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so I'm currently struggling witht the following problem:

Show that the maps $$e_k: \mathbb{N}_0 \rightarrow \mathbb{Z}; x \mapsto \frac{x(x-1) \cdots (x-k+1)}{k!}$$ for $k \in \mathbb{N}_0$ (with $e_0$ identically $1$) form a basis for Int$(\mathbb{Z})$ as a $\mathbb{Z}$-module.

In case that this is not a widespread notation, Int$(\mathbb{Z})$ refers to the functions that map integers to integers.

To be quite honest, I am quite confused by the question, I don't understand the maps. I don't understand why one has a normal arrow and one has an arrow with a little base. what are the maps supposed to mean? Any help would be really appreciated!

NatNat
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    The arrow notation is just the definition of the $e_k$, i.e. $e_k(x) = \frac{x(x-1)\dotsm(x-k+1)}{k!}$. You want to show that the $e_k$ are linearly independent, and that any function $f: \mathbb N_0 \to \mathbb Z$ can be written as $\mathbb Z$-linear combination of the $e_k$. – red_trumpet Mar 10 '21 at 14:12
  • Ooooh that makes so much sense!! yep i can see that now, thanks so much! – NatNat Mar 10 '21 at 14:23
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    The map $\mathbb N_0\to\mathbb Z$ sending $1\mapsto 1$ and everything else to $0$ can't be written as a $\mathbb Z$-linear combination of the $e_k$. – Kenta S Mar 10 '21 at 14:31
  • You're right, Kenta. I then was confused, so just looked it up again, turns out our notes specify that Int$(\mathbb{Z})$ only includes polynomials so that luckily excludes your example! – NatNat Mar 10 '21 at 14:46

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The MSE question 4049277 "Avoiding brute force: determining when a specific polynomial in $\mathbb{Q}[x]$ is an integer for any integer $x$" asks essentially the same question. Some of the answers are for the general case. My answer used the technique of a difference table to solve the general problem of expressing an integer valued polynomial as a $\mathbb{Z}$-linear combination of binomial coefficients. The formula is $$ f(x) = \sum_{k=0}^\infty \Delta^k\!f(0) {x \choose k} $$ where the sum has only a finite number of non-zero terms since $\,f(x)\,$ is a polynomial function.

Somos
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