2
$\begingroup$

I know that there exists a unique injective function $\gamma : \mathbb Q →F$ for any ordered field F.

I don't understand why 'Prove $\gamma(r) = r•1_F$ for every $r\in \mathbb Q$' is an exercise.. Don't we just see $\gamma(r)$ as an element of $\mathbb Q$, hence $\gamma(r)=r$? Am i missing something?

  • 0
    $F$ is$a$vector space over $\bf Q$, and as such $r\cdot1_F$ makes sense.2012-06-20

1 Answers 1

4

If I am interpreting this question correctly:

$\gamma : \mathbb{Q} \rightarrow F$ is probably defined by mapping $0 \mapsto 0_F$ and $1 \mapsto 1_F$. This extends to an injective ring homomorphism into $F$.

Let $\times$ denote the multiplication on $F$. I will define $\cdot$. I think the notation $r \cdot 1_F$ is defined to be by if $r \in \mathbb{Z}$ and $r \geq 0$, then $r\cdot 1_F = 1_F + ... + 1_F$, $r$-times. If $r \in \mathbb{Z}$ and $r < 0$, then $r \cdot 1_F = (-1_F) + ... + (-1_F)$. In general, if $r = \frac{p}{q}$, then $r \cdot 1_F = (p \cdot 1_F)\times (q \cdot 1_F)^{-1}$, recall that $\times$ is the operation on $F$.

Now I think the question is that $\gamma(r) = r \cdot 1_F$ where $\cdot$ is not the multiplication operation on $F$ but the thing I defined above. Now you have a actually question to answer, but it is still pretty easy.