Wikipedia:Reference desk/Archives/Mathematics/2007 May 13

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May 13

Hard-to-Read Formulae

In the image [1](warning: seminaked females), I can't make out most of what's written, and I don't recognize some of the symbols I can see. Could someone translate or transcribe for me? Black Carrot 05:47, 13 May 2007 (UTC)

I see a rotation matrix blocksum a 2x2 identity (IIRC a quaternion rotation matrix?) on the woman to the leftmost (can't make out the rest), Maxwell's equations(?) on the woman second from the left, and SU(3) otimes SU(2) otimes U(1) on the rightmost woman on the bottom line. HTH.

As a quaternion rotation matrix that would be a rotation by 2θ about the i axis. I also see a Feynman diagram on the rightmost woman, Newton's second law on the third woman (first line), and what looks like the definition of the Christoffel symbols on the first woman (third line). [The fourth line there looks like a differential equation involving Christoffel symbols, but I can't place it.] Tesseran 10:32, 13 May 2007 (UTC)

Fourth woman, first line is the Canonical commutation relation. Second line might be a form of Schrodinger's equation. Third woman, fourth line is the wave equation. I think the matrix on the first uses cosh and sinh, making it a Lorentz transformation, which would fit since the rest of the first woman seems to be about relativity: we have something involving m^2c^2, and the last line is the Geodesic equation. Algebraist 12:32, 13 May 2007 (UTC)

You might want to warn people about that pic, as some might be taken aback by it. StuRat 13:47, 13 May 2007 (UTC)

Good call. Label attached. Black Carrot 21:17, 13 May 2007 (UTC)

It's no racier than you'd commmonly see on billboards in liberal countries. —Tamfang 22:22, 15 May 2007 (UTC)

Perhaps not, but if the subjects were turned around, that would definitely be seen as an affront, by some, who would see it as no less than a full frontal assault on their values. StuRat 01:58, 16 May 2007 (UTC)

Not full frontal ... —Tamfang 17:27, 16 May 2007 (UTC)

Modules

I'm trying to do the following: Given R, a commutative ring with 1, and I and J two ideals of R, prove that R/I and R/J are isomorphic as R-modules if and only if I=J. Give an example to show that it is possible to have R/I and R/J isomorphic as rings even if I≠J.

Going from I=J to the isomorphism is trivial, but I have basically no idea how to go the other way. In any discussion of algebraic widgets, as soon as "quotient widgets" arise, my level of understanding plummets through the floor. What does it mean to consider R/I and R/J as R-modules? Maelin (Talk | Contribs) 08:34, 13 May 2007 (UTC)

First, DON'T PANIC. You already know a great deal about (commutative) rings, quotient rings, ideals, and modules. The one (commutative) ring to rule them all (sorry, Tolkien) is the integers. For an ideal, take multiples of 12; this is closed under addition, negation, and multiplication by any integer. Then the quotient ring is "clock arithmetic", the integers modulo 12. A module is just a vector space whose scalars forgot how to divide. (In general, the scalars for a module also need not have a multiplication that commutes; but here that's not the case.)

So what can it mean for the integers modulo 12 to be (almost) a vector space? Since it is called an R-module, where here R = Z, we know that the scalars are the integers. Thus the numbers (or equivalence classes) 0 through 11 must be the "vectors". And, sure enough, we can add them and scale them.

Play with these examples a little, and ask again if you want more specific hints. --KSmrq^T 10:43, 13 May 2007 (UTC)

There is a natural action of R on itself by left multiplication; this action descends to an action of R on R/I by left multiplication. (Explicitly, r acts on x+I by sending it to rx+I, using the additive notation for cosets. You might prefer the notation r acts on [x] sending it to [rx].) An R-module is by definition an object [an abelian group] with an action of R on it. What this means for you is that a proposed isomorphism between two R-modules needs to not only be a bijection between the two underlying objects --- that bijection must preserve the additional structure given by the action of R on the objects. (This is a loose, wordy explanation, since you should have the formal definition written down somewhere. If not, please say so.)

An analogous, perhaps simpler example: consider the group G = Z x Z₂ (where Z₂ is the cyclic group of order 2). Consider two subgroups: H = 2Z x Z₂, and K = Z x 1 (that is, K is simply the first factor of the direct product above). You should convince yourself that G/H and G/K are isomorphic (what familiar group are they isomorphic to?) Yet these quotients both have a natural G-action on them, and as G-sets they are very different.

