3

While working to Apply Lam's theorem to determine all the left ideals of $\begin{pmatrix}\mathbb{Q}&\mathbb{R}\\0&\mathbb{R}\end{pmatrix}$ I have encountered the problem of determining all $\mathbb{Q}$ subspaces of $\mathbb{Q} \times \mathbb{R}$.

A possible subspace will have all first components equal to zero or will contain all rational first components. This is because the subspace condition imposes on the set of first components the property of being an ideal and since $\mathbb{Q}$ is a field it has to be $0$ or $\mathbb{Q}$.

Now I turn to the condition on the set of second components. If $A$ is the subspace we should ensure:

  1. $\forall (q,r),(q',r') \in A.(q,r)+(q',r') = (q+q',r+r') \in A$

  2. $\forall q_1 \in \mathbb{Q},(q,r) \in A.q_1(q,r) = (q_1q,q_1r) \in A$

So first, of all we have that the set of second components has to be a subgroup of $(\mathbb{R},+)$. This is directly related with How can we find and categorize the subgroups of $\mathbb{R}$? and the answers there reveal that it is not easy to classify this family (might be also interesting What do the cosets of $\mathbb{R} / \mathbb{Q}$ look like?)

I wonder if condition 2. is of any help here, in other words, does imposing (call $B$ the set of second components):

$\forall q \in \mathbb{Q},r \in B.qr \in B$

gives us any clarifying condition for the set of subgroups of $\mathbb{R}$ that verify it?

Examples/Properties of solutions: (for reference)

Any extension field K contained in $\mathbb{Q} \subseteq \mathbb{K} \subseteq \mathbb{R}$ works. can I find any that is not of that form?

Using the previous link, it seems that any candidate (excluding $\{0\}$) has to be dense in $\mathbb{R}$.

Any candidate should include $0,1$ (so the set of irrationals is excluded)

I would like to retain for reference, @MorganRodgers example $\langle 1,\sqrt{2},\sqrt{3} \rangle$

I wonder if any topology might be of help here. In the flavour of this.

Jacques Saliba
  • 285
  • 3
    As I tried to convey in the previous question, I think you're chasing ghosts by focusing on this condition. Looking for $\mathbb Q$ subspaces of $\mathbb Q\times \mathbb R$ is the same as finding all subspaces of $\bigoplus_{i\in I}\mathbb Q$ where $I$ is an appropriate uncountable subset. There isn't anything about $\mathbb R$ or the fact you're looking at elements of $\mathbb Q\times \mathbb R$ that mitigates this. – rschwieb Apr 06 '18 at 15:48
  • thanks @rschwieb. anyway this question is more about intellectual curiosity. using the previous link for example, it seems that any candidate (excluding ${0}$) has to be dense in $\mathbb{R}$ – Jacques Saliba Apr 06 '18 at 15:58

1 Answers1

1

All you really need are the $\mathbb{Q}$ subspaces of $\mathbb{R}$ to finish your analysis.

You take any (possibly uncountably infinite) subset of $\mathbb{R}$ and then take the closure under addition and scalar multiplication (by elements of $\mathbb{Q}$). Basically, you are taking their span over $\mathbb{Q}$.

If you take the one generated by $\{1, \sqrt{2}, \sqrt{3}\}$ then this will consist of $\{a+b\sqrt{2}+c\sqrt{3} : a,b,c \in \mathbb{Q}\}$, and is not an extension field of $\mathbb{Q}$ because it is not closed under multiplication (while containing 0 and 1, and having all of its elements algebraic over $\mathbb{Q}$).

xxxxxxxxx
  • 13,302