Here is an attempt for the argument of maximality of $I_a$:

Suppose $I_a$ is not maximal. Then, there exists an ideal $I$ such that, $P(X) \supset I \supset I_a$.

This means that, $I$ contains a set $A$ such that $a \in A$.

Now, note that, $X\setminus \{a\} \in I_a \subset I$.

So, we know that, $X\setminus \{a\} \Delta A \in I$ (because $I$ is closed under addition.)

So, $(X\setminus\{a\})\setminus A \cup A \setminus (X \setminus \{a\})=X\setminus A \cup \{a\} \in I$

Now recall that all singletons in $A$ except $\{a\}$ was in $I_a$ and hence in $I$.

Let's denote $X_a:=X\setminus A \cup \{a\}$. So, our aim is to prove that, $X_A \in I$ because, then we can conclude that, $I=P(X)$ using the closure with respect to "extended" multiplication, here $ \cap$.

But $X_a $ is in $I$!

@MattN Yes. Do you see the argument now?

Because $I$ is an ideal and $X \setminus \{a\}$ is in $I$ and $I$ is closed with respect to $\Delta$ and $A$ is in $I$.

Consider a singleton $\{i\}$ for $a \neq i \in A$, $X_a \setminus \{i\} \cup \{i\}\setminus X_a = X_a \cup \{i\} \in I$

Oh. I see.

So we get $X$ is in $I$. So we're done since then $I = P(X)$.

Now, the argument can be extended with any other singleton in $A$ and $X_{\{a,i\}}$ and this will lead us to the conclusion as you say.

Thumbs up! We are through! : )

Good. : )

But, fortunately or unfortunately, I have an Analysis class at 2.00pm. This was what I had thought during the lunch and so I wanted to put it in here. : (

(15 minutes from now! )

Analysis class sounds like fun. : )

I should be back in two hours from now. Is that fine? @MattN

Sure! I'm doing something else at the moment anyway : )

Shall we start doing this $2$ hours from now? @MattN

Sure.

: )

Oh. Then I might be slightly distracted because teddy might be here. Is that ok?

@MattN That's fine with me. I think I will be distracted too. : )

: )

@KannappanSampath I'm here.

I am here too. : )

I'm thinking about algebras. In particular, the quaternions.

Did you say that every algebra is a ring?

Oh, I see. The term "Algebras" has a precise meaning, right?

I can't exactly remember what you said because I didn't know the definitions yesterday.

Yes. I looked up the definition. It looks as if an algebra is a vector space over a field with an additional operation $\cdot$.

@MattN No, I said that "division algebras" are rings, methinks

Yes! True.

Now I need to look up division algebra.

But, I think I'll have to know the precise definition to be happy. :/

Me too.

Tumeni definitions.

Looks as if a division algebra is an algebra in which we have division, i.e. for every $a,b$ there exists a $c$ such that $a = cb$.

Yes.

Why do you think this is a ring?

No, I think I must have mixed this with a division ring. Sorry!

No problem : )

It's like a vector space with division. But we don't need division to have a ring.

So saying that a division algebra is a ring is like saying an algebra is also a ring (since we don't need division).

Yes. This will need a little more familiarity for me!

But then an algebra is a vector space with one additional operation, the multiplication. Which doesn't give us a ring.

@KannappanSampath Same here. : )

Now then. Shall we proceed to do some more exercises?

@MattN Yes, sure. That counter example thingy, I did not have enough time to think about. Shall we discuss that some other time?

No, let's think about it now.

Let's take the ring of $2 \times 2$ matrices over $\mathbb{R}$.

Ok. Yes.

That's a ring. And it's non-commutative.

Now we need to think about what nilpotent elements in it look like.

All those with precisely one non-zero entry are nilpotent right?

You mean like $$\left ( \begin{array}{cc} \alpha & 0 \\ 0 & 0 \end{array} \right )$$?

I think you are confusing nilpotent with idempotent.

We want matrices $A$ such that $A^n = 0$ for some $n$.

So for example $$A = \left ( \begin{array}{cc} 1 & -1 \\ 1 & - 1 \end{array} \right )$$

$$\left ( \begin{array}{cc} 0 & \alpha \\ 0 & 0 \end{array} \right )\left ( \begin{array}{cc} 0 & \alpha \\ 0 & 0 \end{array} \right )=\left ( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right )$$

I meant something like those. Are they fine?

No.

For these you get $A^2 = A$.

Did I multiply them rightly? Let me check. :/

Sure, I think I did.

You're right : D My mistake.

I should've done it on paper : / Sorry.

Ok. So we have already found $2$ nilpotent elements.

@MattN never mind!

What confused me is that you wrote "all those with precisely one non-zero entry are nilpotent" because the following satisfies that but isn't nilpotent:

I realise that that was my mistake. Sorry. : )

Np : )

$$\left ( \begin{array}{cc} 0 & 0\\ \alpha & 0 \end{array} \right ) \left ( \begin{array}{cc} 0 & 0\\ \alpha & 0 \end{array} \right )=\left ( \begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array} \right )$$

I wonder how we know when we have found all of them.

Well, I think we have enough to prove that $N$ cannot be an ideal.

For instance, the sum $ \left ( \begin{array}{cc} \alpha & 0 \\ 0 & 0 \end{array} \right )+ \left ( \begin{array}{cc} 0 & 0 \\ \beta & 0 \end{array} \right )= \left ( \begin{array}{cc} \alpha & 0 \\ \beta & 0 \end{array} \right )$ can never a nilpotent element if $\alpha \neq 0$. Is this true?

(Lesson: Never copy TeX codes! : D)

@DylanMoreland How do we know that we found all nilpotent elements of a ring $R$?

In general, who knows.

@MattN I think that is a nice question.

But for matrices, the minimal polynomial of a nilpotent matrix splits.

Because it's just $X^n$.

Ooh.

So Jordan form is always available. I would try to describe some nice representatives of the similarity classes.

@DylanMoreland I don't know about this. I haven't learnt enough. : /

@DylanMoreland But yes, you can tell Matt and I'll get back once I have learnt these things. Sorry if I was responsible for this halt.

Holy cannoli, I don't know all these things either!

: D

I'm not sure what to make of this. What do we gain if we write a matrix in Jordan normal form?

Welcome Prof. Gaillard

@KannappanSampath Sorry for being easily distracted.

@MattN Never mind. I was distracted now.

(I had typed in half the message and someone knocking my door here!)

: )

Hi Kannappan!

Hello Pierre-Yves. : ) Kannappan will be back later or tomorrow, he said he was going to do some integrals. In here we're doing Commutative Algebra together.