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Infinite ascending chains of Q subspaces of C?

  1. Oct 27, 2013 #1
    http://img826.imageshack.us/img826/6069/kj13.png [Broken]

    The proof is showing that particular matrix ring is not left artinian or noetherian. For any [itex]\mathbb{Q}[/itex] subspace [itex]V[/itex] of [itex]\mathbb{C}[/itex] that certainly is a left ideal of [itex]R[/itex]. But I can't seem to find an infinite ascending (or descending) chain of left ideals. This would involve finding ascending or descending [itex]\mathbb{Q}[/itex] subspaces [itex]V[/itex] of [itex]\mathbb{C}[/itex] .

    For example [itex]\mathbb{Q}(1+i)[/itex] is a [itex]\mathbb{Q}[/itex] subspace [itex]V[/itex] of [itex]\mathbb{C}[/itex]. [itex]\mathbb{Q}(1+i) + \mathbb{Q}(1+2i)[/itex] is another [itex]\mathbb{Q}[/itex] subspace [itex]V[/itex] of [itex]\mathbb{C}[/itex] such that the first one is a strict subset of this. However after this I couldn't think how to continue the pattern. Can anyone help with constructing these infinite ascending or descending subspaces?

    Edit: After searching around I realised this is quite obvious. [itex]\mathbb{C}[/itex] treated as a vector space over [itex]\mathbb{Q}[/itex] is infinite dimensional.

    So take a random element [itex]a[/itex] of [itex]\mathbb{C}[/itex]. [itex]\mathbb{Q}a[/itex] is a [itex]\mathbb{Q}[/itex] subspace [itex]V[/itex] of [itex]\mathbb{C}[/itex]. Then take an element [itex]b[/itex] from [itex]\mathbb{C}[/itex] not in [itex]\mathbb{Q}a[/itex]. [itex]\mathbb{Q}a+\mathbb{Q}b[/itex] is another [itex]\mathbb{Q}[/itex] subspace [itex]V[/itex] of [itex]\mathbb{C}[/itex] that strictly contains [itex]\mathbb{Q}a[/itex]. Take [itex]c[/itex] from [itex]\mathbb{C}[/itex] not in [itex]{\mathbb{Q}a+\mathbb{Q}b}[/itex] to obtain [itex]\mathbb{Q}a+\mathbb{Q}b + \mathbb{Q}c[/itex] and so on and so on.

    However, this shows it is not Noetherian. How do I show it is not Artinian?

    Edit 2: I guess since [itex]\mathbb{C}[/itex] treated as a vector space over [itex]\mathbb{Q}[/itex] is infinite dimensional, means [itex]\mathbb{C}[/itex] has basis [itex] {r_1, r_2, r_3, ... } [/itex]. Hence I can take [itex] {r_2, r_3, ... } [/itex] to form a new [itex]\mathbb{Q}[/itex] subspace [itex]V[/itex] of [itex]\mathbb{C}[/itex] which is strictly smaller than [itex]\mathbb{C}[/itex]. Keep doing this to obtain a infinite strictly descending chain of ideals.
    Last edited by a moderator: May 6, 2017
  2. jcsd
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