Commutative Algebra - 2012-03-04

Exercise 1, page 98: Prove the binomial identity, i.e. if $R$ is a commutative ring then $$(a + b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}$$ for $a,b \in R$.

By induction:

$n=1$: $$ (a + b)^1 = {1 \choose 0} b + {1 \choose 1} a = a + b$$

Assume that $$(a + b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}$$

We claim that $(a + b)^{n+1} = \sum_{k=0}^{n+1} {n + 1 \choose k} a^k b^{n +1 -k}$:

$$ (a + b)^{n+1} = (a + b)(a + b)^n = (a + b) \sum_{k=0}^n {n \choose k} a^k b^{n-k} = $$

$$ = a \sum_{k=0}^n {n \choose k} a^k b^{n-k} + b \sum_{k=0}^n {n \choose k} a^k b^{n-k} = \sum_{k=0}^n {n \choose k} a^{k+1} b^{n-k} + \sum_{k=0}^n {n \choose k} a^k b^{n + 1 - k} $$

Matt N.

4:22 PM

$$ = \sum_{k=1}^n {n \choose k - 1} a^{k} b^{n + 1 - k} + \sum_{k=0}^n {n \choose k} a^k b^{n + 1 - k}$$

We want to use ${n \choose k - 1 } + {n \choose k} = {n + 1 \choose k}$ so we rewrite the above as follows:

$$ = \sum_{k=1}^n {n \choose k - 1} a^{k} b^{n + 1 - k} + {n \choose 0} b^{n+1} + \sum_{k=1}^n {n \choose k} a^k b^{n + 1 - k} $$

$$ = \sum_{k=1}^n {n + 1 \choose k} a^k b^{n+1-k} + b^{n+1}$$

Kannappan Sampath

Now make a change of variable. $l=k-1$.

Matt N.

But we want to keep ${n+1 \choose k}$

$$ = - {n + 1 \choose 0} b^{n + 1} + \sum_{k=0}^n {n + 1 \choose k} a^k b^{n+1-k} + b^{n+1}$$

Kannappan Sampath

Oh, sorry, we just combine the two terms. The last stand alone term corresponds to $k=0$.

Matt N.

$$ = \sum_{k=0}^n {n + 1 \choose k} a^k b^{n+1-k} $$
$\Box$

That was also not very exciting.

: )

Kannappan Sampath

@MattN I don't understand the remark. What do you mean?

Matt N.

4:33 PM

@KannappanSampath Like the group question about $P(X)$ and associativity of $\Delta$.

I don't like fiddly questions. They cost a lot of time (doing symbol manipulation) and don't teach you understanding of what you want to understand.

Kannappan Sampath

Agreed! :-)

Matt N.

I mean, from this we learned that the binomial coefficient formula only works in commutative rings. But otherwise we have not learnt much about rings, unfortunately.

Maybe we should be selective about which exercises we do.

Kannappan Sampath

@MattN Yes, I think our criterion should be those that are relatively deep and new.

Matt N.

Yes.

Kannappan Sampath

But, however, someone will solve all the problems and TeX them up. So, it gets more useful.

Matt N.

4:39 PM

You want to do that? Like make a solution manual for the book?

Kannappan Sampath

@MattN Yes. That would be good. No?

Matt N.

If you do we'll just have to put in a little more work then we might as well do them all, no matter how boring . )

@KannappanSampath Yes!

The only reason why I suggested to be selective up there^ is because I have an exam and I need to progress as fast as possible.

Have you texed up the ones we did yesterday?

Kannappan Sampath

@MattN No. Not really. I'll do it as I find time.

I am sorry though. : (

Matt N.

Sure.

No problem.

Kannappan Sampath

@MattN For the exam, I think, we'll have to do interesting and insightful problems.

So, I suggest we skip boring problems.

Matt N.

4:42 PM

@KannappanSampath For the exam I need to know all of Atyiah and some extra.

@KannappanSampath Yes.

Kannappan Sampath

I think we'll move to exercise 2 on pg. 98

Matt N.

Ok.