To start with, you could look at the "kernel" of an action, which is in this case those elements of G that act trivially in a given action [on a given G-set]. If the kernels of two actions are different, then the two actions cannot be the same. As an example, the group of integers acts on the set of real numbers as follows: let the integer n act by multiplication by (-1)ⁿ, sending the real number x to the real number (-1)ⁿx. First check that this is a well-defined action; then see if you can identify which integers act trivially. Tesseran 10:53, 13 May 2007 (UTC)

(after edit conflict) R/I is a ring, and therefore an abelian group (by forgetting multiplication). We can define a "left" scalar multiplication operation • : R × R/I → R/I by r•(a+I) = (ra)+I. That is all we need by way of ingredients for R/I to be a module over R, and it is easy and straightforward to verify that the four listed conditions are met. (Actually we can likewise define a "right" scalar multiplication (a+I)•r = (ar)+I, giving us a bimodule.) For a homomorphism h : M → N to be also a module homomorphism (where M and N are R-modules over the same ring R), it has to respect scalar multiplication: h(r•x) = r•h(x).  --Lambiam^Talk 10:54, 13 May 2007 (UTC)

Okay, I think I can see R/I and R/J as R-modules now, but I have no idea how to prove that the isomorphism implies I=J. I tried using the fact that, since R/I and R/J are isomorphic, there exists a bijective homomorphism φ : _R(R/I) -> _R(R/J) , and then by the first isomorphism theorem, _R(R/I) / ker(φ) ≈ im(φ), and since φ is bijective, ker(φ) = I and im(φ) = _R(R/J), so _R(R/I) / I ≈ R/J, but this looks very suspiciously like not telling me anything useful. Am I going in the complete wrong direction? Maelin (Talk | Contribs) 13:08, 13 May 2007 (UTC)

Tesseran has already given you the idea, I think. Consider the kernel of the R-action on R/I, that is, the set of r in R such that r(a + I) = 0 + I for all a in R. Can you work out what this kernel is? Now use the isomorphism between R/I and R/J as R-modules to show that both actions must have the same kernel. Algebraist 13:14, 13 May 2007 (UTC)

The kernel is {0}, but so what? What does that have to do with the isomorphism theorem? And what does it matter whether the actions have the same kernel when they're different actions? *getting increasingly frustrated at impenetrable, unintuitive concepts using impenetrable, unintuitive notation* Maelin (Talk | Contribs) 13:34, 13 May 2007 (UTC)

The kernel is not {0}! Consider a special case: R=Z, I = 4Z say. Z acts on Z/4Z by multiplication. Which elements of Z send everything to 0 by this action? Algebraist 13:40, 13 May 2007 (UTC)

This is why I hate this damn coset notation and in fact the entire concept. Is the kernel I then? Maelin (Talk | Contribs) 13:45, 13 May 2007 (UTC)

If you think so, prove it. Aside: I myself have always found ring quotients (unlike group quotients) very intuitive, since they generalise the natural idea of modular arithmetic on Z. To quotient by an ideal I is just to declare all elements of I to equal 0, and accept all consequences. Modules, on the other hand, have always baffled me. Algebraist 13:55, 13 May 2007 (UTC)

Suppose r is in I. Then ra is in I, so ra+I = 0+I, so r is in the kernel. Conversely, suppose r is not in I. Consider a=1. Then ra+I = r+I ≠ 0+I since r is not in I, so r is not in the kernel. Therefore the kernel must be I. QED. Right? Maelin (Talk | Contribs) 14:03, 13 May 2007 (UTC)

Correct Algebraist 14:19, 13 May 2007 (UTC)

Okay, but I don't understand how I can use the isomorphism theorems. I haven't seen any results anywhere that allow me to derive an equality from any kind of isomorphism that look even remotely related to this problem. Should this be a proof by contradiction, maybe? I can't even see how finding the kernel of the R-action on R/I is relevant. Maelin (Talk | Contribs) 14:23, 13 May 2007 (UTC)

Because as I claimed above, the isomorphism between R/I and R/J (as modules) implies the actions have the same kernel. The isomorphism theorems are (I think) useless here, which is unsurprising since you are given an isomorphism and asked to prove an equality, which the isom. theorems can't give you. Algebraist 14:59, 13 May 2007 (UTC)

So you're saying that the kernel of the R-action on R/J must also be I, because of the isomorphism? That isn't obvious to me at all. Maelin (Talk | Contribs) 21:21, 13 May 2007 (UTC)

(cancelling indents) It wasn't obvious to me either (though it is now). As I said earlier, I have little intuition for modules. It is however, obviously a result that might be true. So what you do is, you try to prove it. If you succeed, you've learned something, and hopefully improved your understanding of modules. If you fail, then work out why you failed, use this to construct a counterexample, and you've improved your understanding of modules. Algebraist 09:19, 14 May 2007 (UTC)

