Infinite ascending chains of Q subspaces of C?

1. Oct 27, 2013

Silversonic

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 $\mathbb{Q}$ subspace $V$ of $\mathbb{C}$ that certainly is a left ideal of $R$. But I can't seem to find an infinite ascending (or descending) chain of left ideals. This would involve finding ascending or descending $\mathbb{Q}$ subspaces $V$ of $\mathbb{C}$ .

For example $\mathbb{Q}(1+i)$ is a $\mathbb{Q}$ subspace $V$ of $\mathbb{C}$. $\mathbb{Q}(1+i) + \mathbb{Q}(1+2i)$ is another $\mathbb{Q}$ subspace $V$ of $\mathbb{C}$ 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. $\mathbb{C}$ treated as a vector space over $\mathbb{Q}$ is infinite dimensional.

So take a random element $a$ of $\mathbb{C}$. $\mathbb{Q}a$ is a $\mathbb{Q}$ subspace $V$ of $\mathbb{C}$. Then take an element $b$ from $\mathbb{C}$ not in $\mathbb{Q}a$. $\mathbb{Q}a+\mathbb{Q}b$ is another $\mathbb{Q}$ subspace $V$ of $\mathbb{C}$ that strictly contains $\mathbb{Q}a$. Take $c$ from $\mathbb{C}$ not in ${\mathbb{Q}a+\mathbb{Q}b}$ to obtain $\mathbb{Q}a+\mathbb{Q}b + \mathbb{Q}c$ 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 $\mathbb{C}$ treated as a vector space over $\mathbb{Q}$ is infinite dimensional, means $\mathbb{C}$ has basis ${r_1, r_2, r_3, ... }$. Hence I can take ${r_2, r_3, ... }$ to form a new $\mathbb{Q}$ subspace $V$ of $\mathbb{C}$ which is strictly smaller than $\mathbb{C}$. Keep doing this to obtain a infinite strictly descending chain of ideals.

Last edited by a moderator: May 6, 2017