Exercise 2, page 98: Let $R$ be a ring with identity and let $a \in R$. Prove the following:
(a) If $R$ has no zero divisors then the only nilpotent element of $R$ is $0$ and the only idempotent elements are $0$ and $1$.

Kannappan Sampath

Solution:

Suppose $a \neq 0$ be a nilpotent element. i.e. $a^n=0$ for some $n \in \Bbb N$.

Matt N.

Then $a \cdot a^{n-1} = 0$ and both $a \neq 0$ and $a^{n-1} \neq 0$. Hence we have a zero divisor. Contradiction.

What do you think?

Kannappan Sampath

@MattN Yeah, right. If we wanted to be even more rigorous, we should mention that $n$ was chosen to be smallest of positive numbers $k$ for which $a^k=0$. But, never mind.

Matt N.

4:50 PM

Yes, I tacitly assumed that, you're absolutely right.

Now the second claim: $0$ and $1$ are the only idempotent elements.

Kannappan Sampath

Yes, for this note that, if $a^2=a$, and if $a \neq 0$, we must have that, $a(a-1)=0$.

Now, as there are no zero divisors, one of the factors must be $0$, but as $a \neq 0$, we should conclude $a=1$.

Matt N.

Yes.

(b) No unit of $R$ is nilpotent. The only idempotent unit in $R$ is $1$.

Kannappan Sampath

(b) No unit of $R$ is nilpotent. The only idempotent unit of $R$ is $1$.

JINX

Matt N.

: )

Reminder: a unit in a ring is an element with an inverse.

Kannappan Sampath

Yes. So, let $a$ be a unit. We have to prove it is not nilpotent.

We have $b$ such that $ab=1$.

Looks like proving that: Nilpotent elements cannot be unit is easier. What do you think?

Matt N.

4:59 PM

I think I have it by contradiction.

Assume by contradiction that we have $a^n = 0$ for some $n$ then $1 = ab = a^2 b^2 = \dots = a^n b^n = 0 \cdot b^n = 0$. Contradiction.

Kannappan Sampath

Looks like you have assumed commutativity. No?

Matt N.

Could be. Where do you mean?

Kannappan Sampath

Why is $ab=a^2b^2$ ?

Matt N.

$ab = a \cdot 1 \cdot b = a (ab) b = aabb = a^2 b^2$.

Kannappan Sampath

Hah. Sorry!

Matt N.

5:02 PM

I don't think I'm using commutativity there. But maybe there's something else I'm doing wrong.

: )

@KannappanSampath No, scrutiny is a good thing!

Kannappan Sampath

@MattN No mistakes. Just not enough practice with these tricks for me. : (

Matt N.

@KannappanSampath That's why we're practicing : ) I don't have enough practice either : )

Hold on, give me 5 minutes to comment on a question on main.

BRB

Kannappan Sampath

Oh. Sure.

Matt N.

Back : )

Kannappan Sampath

We have second part of (b) still left.

The only idempotent unit of $R$ is $1$.

So, let $a$ be a unit.

then, there exists $b$ such that $ab=1$.

If $a$ is idempotent, then, $a^2=a \implies a^2b=ab=1 \implies a ab=1 \implies a=1$.

Matt N.

5:13 PM

Let $a$ be an idempotent unit of $R$ that is there exists a $b$ such that $ab = 1$ and $a^2 = a$. Then $1 = ab = a^2 b = a \cdot 1 = 1$.

Well done : )

Kannappan Sampath

Yes, next up is (c) If $a$ is nilpotent $1-a$ is unit.

Matt N.

(c) If $a$ is nilpotent then $1-a$ is a unit. If $a$ is idempotent then $1-a$ is idempotent.

^I had typed it already : )

Kannappan Sampath

Oh, no problems. : )

Matt N.

Let $n$ be the $n$ such that $a^n = 0$.

Now we want to find an inverse of $1-a$.

Kannappan Sampath

(let $n$ be the smallest $n$, you mean...?)

Matt N.

5:16 PM

Yes.

Actually, I haven't thought about whether we need the smallest one : )

Kannappan Sampath

$\dfrac{1}{1-a}=1+a+a^2+a^3+a^4+a^5+ \cdots$?

