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Proposition: If a polynomial $a_0+a_1z+\dots +a_nz^n=0$ for all $z\in F$, then $a_0=a_1=\dots=a_n=0$.

Where $F$ denotes the field of real or complex numbers.

Proof: Let $P_n(F)$ denote the vector space of all polynomials in $F$. We prove that $a_0,z,z^2,\dots,z^n$ is a basis of $P_n(F)$.

Clearly $a_0,z,z^2,\dots,z^n$ spans $P_n(F)$ by definition of $P_n(F)$. We just need to show that this list is linearly independent. To do this, it suffices to show that $z^j\not\in$ span$(a_0,z,\dots,z^{j-1})$.

Suppose there exist scalars $a_1,\dots,a_{j-1}\in F$ such that $z^j=a_0+a_1z+\dots+a_{j-1}a^{j-1}$. Then differentiating both sides $j$ times shows that the L.H.S is non-zero whereas the R.H.S is zero. A contradiction. Thus, the list $a_0,z,z^2,\dots,z^n$ is linearly independent.

Because the list $a_0,z,z^2,\dots,z^n$ is linearly independent, a linear combination of this equals zero if and only if $a_0=a_1\dots=a_n=0$. Hence, if a polynomial is the zero function, then all the coefficients are zero.

Is this proof correct?

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Your argument is fine, but it is unnecessarily complicated. There is no need to use linear independence.

  • if you are going to use differentiation anyway, just evaluate at zero; then differentiate, and repeat. In each iteration you get respectively that $a_0=0$, $a_1=0$, etc.

  • but there is no need to use differentiation, either. Taking $z=0$, you get $a_0=0$. Now $$ 0=a_1z+\cdots+a_nz^n=z (a_1+a_2z+\cdots+a_nz^{n-1}). $$ Since this equality holds for all $z$, we get that the polynomial inside the brackets is zero for all nonzero $z$, and hence for all $z$ by continuity. Now we can iterate the argument.

  • a third argument can be obtained using linear algebra. If we consider the $n$ equations $$ a_0+a_1\,k+a_2\,k ^2+\cdots+a_n\,k ^n=0,\qquad\qquad k=1,\ldots,n, $$ this is a homogeneous system of linear equations with matrix $[k^{j-1}]_{k,j}$. This is a Vandermonde matrix and hence its determinant is nonzero. Which means that the only solution to the system is $a_0=\cdots=a_n=0$.

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Your proof is essentially correct, but you must replace $a_0$ by $1$ because it is possible that $a_0 = 0$ in which case $a_0,z,z^2,\dots,z^n$ is not a basis. Also you should not write

Suppose there exist scalars $a_1,\dots,a_{j-1}\in F$ such that $z^j=a_0+a_1z+\dots+a_{j-1}a^{j-1}$.

  1. The $a_k$ could be confused with the given coefficients of the polynomial in your question. But this is just a question of presentation.

  2. The $a_0$ here is not the same as the $a_0$ in your "basis" $a_0,z,z^2,\dots,z^n$ (but as I said, you should use $1$ here).

  3. You must suppose that there exist $b_0,\ldots,b_n \in F$ not all zero such that $b_01 + b_1z + \ldots + b_nz^n = 0$ for all $z \in F$. Then let $j$ be the largest index such that $b_j \ne 0$. You get $$z^j = \frac{b_0}{b_j} + \frac{b_1}{b_j}z + \ldots +\frac{b_{j-1}}{b_j}z^{j-1} .$$ This allows to proceed as you did.

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I think there is a much simpler proof. Assume that $$\sum_{k=0}^n a_k z^k = 0.$$ Differentiating $n$ times, we get $n!a_n = 0$, so $a_n = 0$. Repeat this process to find that $a_i = 0$ for $i = 0,1,\dots,n$. Therefore, $f(z) = 0 + 0z + \dots + 0z^n$ is clearly the additive identity polynomial function in a field. Note that this also proves that $(1,z,z^2,\dots,z^n)$ form a basis for all polynomials of degree less than or equal to $n$, however the converse was unnecessary.

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The fundamental theorem of algebra is often stated as: an $n$-th degree polynomial over $\Bbb C$ has exactly $n$-roots.

The easy direction is that it can have at most $n$. That's also called the factor theorem: if $p(a)=0$, then $(z-a)\mid p(z)$.

Now to your problem: if a polynomial equals zero, then in particular it has more than $n$ roots. This implies that it's the zero polynomial (by the factor theorem).

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