# Homomorphism from $\mathbb{Q}$ to an ordered field F

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?

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@Qiaochu I don't understand.. If what i said is what i need to prove, then you mean $r$ is an element of intersection of $Q and F$? – Katlus Jun 20 '12 at 7:17
@Dylan I showed that $r$ is a homomorphism. This exercise follows right after showing that $r$ is a homomorphism. – Katlus Jun 20 '12 at 7:19
I'm just taking issue with the use of the word "function", which doesn't mention any structure being respected. But now I'm confused. What does $r \cdot 1_F$ mean to you? Certainly $na$ makes sense where $n \in \mathbf Z$ and $a$ is an element of some ring. I do maintain that there are no completions in sight :) – Dylan Moreland Jun 20 '12 at 7:19
It may be convenient for some purposes to make believe that $\bf Q$ is a subset of $F$, but that doesn't make it one, and the symbol $r$ has no meaning in $F$ until you give it one. So $\gamma(r)=r$ makes no sense. – Gerry Myerson Jun 20 '12 at 7:25
@Katlus: The book's notation is unfortunate. In the expression $r\cdot 1_F$, we are not dealing with a product of elements of $F$, since $\mathbb{Q}$ is not necessarily a subset of $F$. Instead, $r$ is an operator on $F$. The definition of this operator is not hard. The quickest way of describing it is that $F$ has a natural vector space structure, with scalars in $\mathbb{Q}$. – André Nicolas Jun 20 '12 at 7:30

$\gamma : \mathbb{Q} \rightarrow F$ 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.