Matt N.

Yes.

Now we apply nilpotent to that.

Kannappan Sampath

So, the series trails of at a finite stage.

Matt N.

Yes.

Then we get $(1-a)(1 + a + \dots + a^n) = 1 + a + \dots + a^n - a - a^2 - \dots -a^n - a^{n + 1} = 1 - a^{n+1}$.

Oh. Stuck.

Kannappan Sampath

I think it suffices to take the sum till $n-1$, so that that term will be the inverse of $1-a$.

Matt N.

5:22 PM

$a^{n+ 1}=0$

Ooh, I've already forgotten what we want to show. Of course, we want it to be $1$, not $0$.

Kannappan Sampath

Yes, that is also true.

Matt N.

: )

So we're not stuck but we're done : )

Yay!

Kannappan Sampath

Yes. I wanted to point out $a^n=0$ and does not contribute to the sum.

Sure, we are through. : )

Matt N.

You're right. : )

Next one : )

(b) (part 2) If $a$ is idempotent then $1-a$ is too.

So we have $a^2 = a$.

And $(1 -a) (1 -a) = 1 - 2 a + a^2 = 1 - a$

Kannappan Sampath

So, $(1-a)^2=(1-a)(1-a)=1-a-a+a^2=1-a-a+a=1-a$

Matt N.

5:25 PM

Done.

Kannappan Sampath

Through to the next.

(d) Nilpotent elements form an ideal in a commutative ring $R$

Matt N.

Let $N$ denote the set of all nilpotent elements of $R$.

We want to show that for $n_1, n_2 \in N$ we have $n_1 + n_2 \in N$ and for $r \in R$ we have $rn_1 \in N$.

Kannappan Sampath

We are required to show it is non-empty, closed under subtraction and under "extended" multiplication.

Matt N.

Very good^

Kannappan Sampath

Thank you. : )

Matt N.

5:28 PM

First we observe that it'sn non-empty because $0$ is nilpotent.

Kannappan Sampath

Yes.

Matt N.

Next we want to show that $n_1 - n_2$ is in $N$ again, that is, for some $k$ we have $(n_1 - n_2)^k = 0$.

We can use the binomial theorem for that.

Or not? Hm....

Kannappan Sampath

Yes. And, Let $k=N_1+N_2$ where $n_1^{N_1}=n_2^{N_2}=0$

@MattN We can as the ring is explicitly given to be commutative.

Matt N.

Yes. But I was wondering whether that leads anywhere.

Kannappan Sampath

It will as exponent of one or the other of $n_1$ and $n_2$ will exceed $N_1$ or $N_2$ respectively. No?

Matt N.

5:33 PM

Let's write it down and see:

$$(n_1 - n_2)^k = \sum_{i=0}^k {k \choose i} n_1^i (-1)^{k-i}n_2^{k-i}$$

Kannappan Sampath

Yes, when $i$ ranges from $N_1$ to $N_2$, we have that, $n_1^i$ will vanish. Right?

Matt N.

Yes.

Kannappan Sampath

Now, as $i$ ranges from $0$ to $N_1-1$, we have that $N_1+N_2-i$ will range down from $N_1+N_2$ to $N_2+1$, and hence $n_2^{k-i}$ will have to vanish.

Is this OK or I have cheated somewhere?

Matt N.

No that's right.

Very good.

Kannappan Sampath

In fact, I think, for $k=N_1 \vee N_2$ is sufficient. What do you think?

Matt N.

5:40 PM

Now last but not least we need to show $r n_1 \in N$. : )

@KannappanSampath Yes, that's what I was going to write originally but then $N_1 + N_2$ works just fine. : )

Kannappan Sampath

@MattN Oh, OK. Thank you.

Matt N.

The last one is easy: $(rn_1)^{N_1} = r^{N_1} n_1^{N_1} = r^{N_1} \cdot 0 = 0$.

and we're done : )

Note: ^here I used commutativity.

Kannappan Sampath

Thanks to commutativity. We have relied on that heavily.

(I mean in both the parts.)

Matt N.

