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I am studying real algebraic geometry. Let $K$ be an Archimedean Ordered field.

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While trying to prove this lemma, I am not able to understand how can we choose a minimal $ n \in \mathbb{Z}$ which satisfies $mb \leq n+1$.

$K$ being an Archimedean field satisfies only the property that for any element $mb \in K$, we have a natural no. $n \in \mathbb{N}$ that satisfies $mb \leq n$ (definition given for the Archimedean ordering). Now, we don't even know how $x \leq y$ is defined in $K$. How do we prove the existence of such minimal element in $\mathbb{Z}$?

MUH
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  • we knows that $mb<0$ or $mb\ge 0$ if the order is linear. The Archimedean condition seems to imply that the order is linear because every $v\in K$ is comparable with a non-negative integer – Masacroso Mar 01 '19 at 14:58

1 Answers1

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Theorem. Let $(K,\le)$ be an Archimedean field. Then for $x\in K$, there exists $n\in \Bbb Z$ with $n-1<x\le n$.

Proof. By the Archimedean property, there exists $n_0\in \Bbb N$ with $-x< n_0$. Then $$\tag1x+n_0>0.$$ Also, there exists $n_1\in\Bbb N$ with $x+n_0\le n_1$. As $\Bbb N$ is well-ordered, there exists a minimal such $n_1$. From $(1)$ we see that $n_1\ne0$, hence $n_1-1\in\Bbb N$ and by minimality of $n_1$, $x+n_0\not\le n_1-1$. So with $n:=n_1-n_0\in \Bbb Z$, we conclude $$n-1<x\le n $$ as desired. $\square$

Corollary. Let $(K,\le)$ be an Archimedean field. Then for $x\in K$, there exists a minimal $n\in \Bbb Z$ with $x\le n$.

Proof. The $n$ of the theorem is already minimal: If $m\in \Bbb Z$ and $m<n$, then $m\le n-1<x$. $\square$

Definition. In the situation of the Theorem, we call the uniquely (according to the corollary) determined $n$ the ceiling of $x$, denoted as $\lceil x\rceil$.

Hagen von Eitzen
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