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According to my interpretation to one of the answers in Splitting in Short exact sequence, $$\Bbb R \cong \Bbb Q \oplus \Bbb R / \Bbb Q$$

also, according to What is known about the quotient group $\mathbb{R} / \mathbb{Q}$?

$$\Bbb R \cong \Bbb R / \Bbb Q$$

as groups.

This seems very strange to me as if the direct sum with $\Bbb Q$ adds nothing to the group structure at all in any way even "attacted" on the side separately. First, I was wondering if this is indeed true, and if so if anybody could construct a nice isomorphism of groups so that we can see explicitly that:

$$\Bbb Q \oplus \Bbb R / \Bbb Q \cong \Bbb R / \Bbb Q$$

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    $\begingroup$ The isomorphism would be the one from $Q\oplus R/Q \to R$ composed with the isomorphism $R\to R/Q$, no? $\endgroup$
    – Braindead
    Dec 24, 2014 at 3:36
  • $\begingroup$ Yeah, but there was some discussion along the lines of not being able to write down one or both of these isomorphisms for one reason or another I'm not sure. Was wondering if somebody could come up with something explicit to show the last one. $\endgroup$ Dec 24, 2014 at 3:41
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    $\begingroup$ Wanna really blow your mind? $\mathbb R/\mathbb Q\cong \mathbb R$. $\endgroup$ Dec 24, 2014 at 4:33
  • $\begingroup$ @ThomasAndrews didn't I already state that? $\endgroup$ Dec 24, 2014 at 17:14
  • $\begingroup$ Yep, reading comprehension failure on my part. @MatthewLevy $\endgroup$ Dec 24, 2014 at 17:18

2 Answers 2

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The reason why this may seem wierd is because we tend to think of $\mathbb{R}$ as a one dimensional, single object, so the idea that quotienting out by some non-trivial subobject doesn't change it is naturally somewhat odd. However, let me give a more intuitive example. Suppose you have $A = \mathbb{Z}$ (or $\mathbb{Q}$, ... or really any other abelian group). Let $$A = A_0 = A_1 = A_2 = ... $$ Now consider $$B = A_0 \oplus A_1 \oplus A_2 \oplus A_3...$$ $$B' = A_1 \oplus A_2 \oplus A_3 \oplus A_4 ... $$ Well clearly by just relabelling the indices we have $B \cong B'$. But at the same time $B' \cong B/A_0$, or thinking the other way $B \cong A \oplus B'$. So clearly, when we paste together infinitely many copies of the same group, adding another copy doesn't change anything about the structure. Note that while the above used countably many indices for simplicity, it works for any infinite direct sum.

Now, the thing is that once we discard the order and field structures of the reals, and only think of them as a $\mathbb{Q}$-vector space, they are no longer one dimensional. They are a vector space, but by cardinality concerns it must be of uncountably many dimensions. So the reals look like

$$\bigoplus_{i\in I} \mathbb{Q}$$ Where $I$ has the same cardinality of the reals (I'm not entirely sure, but I think this is where we start using the axiom of choice). So it's rather reasonable that cutting off a single copy of the rationals doesn't change the isomorphism class. Exhibiting an explicit isomorphism is hard, as it would require giving a basis for the reals as a vector space over the rationals, which cannot be done constructively.

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  • $\begingroup$ Okay, intuitively this makes a lot of sense, thanks. Let's see if anybody else has anything to add to this topic. $\endgroup$ Dec 24, 2014 at 3:47
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    $\begingroup$ You're right, axiom of choice is needed for existence of a Hamel basis of reals. Existence of basis in an arbitrary vector space is actually equivalent to axiom of choice (notice that you can define a Vitali set in terms of a Hamel basis of reals). $\endgroup$
    – tomasz
    Dec 24, 2014 at 11:03
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Let $V_1, V_2$ be two $\mathbb{Q}$-vector spaces that have dimension $\mathcal{c}$ over $\mathbb{Q}$. Then $V_1\cong V_2$ by choosing an explicit set of basis elements for both vector spaces. In this case $V_1=\mathbb{R}/\mathbb{Q}$, and $V_2=\mathbb{Q}\oplus \mathbb{R/Q}$. Similarly you construct $V_3=V_1\oplus V_2\cong V_2\cong V_1$, etc.

Let us try to be explicit. If you know how to construct a bijective map $$ \mathbb{R}\cup \{i\}\leftrightarrow \mathbb{R} $$ then you can construct the isomorphism between $V_1$ and $V_2$. If you cannot do that, yoy can try to construct a bijective map $$ \mathbb{N}\cup \{i\}\leftrightarrow \mathbb{N} $$first, then leave other elements aside.

This also showed that they are isomorphic as abelian groups, because as the other post you cited point out, they are all divisible torision-free abelian groups. So they are isomorphic if they are isomorphic as $\mathbb{Q}$-vector spaces.

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