(e) Provide a counterexample to part (d) if $R$ is not commutative.

Kannappan Sampath

@MattN Yes, sure.

Matt N.

5:44 PM

Now first we need to come up with a non-commutative ring. The quaternions come to mind.

Let me look up the definition and post it here.

Kannappan Sampath

I think I don't have a good stock of easy non-commutative rings.

Matt N.

Matrices are also non-commutative.

Kannappan Sampath

The list of nilpotent elements is just informidable. No?

Matt N.

Ah, the quaternions are not actually a ring apparently. They are a non-commutative division algebra.

Kannappan Sampath

I think division algebra is a unital ring without non-zero zero divisors.

Matt N.

5:48 PM

@KannappanSampath informidable = not to be dreaded?

Kannappan Sampath

@MattN Oops. I meant the opposite! : )

Matt N.

@KannappanSampath I had not heard the word before : )

I'm just looking up the definition of algebra.

Kannappan Sampath

@MattN The free dictionary says it exists though. :-)

Matt N.

Yes : )

Kannappan Sampath

Well, I'd like to bring your attention to one subtelty.

Matt N.

5:53 PM

Yes?

Kannappan Sampath

In our proof for part 1 of (b), we had said $1=0$ is a contradiction. Remember?

54 mins ago, by Matt N.

Assume by contradiction that we have $a^n = 0$ for some $n$ then $1 = ab = a^2 b^2 = \dots = a^n b^n = 0 \cdot b^n = 0$. Contradiction.

Matt N.

Yes.

Kannappan Sampath

But, in a zero ring, where identities coincide, this is not a contradiction.

(and only there is that not a contradiction.)

Matt N.

Hm. But we implicitly have that $R$ is not the zero ring because the question assumes that $1 \in R$.

Kannappan Sampath

And, this book has a proposition that: In a ring with identity, $1 \neq 0$.

Matt N.

5:56 PM

Yes.

Kannappan Sampath

@MattN But $1$ could equal $0$ right?

I think this proposition is not true. What do you think?

Matt N.

@KannappanSampath I'm not an algebraist (unfortunately, or at least not yet) but I think if you write $1$ then you mean $1 \neq 0$.

Kannappan Sampath

Well, I asked Dylan he said there is no reason why we should get rid of the zero ring.

(I mean we need not assume, $1 \neq 0$)

Matt N.

But the exercise starts by assuming that $R$ is a ring with identity and on page 49 they say that the identity is non-zero.

Kannappan Sampath

This exercise fails for zero ring, right?

Matt N.

6:02 PM

It would but by assuming that $1$ is in $R$ and previously requiring that $1\neq 0$ (page 49) they have excluded that case.

Kannappan Sampath

@MattN Yes. So, we are right. But this is a bit of a small thing that is worth keeping in mind.

Matt N.

Yes, I agree.

Kannappan Sampath

Anyway, we can move on. I requested Dylan to come over and shed some light on this issue.

@MattN The counter example thingy. Does anything strike your mind?

Matt N.

I'm working on that.

I need a break. I'm starved. Makes thinking difficult.

Kannappan Sampath

@MattN Sure. Have a good snack! Dinner time?

Matt N.

6:08 PM

Unfortunately, dinner is only ready in 40 minutes.

Kannappan Sampath

@MattN Oh, then after dinner?

Matt N.

@KannappanSampath Ok. How about an hour of break, or so?

Kannappan Sampath

@MattN Yes. Sure!

Kannappan Sampath

6:43 PM

Summary of exercise (2):

1. Nilpotent elements are either zero or zero divisors.$^\dagger$

2. In a ring with no zero divisors$^\dagger$, the idempotent elements are $0$ and $1$.

3. Units are not nilpotent.

4. The only idempotent unit is $1$.

5. If $a$ is nilpotent, $1-a$ is a unit.

6. If $a$ is idempotent, $1-a$ is idempotent.

7. The set of nilpotent elements of a commutative ring form an ideal.

$\dagger$ According to Adkins, $0$ is not a zero divisor. But, Atiyah defines $0$ also to be zero divisors.

The (3) requires the assumption that $R \neq \{0\}$.

