 19 at 15:48
• $0_{\mathbb{R}[X]}$ is the polynomial where each coefficient is equal to $0$ here. Thank you for your answers, I just wanted to be sure ! – deeppinkwater Mar 19 at 15:51

2 Answers

The polynomial $$P$$ has at most $$n$$ roots, unless it is equal to the $$0_{\Bbb{R[X]}}$$ . The last by definition means $$a_i=0$$. That $$P$$ has infinite number of roots, we conclude that $$a_i=0$$ .

About the second question we may conclude that $$a_2+a'_2=0$$ .

Do you know when are two polynomials the same? Say we have $$p(x)=a_nx^n+...+a_2x^2+a_1x+a_0$$ and

$$q(x)=b_mx^m+...+b_2x^2+b_1x+b_0$$

then $$p(x)= q(x)$$ iff $$m=n$$ and $$a_0=b_0;$$ $$a_1=b_1;$$ $$a_2=b_2;$$ $$\vdots$$ $$a_n=b_n;$$

So if $$q(x)=0$$ then $$a_0=a_1=a_2=...=0$$ and you can not say $$a_2=a_2'=0$$. You can and must say $$a_2+a_2'=0$$.