You are on the right track when you invoke the fact that any polynomial of degree $n$ over a field $\mathbb{K}$ has at most $n$ roots. Call the map given $f$ and suppose that $p \in \mathbb{K}[x]_{\leq n}$ is such that $f(p) = (0, 0, ..., 0)$. In other words, $p(\alpha_i) = 0$ for all $i = 0, 1, ..., n$. In other words, $p$ has $n+1$ distinct roots. Is this possible if $p$ is a non-zero polynomial?

For surjectivity, consider the case of $(1, 0, 0,..., 0)$. Can you find a polynomial $p$ of degree $n$ such that $p(\alpha_0) = 1$ and $p(\alpha_i) = 0$ for all $i = 1, ..., n$? Once you have found such a polynomial $p$, can you extend it to other vectors of the form $(0,0,..,0,1,0,...0)$? Note that these vectors form a basis for $\mathbb{K}^{n+1}$, how does the linearity of $f$ help you then?

@HoWeiHaw Thank you! For the injective part, I would conclude it's not possible, I mean it's a contradiction. $p$ cannot have at most $n$ roots and $n+1$ distinct roots at the same time. Though I am missing the bridge that connects this fact, to the fact that only the zero polynomial lies in the kernel, if we require $p$ to have $n$ and $n+1$ roots at the same time...

@HoWeiHaw For surjectivity if I understood well, I can for example find $p = (x-1)x^n$ for the first idea, and ... like $p = x^{n-k} \cdot (x-1) \cdot x^{k}$? I will find a basis for $\mathbb{K}^{n+1}$ hence the image vectors fulfill $\mathbb{K}^{n+1}$ hence we have the surjectivity. The linearity though I don't know how it may help

We know that any nonzero polynomial $p \in \mathbb{K}[x]_{\leq n}$ has at most $\deg(p)$ roots where $0 \leq \deg(p) \leq n$. So, any nonzero polynomial $p$ cannot be mapped to the vector $(0,0, ..., 0)$ as this would mean that it has at least $n+1$ distinct roots, which is a contradiction. Hence, the only polynomial in $\mathbb{K}_{\leq n}$ that can be mapped to the vector $(0, 0, ..., 0)$ is?

For the surjectivity part, $p = (x-1)x^n$ would not work for 2 reasons. Firstly, $p$ has degree $n + 1$ here and so $p \notin \mathbb{K}[x]_{\leq n}$. Secondly, $p(\alpha_0) = (\alpha^n - 1)\alpha_0^n$ which need not be $1$ (the same reasoning applies for the other $\alpha_i$). Further hints: How should a polynomial $p$ with roots $\alpha_1, \alpha_2, ..., \alpha_n$ look like first? If we want $p(\alpha_0)$ to be $1$, how do we enforce that? Now, repeat this for the other vectors.

Finally, given any $(c_0, c_1, ..., c_n) \in \mathbb{K}^{n+1}$, we can write it as $c_0(1, 0, ..., 0) + c_1(0, 1, ..., 0) + ... + c_n(0, 0, ..., 0, 1)$. Since $f$ is linear, what can we do with the polynomials that we have found earlier?

Ok, so for the first thing: I understood only the null polynomial can be mapped to the null vector. About the surjectivity, a polynomial $p$ with such roots would be written as $$p = (x - \alpha_0)(x - \alpha_1) \ldots$$ hence if we want $p(\alpha_0) = 1$ we must require $p = (x - \alpha_0)^{n} + 1$ right? Or also $p = (x - \alpha_0)(x - \alpha_1) \ldots (x - \alpha_n) + 1$ Then we repeat. Finally, I can say $f(c_k) = f(c_k(1, 0, \ldots )) = c_k f(1, 0, \ldots )$ for all $k$ from $1$ o $n$ hence the image is fulfilled (I know I say it bad, but I might have got it)

You are getting there!

Yay! Slowly haha. I'm sorry for the disturb, I am not still able to do proofs

We want $p$ to be a degree $n$ polynomial. Since $p$ has $n$ distinct roots $\alpha_1, ..., \alpha_n$, $p$ must be the polynomial $p(x) = (x - \alpha_1)...(x - \alpha_n)$. However, this is not close to what we want since $p(\alpha_0) = (\alpha_0 - \alpha_1) ... (\alpha_0 - \alpha_n)$. Fortunately, $\alpha_0$ and all the other $\alpha$'s are distinct, so how do we modifiy $p(x)$ to get $p(\alpha_0) = 1$?

I would modify the polynomial as $p(x) = b(x-\alpha1)...(x-\alpha_n) + 1$ where $b = 1/((\alpha_0 - \alpha_1)...(\alpha_0 - \alpha_n))$

Close! But do you need the +1 there?

Oh, right I don't! Since the division would itself make $1$

Don't know why I thought it made zero, lol

Yup! You got the first part. So, you can actually do this for the other vectors of the form (0,1,0,...,0) etc.

Yess!

Thank you!

You're welcome! I think you should be able to figure out how the linearity of $f$ helps you from here.

Indeed I am, thank yoy!!

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