Note that the converse of (1) is false.

That is there are zero divisors that are not nilpotent!

Kannappan Sampath

7:14 PM

@Dylan Welcome!

Dylan Moreland

Hi. Just poking my head in.

Kannappan Sampath

Did you just see where we needed your help?

Dylan Moreland

No, what's the matter?

Kannappan Sampath

In the list above, the item $3$ is not clearly true for a $0$ ring, right?

Dylan Moreland

I would not worry about this.

Kannappan Sampath

7:24 PM

This has always been a problem with me! I get torn between definitions! : (

@DylanMoreland Anyway, thanks for saying! I won't worry too. : )

Dylan Moreland

Please try :)

Kannappan Sampath

@DylanMoreland Was that a sarcastic remark or real advice? : )

Matt N.

Real advice, I'd assume. I'm back.

Kannappan Sampath

Well, I put down all those what we had proved, saw that?

Matt N.

Just looking at it.

Lots of information.

Kannappan Sampath

7:42 PM

Yes, and they seem to be of not much later use. This exercise probably tests if we know defns well! What do you think here?

Matt N.

I'm not sure. These could pop up in later exercises.

Kannappan Sampath

Oh, well. I anyway remember sth if I think I shoud not remember that. : )

Matt N.

: )

So how do you make the quaternions into a ring?

Kannappan Sampath

Well, the multiplication is defined "like" cross product of vectors. I think it is discussed in the book at pg. 54.

I am wondering what the nilpotent elements of $End(A)$ will be?

Matt N.

Ah yes, right.

Kannappan Sampath

7:55 PM

I have a silly thing to ask: Endomorphisms are group homomorphisms, right?

Matt N.

Yes. But apparently it's more general: here

Taken from here, page 178.

Kannappan Sampath

I'll be right back. I am to fetch a bottle of water.

Matt N.

So the quaternions have the nilpotent elements $N = \{ 0, 1 + i, 1 + j, 1 + k, i + j, i + k, j + k, 1 + i + j + k\}$

@KannappanSampath Sure!

Now all we need to do is show that $N$ is not an ideal of $Q$. Easy enough, since $1 + i + j$ is not in it.

Kannappan Sampath

8:11 PM

I am in. : )

Dylan Moreland

@KannappanSampath Real advice.

Kannappan Sampath

@MattN I believe the grab has nothing to do with the question in hand except that $N$ information there, right?

@DylanMoreland Sure. Thank you. But I thought you told me that because this is the third time I'm troubled on a relatively unimportant point!

@MattN Yes, got it! :-)

Matt N.

@KannappanSampath The grab just tells us the nilpotent elements of $Q$ : )

Kannappan Sampath

@MattN Clever search. And, a good grab!

Matt N.

So we're done with exercise 2.

Kannappan Sampath

8:21 PM

Yes. Sure. (But, are you convinced with the cheating in last part?)

I am not sure if I'll be able to justify $N(Q)$ is that set there. : (

Matt N.

What cheating?

Oh I see.

Well, let's find a different counterexample then.

Kannappan Sampath

Or may be think about it for today and discuss tomm. ?

Matt N.

Yes. It's late there, isn't it?

Kannappan Sampath

@MattN Yes, it is. But I can stay up for 2/3 more exercises. : )

Matt N.

Ok. In exercise 3 we have $(P(X), \Delta, \cap)$ is a commutative ring with identity.

Kannappan Sampath

8:28 PM

Yes.

Matt N.

Now we are supposed to show that $A^2 = A$.

That's true, since $A^2 = A \cap A = A$.

Kannappan Sampath

But $A^2=A \cap A=A$. So, done. Right?

Matt N.

Yes.

Next we want to show that if $P(X)$ is an integral domain then $X$ is a singleton set.

Kannappan Sampath

Yes.

Matt N.

Integral domain means we don't have zero divisors i.e. any two sets have non-empty intersection.

Kannappan Sampath

8:31 PM

any two non-empty sets...

Matt N.

Yes, right.

