0

The motivation to this question can be found in

How I can express $(x,y)∈G$ by using the $r$ independent points $P_1,P_2,\ldots,P_r$

We know that there is an isomorphism $$φ:ℤ^{r}×C(ℚ)^{\text{tors}}→ℤ^{r}⊕C(ℚ)^{\text{tors}}=C(ℚ)$$ defined by $$(Q,T)→φ(Q,T)=Q+T$$

where $$Q=∑_{k=1}^{r}α_{k}P_{k}$$ and $T$ is a torsion point.

My question is: Can we extend the map $φ$ to $ℝ^{r}×C(ℚ)^{\text{tors}}→C(ℚ)$ as an isomorphism or not?

Community
  • 1
Safwane
  • 3,840
  • First of all, is there a bijection between these sets? –  Apr 03 '13 at 17:48
  • Yes, see http://math.stackexchange.com/questions/346406/how-i-can-doing-a-sum-of-a-vector-in-r-with-a-equivalence-class-in-n – Safwane Apr 03 '13 at 17:50
  • 1
    I don't see the connection between this question and the link. Here you have $\mathbb{R}^r$, which never appears there. Let me ask you this: is there a bijection between $\mathbb{R}$ and $\mathbb{Z}$? –  Apr 03 '13 at 17:56
  • In fact I do not know. – Safwane Apr 03 '13 at 17:58
  • Well, the answer is no. There is a basic distinction between countable sets (such as the natural numbers and integers) and uncountable sets (such as the real numbers) that every student of math (no matter their interests) needs to be completely clear on. Wikipedia can get you started: http://en.wikipedia.org/wiki/Countable_set –  Apr 03 '13 at 18:03
  • And what would $rP$ mean, when $r$ is a real number and $P$ is a point on $E$? – Álvaro Lozano-Robledo Apr 04 '13 at 02:13
  • Yes. You are right, that element has no sense unless $r$ is an integer. – Safwane Apr 04 '13 at 07:18

0 Answers0