Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
@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
1  
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
2  
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
2  
@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

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.