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For the upper triangular matrix ring given by $\begin{pmatrix}\mathbb{Q}&\mathbb{R}\\0&\mathbb{R}\end{pmatrix}$ I want to determine its left ideals. For example, this link refers to Lam's "A first course in non-commutative rings" applied to a generic matrix ring $\begin{pmatrix}R&M\\0&S\end{pmatrix}$:

The left ideals are all of the form $I_1\oplus I_2$ where $I_2$ is a left ideal of $S$, and $I_1$ is a left $R$ submodule of $R\oplus M$ which contains $MI_2$.

I want to translate this to explicitly determine all the left-ideals of my matrix ring above. So this is just a mere translation exercise. Let's see if I can do it correctly.

$I_2$ is a left ideal of $\mathbb{R} \implies I_2 = 0$ or $I_2 = \mathbb{R}$

$I_1$ is a left $\mathbb{Q}$-submodule of $\mathbb{Q} \oplus \mathbb{R}$ which contains $\mathbb{R}I_2$

So now, I unroll this complex definition.

A $\mathbb{Q}$-module is a vector space over $\mathbb{Q}$. The underlying set for this vector space is $\mathbb{Q} \oplus \mathbb{R}$ which I identify with $\mathbb{Q} \times \mathbb{R}$. So here I'm considering operations of the kind $q_1(q,r) = (q_1q,q_1r)$ and I have to determine the subspaces of this vector space. Is it as easy as taking a subspace of $\mathbb{Q}$ (probably $\{0\}$ or $\mathbb{Q}$ and a subspace of $\mathbb{R}$ over $\mathbb{Q}$ (giving a field extension?) and then doing the product? But then, how can this contain $\{0\},\mathbb{R}$ if it is bidimensional?

Can you help me to figure out this situation?

Thoughts

There is some sort of identification in the statement of the theorem because $MI_2$ should be $\mathbb{R}$ not $\{0\} \times \mathbb{R}$

ok, now I see the identification, it corresponds to $\begin{bmatrix}0 & M \\ 0 & 0\end{bmatrix}\begin{bmatrix}0 & 0 \\ 0 & I_2\end{bmatrix} = \begin{bmatrix}0 & MI_2 \\ 0 & 0\end{bmatrix}$

However I still don't see very well what should be explicitely the sets in @rschwieb answer. It seems to me that for the second one the possibilities are just $\mathbb{Q} \times \mathbb{R}$ and $\{0\} \times \mathbb{R}$. What about the first one?

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There are just two cases:

$I_2=\{0\}$, and $I_1$ is any $\mathbb Q$ subspace of $\mathbb Q\times \mathbb R$

$I_2=\mathbb R$ and $I_1$ is any subspace of $\mathbb Q\times \mathbb R$ containing $\{0\}\times\mathbb R$.

Then

$\left\{\begin{bmatrix}x&y\\0&z\end{bmatrix}\,\middle|\, (x,y)\in I_1, z\in I_2\right\}$ is a left ideal.

As you can see, there are only two left ideals of the second type, and there are infinitely many left ideals of the first type.

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  • $\begingroup$ @JacquesSaliba Like it says... any $\mathbb Q$ subspace of $\mathbb Q\times \mathbb R$. What else is there to say? $\mathbb R$ is an infinite dimensional $\mathbb Q$ space, so there are lots of things that could happen. $\endgroup$ – rschwieb Apr 6 '18 at 14:56
  • $\begingroup$ @JacquesSaliba I think you are making a mental mistake in the first sentence of your last comment. YES, either all entries of the left hand size will be $0$, or else you will see all of $\mathbb Q$. On the other hand, $\mathbb Q\times\{0\}$ is not necessarily contained in $I_1$. So you could have, for example, $\langle(1,\pi)\rangle$ which is not of the form $A\times B$ for $A\subseteq \mathbb Q$ and $B\subseteq \mathbb R$. There are lots and LOTS of $\mathbb Q$ subspaces of $\mathbb Q\times \mathbb R$ with no easy description. $\endgroup$ – rschwieb Apr 6 '18 at 15:04
  • $\begingroup$ @JacquesSaliba You may as well just identify $\mathbb Q\times \mathbb R$ with $\bigoplus _{i\in I}\mathbb Q$ where $I$ is an index set (necessarily uncountable.) $\endgroup$ – rschwieb Apr 6 '18 at 15:06

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