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I want to prove the following: Let $V=\mathbb F _{n}[x]$.

For $p(x)\in V$, we define $f_i(p(x))= p^{(i)}(1)$, where $f_i \in V^*$ for $i =0,...,n$.

Proof that $\mathcal B=(f_0,...,f_n)$ is a basis of $V^*$ if and only if $char(\mathbb F)=0$ or $p>n$, and find the basis of $V$ such that $\mathcal B$ is its dual basis.

I really can't work out how to introduce the characteristic of $\mathbb F$ in the proof. As far as I know, the characteristic of a field is something like the smallest natural number $n$ such that $n×1=0$, but i can't figure out how to arrive to that conclusion in my proof.

As for finding a basis,so that $(f_0,...,f_n)$ is its dual, if I let the basis be $\{ p_0(x),p_1(x),...,p_n(x) \}$, where I define $p_i(x)=\sum_{k=0}^{n}\lambda_{k,i}x^k$, and taking into account that $f_i(p_j(x))= \delta _{ij}$. I get a system of a lot of equations that can be expressed as a matrix system like this: $$ \pmatrix{ \lambda_{0,0} & \lambda_{1,0} & \cdots & \lambda_{n,0} \\ \lambda_{0,1} & \lambda_{1,1} & \cdots & \lambda_{n,1}\\ \vdots &&&\\ \lambda_{0,n} & \lambda_{1,n} & \cdots & \lambda_{n,n}} \cdot \pmatrix{1&0&0& \cdots&0\\ 1&1&0&\cdots&0\\ 1&2&2&\cdots&0\\ 1&3&6&\cdots&0\\ \vdots&\vdots&\vdots&&0\\ 1&n&n(n-1)&\cdots&n!}= \pmatrix{1&0&\cdots&0\\ 0&1&&\\ \vdots&&\ddots&\vdots\\ 0&\cdots&&1}$$ So basically the solutions I want are the coefficients of the polynomials that form the basis I'm looking for, and to find them I would just have to compute the inverse of the second matrix but I dont really know how to do it. Can someone help me with that, or tell me if there is any easier way of solving the problem?

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If the characteristic of $\mathbb{F}$ is $q$, with $0<q\le n$ (using $q$ not to confuse it with $p$ for polynomials), then $x^q$ has all derivatives equal to $0$.

If $\{\xi_0,\xi_1,\dots,\xi_n\}$ is any basis for $V^*$, then, for every $v\in V$, if $v\ne0$ there exists $0\le i\le n$ such that $\xi_i(v)\ne0$.

In order to show the given set is a basis, it's sufficient to show it is linearly independent.

Suppose $c_0f_0+c_1f_1+\dots+c_nf_n=0$. In particular, $$ (c_0f_0+c_1f_1+\dots+c_nf_n)(1)=0 $$ implying $c_0=0$. Then also $$ (c_1f_1+\dots+c_nf_n)(x)=0 $$ implying $c_1=0$. Proceed by induction, using the fact that no derivative of $x^k$ is the zero polynomial provided the characteristic is $0$ or is $>n$.

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  • $\begingroup$ Just to know if i am getting it right. . so for each step in the induction, you are taking into account the result from the previous step? Because when evaluating 1 to the linear combination and you get that c0=0; then in the next step of evaluating x, you are using that c0=0 to conclude c1 must equal 0 and so on. Is that right? $\endgroup$ – L. Sandoval Mar 18 '18 at 22:40
  • $\begingroup$ @L.Sandoval More or less. You can prove that the $k$-th derivative of $x^k$ is $k!$, which is different from $0$ under the stated conditions and assume by induction that $c_i=0$ for $0\le i<k$. $\endgroup$ – egreg Mar 18 '18 at 22:47

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