1
$\begingroup$

Hartshorne book "Algebraic geometry" Proposition I.7.3 (b),

Proposition I.7.3.(b) : If $f: \mathbb Z \rightarrow \mathbb Z$ is any function, and $Q$ is a numerical polynomial such that $\Delta f=f(n+1)-f(n)=Q(n)$ for all $n\gg 0$, then there exists a numerical polynomial $P$ such that $f(n)=P(n)$ for all $n\gg 0$.

In proof of this proposition, If $\Delta(f-P)=0 \text{ for } n\gg 0$, then $(f-P)(n)$ is constant for $n\gg 0$ and so $f(n)=P(n)$ for $n\gg 0$.

But I can't understand this proof. May be I see more in detail.?

  • 0
    You don't need to write $n>>0$. You can write $n\gg 0$. (I changed it.)2012-09-06
  • 1
    Should be $f(n) = P(n) + c$ for some constant $c$ in the last part of the statement.2012-09-06
  • 0
    Not having Hartshorne handy, I don't know 1. what $\Delta$ means, 2. what kind of things $f$ and $P$ are. Maybe you could edit this information into the body of the question.2012-09-06
  • 1
    Let $g: \mathbb Z \rightarrow \mathbb Z$ be any function. If $\Delta(g)=0 \text{ for } n\gg 0$, then $g(n) = g(n+1) = g(n+2)=\cdots$ for $n\gg 0$. So $g(n)$ is constant for $n\gg 0$.2012-09-06

0 Answers 0