Holiday puzzles

[Ben]

So, to drag you away from the TV this holiday period, try this easy one.

 

Let [math]n\mathbb{Z}[/math] denote the set of integers exactly divisible by [math]n \in \mathbb{Z}[/math] . Let [math]\mathbb{Z}_n \equiv \mathbb{Z}/ n \mathbb{Z}[/math] be the set of integers modulo [math]n[/math].

 

Show that [math]\mathbb{Z}_n [/math] is a field iff [math]n[/math] is prime.

 

(Recall that, in math, a field is a set equipped with a pair of closed commutative operations, usually written as [math]+[/math] and [math]\times[/math] , each of which admits of an inverse and an identity).

 

I pose this rather simple test, as I remember that when I was set this as an assignment some years back I was told that my proof was "not exactly wrong, but extremely inelegant"

 

Hint 1. First show that Z_n is a ring for any n

 

Hint 2. Find an ideal for the ring Z

 

Hint 3. Naughty, you're peeking!!

 

PS. nootropic is barred from this puzzle, at least for the time being.

 

PPS. First person to get a better proof than mine may nominate their favourite charity, and I will donate 100 euros. That's a promise. Would I Lie to you? (Drunken chuckle - hic)

 

Have a good holiday, one and all!

[Nootropic]

PS. nootropic is barred from this puzzle, at least for the time being.

 

Okay, I guess I won't be a scrooge and ruin everyone's fun

 

But I pose a generalization (which I doubt anyone else will be able to answer...maybe) if you figure out the previous one (which is a corollary of this result!).

 

Let A be a commutative ring with [multiplicative] identity and I an ideal contained in A. Then I is a maximal ideal (that is, maximal with respect to inclusion) if and only if A/I is a field.

 

There's multiple ways to go about this. In particular one route is to show that the only ideals of a field are the zero ideal and the whole field itself.

 

Perhaps this is a good chance to try to introduce Zorn's Lemma for the unfamiliar to derive a few extremely useful results in commutative ring theory. Nothing like an exercise in abstraction! Ben?

[Ben]

Okay, I guess I won't be a scrooge and ruin everyone's fun

Except that this sort of thing seems not to be the sort of "fun" we have here, generally. Ho, hum....

[

But I pose a generalization ......if you figure out the previous one (which is a corollary of this result!).

Yikes! I'm a naughty boy; asking you to prove a corollary before the main theorem!

 

But of course I have a proof of your theorem - it's in all the texts, so I debar myself as a CHEAT thereby

 

Perhaps this is a good chance to try to introduce Zorn's Lemma for the unfamiliar to derive a few extremely useful results in commutative ring theory. Nothing like an exercise in abstraction! Ben?

Well, ahem, first I seriously doubt you can say anything meaningful about Zorn without an introduction to order theory.

 

Second, I doubt that, even if you COULD, there would be much interest here. Do you think, for example, someone you likes to post like this would be interested?

 

Third, don't get me wrong - there are some very talented physicists here, who seem to understand a lot of math. Just that they are not interested in (or unfamiliar with) the level of abstraction that interests me (and possibly you)

 

Finally, I enjoy our chats and arguments, but it's like we may as well be sitting over beer, with no others present. Is this what a chat-room should be about? In all my self-aggrandizing "tutorial" threads, I have tried to carry all others along with me, simply for the sake of getting a "discussion" going.

 

It seems I failed, and the failure is mine, no-one else's.

 

Sorry to strike a seemingly self-pitying note - that was not how I intended it to come out.

[Ben]

Let A be a commutative ring with [multiplicative] identity and I an ideal contained in A. Then I is a maximal ideal (that is, maximal with respect to inclusion) if and only if A/I is a field.

No takers, huh? So lemme step through a proof which might educate some of you about rings, fields and ideals.

 

First note that if [math]A[/math] is a commutative ring, we must have that [math]ab = ba[/math] and [math]a+b = b+a \in A[/math]. And, most specifically, that if [math]a+a' = 0[/math] then [math]a' + a = 0[/math]

 

Second, an ideal [math]I \in A[/math] is a subring such that, for some [math]i \in I[/math] and any [math]a \in A[/math] that [math]ia \in I[/math] and [math]ai \in I[/math]. If [math] A [/math] is a commutative ring, obviously these are equal.

 

The ideal [math]I[/math] will be maximal in [math]A[/math] iff, for any other ideal [math]J \in A[/math] then either[math] I = J[/math] OR [math] J = A[/math]. Note the following important properties of ideals: [math]II = I,\,\,\, I+I =I[/math].

 

The quotient ring [math]A/I[/math] has as elements [math]I[/math] and [math]a + I = I +a,\,\,(a \in A)[/math] by the above (A is commutative by construction, I is so by definition). Note also by the above that [math]I[/math] is a zero element in [math]A/I[/math]

 

So to the proof.

 

Suppose that [math]A/I[/math] is a field, and let [math]I,\,\,J[/math] be ideals in [math]A[/math] such that [math]I \subsetneq J \subseteq A[/math]. Let [math]j \in J,\,\,\, j \notin I[/math]. Then, evidently, [math]j+I[/math] is not a zero element in [math]A/I[/math].

 

Iff [math] A/I[/math] is a field, I can always find some element [math]k \in A[/math] such that [math](j+I)(k+I) = 1 + I[/math] ( since in any field [math]F[/math], we must have, for any [math]x \in F[/math], some [math]x' \in F[/math] where [math] xx' = 1[/math], so by the above [math]jk = 1,\,\, II = I[/math] and therefore [math]jk+I = 1+I[/math] ).

 

So if [math]jk+I = 1 +I[/math] by the forgoing, we must have that [math]jk-1 \in I[/math] using the cancellation law for addition in a field (strictly, it only needs to be an integral domain, but every field IS an integral domain)

 

I will set [math]n = jk-1[/math], where, self evidently [math] n \in I \subsetneq J[/math] so I will have that [math]n = jk-1 \Rightarrow 1 = jk-n \in I[/math].], again using the cancellation law.

 

But [math]I \subsetneq J[/math] and [math]j \in J,\,\,j \notin I[/math] and, further that [math]jk \in J[/math] by construction, so that [math] n = jk - 1 \in J[/math], then I conclude that [math]jk-n=1 \in J[/math]

 

Thus, for any [math] a \in A[/math], I have that [math]1a \in J[/math] which is quite simply [math]a \in A[/math], thus [math]J = A[/math], and therefore [math] I [/math] is a maximal ideal in [math]A[/math].

 

PS by edit: Ya, well, I know... the neo-physicists here (id est those with a "new theory") feel no need to understand such elementary mathematics. I strongly suspect they are wrong, but then, what do I know of physics?

[zeus]

"Ben"???I did not think it was called that name, do you know what is meant by do? Is a stubide

[zeus]

stupid

[sanctus]

Zeus, what is the point of your posts? Are they insulting or do I not get something?

 

Ben, you said:

Finally, I enjoy our chats and arguments, but it's like we may as well be sitting over beer, with no others present. Is this what a chat-room should be about? In all my self-aggrandizing "tutorial" threads, I have tried to carry all others along with me, simply for the sake of getting a "discussion" going.

 

It seems I failed, and the failure is mine, no-one else's.

 

Sorry to strike a seemingly self-pitying note - that was not how I intended it to come out.

 

Don't agree. i think they are really interesting and I re-learn a lot of stuff I forgot. So, keep them coming, please.

It is just a matter of how much time I have if I join the discussions or not. And if you keep it up, it will attract more and more members interested in this kind/level of math.