Kannappan Sampath

If there were two elements $\{a,b\}$, the sets, $\{a\}$ and $\{b\}$ intersect trivially. So, the cardinality is forced to be less than or equal to $1$.

Matt N.

I assume we'd have to be more general and say "Assume that the cardinality is greater than one..."

Kannappan Sampath

Yes, you're right.

Matt N.

But yes, you're right.

Kannappan Sampath

8:34 PM

: )|( ?

Matt N.

What's that^? : )

Kannappan Sampath

I don't know how I should react. You were right and you say, "But, yes, you were right." I am puzzled.

Matt N.

Well what you said is not wrong, so...

: )

Yay. Next one!

Our speed is quite reasonable, what do you think?

Kannappan Sampath

@MattN Definitely yes.

So, if $P(X)$ is an integral domain, then, it is isomorphic to $\Bbb Z_2$.

Is that OK?

Matt N.

Yes.

Kannappan Sampath

8:39 PM

Now to next problem.

Matt N.

Hold on. What if $X = \varnothing$.

Kannappan Sampath

Then, we have a zero ring?

Matt N.

Yes. How is that automatically excluded from this exercise?

Kannappan Sampath

Because, an integral domain is a commutative ring with identity.

Matt N.

I think in the example it should say "Let $X$ be a non-empty set..."

Kannappan Sampath

8:43 PM

:3687158 Yes. Then and only then, is $(P(X), \Delta, \cap)$ is a ring with identity.

Matt N.

Unless I'm getting confused.

But we just saw that we get the zero ring if $X$ is empty.

Kannappan Sampath

Yes, that's a typo!

Matt N.

Now we ended up worrying about what we decided not to worry about.

Kannappan Sampath

As in: The example claims $P(X)$ is a ring with identity. So, it should have also said, that $X$ is non-empty.

@MattN Once in a while, it's OK, methinks! : )

Matt N.

: )

Exercise 4, page 98: Define $I_a := \{ A \in P(X) \mid a \notin A \}$. Prove that $I_a$ is a maximal ideal in $P(X)$.

Kannappan Sampath

8:47 PM

Then, we'll have to prove that $P(X)/I_a$ is a field, right?

Matt N.

I'm not sure they ask for a proof but let's do the maximal ideal proof first.

Kannappan Sampath

I mean the strategy for the problem would be to show that $P(X)/I_a$ is a field and hence $I_a$ is a maximal ideal right?

Matt N.

Oh, I see.

That is also a possibility. I was thinking of proving that it's maximal by contradiction.

Kannappan Sampath

Sure, I'll attempt that now.

Matt N.

I'm not sure which is better here though.

Kannappan Sampath

8:52 PM

Let's attempt contradiction because, anyway, we'll have to answer the next question what is $P(X)/I_a$. OK?

Matt N.

Ok!

Kannappan Sampath

So, there is a proper ideal,$ I$ that strictly contains $I_a$.

So, $I$ contains a set that contains $a$. Am I right here?

Matt N.

Yes. I haven't finished the proof though.

I'm confused. What's the inverse of a set with respect to $\cap$?

Oh. We don't have inverses.

Bleh.

Kannappan Sampath

We need not have inverses, right?

Matt N.

That breaks my proof idea.

@KannappanSampath Well for the proof I had in mind, yes.

Kannappan Sampath

8:58 PM

So, can we conclude that $I$ has to be $P(X)$?

Matt N.

I don't see how to do that without inverses : /

I'm starting to be sleepy.

Will you be doing this tomorrow?

Kannappan Sampath

@MattN Same here, a few minutes back! :/

Matt N.

I would think so! Given how late it is there : )

Kannappan Sampath

@MattN Sure, I plan to spend an hour or so everyday for reading and an hour for exercises.

Matt N.

Sounds good. Then we'll continue tomorrow : )

Kannappan Sampath

9:01 PM

And the counter example thingy also. (A reminder)

Matt N.

Yes.

Kannappan Sampath

Bye @Matt. Have a good night's sleep.

(I am off to map to $\Bbb R^2$. Bye.)

Matt N.

@KannappanSampath Thanks, good night to you too!

Transcript for

$Mathematics$ Commutative Algebra