That's really the issue, you see. I haven't the faintest idea how I might prove that. I can't see any way to connect the kernel of the R-action on R/I and the R-action on R/J even with the isomorphism. What is the basic idea of the proof that the kernels are the same? Maelin (Talk | Contribs) 11:46, 14 May 2007 (UTC)

Okay, I've found a proof that goes via another way. Now I'm trying to find the counterexample to show that it's not true for R/I ≈ R/J as rings. I know the integers can't offer a counterexample, since the only ideals in Z are nZ, and Z / nZ = Z_n, and no two of them are isomorphic without having the same n. I also know the rationals, reals, etc won't, since they're all fields and fields don't contain any nontrivial ideals. I suspect it will be some polynomial ring, but I have only a vague grasp of ideals in them and virtually no idea what quotient rings look like or how they might be isomorphic. Can anybody help? Maelin (Talk | Contribs) 14:21, 14 May 2007 (UTC)

Again, Tesseran has done this already up there (more or less). Aside:proof that the actions of R on A and B have the same kernel, given that A and B are isomorphic R-modules: let r be such that ra=0 for all a in A. Then Φ(ra)=0 for all a. But Φ(ra)=rΦ(a) and Φ is surjective so rb=0 for all b in B. Thus (r in kernel of action on A) implies (r in kernel of action on B) and the reverse holds by the same argument using Φ^-1. Thus the kernels are equal. Algebraist 16:34, 14 May 2007 (UTC)

To answer your earlier question, the idea, as so often when doing proving very low-level facts in algebra, is not to have any ideas. The closest thing to an idea needed is to prove two sets are equal by showing each contains the other, but since that's the definition of set equality, I'm not sure that counts. Algebraist 16:38, 14 May 2007 (UTC)

Transformation

Hi there, I have a question on transformation.

What transformations must you apply to y=x^2 to create the new graph? List the transformations in the order you would apply them.

a) y=-x^2+9

b) y=(x-3)^2

c) y=(x+2)^2-1

d)y=-2(x-4)^2+16

this is not for homework. I just want to know.

thank you —The preceding unsigned comment was added by 76.64.55.160 (talk) 14:27, 13 May 2007 (UTC).

Homework or not, this is straight out of some textbook. I suggest the book's section on curve sketching and transformations as a fruitful source of information. Alternatively, you might find playing around with an online graphing calculator (there are loads of them) aids your understanding. Algebraist 15:11, 13 May 2007 (UTC)

Try GCalc 2: [2]. StuRat 18:08, 13 May 2007 (UTC)

What's the graph in the first place? You'll have to use the concepts of transformations taught in your textbook. Splintercellguy 03:52, 14 May 2007 (UTC)

Minimum value

Hi there, I have a question about minimum value.

How can you find the minimum value of the graph of y=3x^2-9x-30 and use it to express the equation in vertex form and can you show the work so I can understand. thanks. by the way, it is not homework. —Preceding unsigned comment added by 76.64.55.160 (talk • contribs)

What do you mean by vertex form? I've never heard the term before. Algebraist 15:19, 13 May 2007 (UTC)

Don't worry, Google told me. This is what was taught to me under the name Completing the square. Algebraist 15:23, 13 May 2007 (UTC)

Transformation 2

What transformation must be applied to the graph of y=x^2 to produce the graph y=2x^2-12x+7 and can you justify you reason for the question? thanks. this is not homework. —The preceding unsigned comment was added by 76.64.55.160 (talk) 14:38, 13 May 2007 (UTC).

Observe that 2x²-12x+7=2(x-3)²-11. This has the same form as one of your previous questions. Algebraist 15:17, 13 May 2007 (UTC)

As an aside, can someone explain why some usernames are blue and some are red? I used to think it was denoting status, in that some had earned the right to be in some sort of in group, but this questioner's ID is in both colours in the various posts. -- 86.132.167.85 19:03, 13 May 2007 (UTC)

One links to Special:Contributions/76.64.55.160, which is not empty, the other to User:76.64.55.160, which does not exist.  --Lambiam^Talk 20:08, 13 May 2007 (UTC)

To be precise, the "Special:Contributions" link would be blue even if the list was empty. In fact, all links to "Special:" pages (even Special:Therereallyisnosuchpage) are always blue, period. But yes, usually blue links mean the target page exists, and red links that it doesn't. —Ilmari Karonen (talk) 20:20, 13 May 2007 (UTC)

In my case, it's because I don't write on my User page. The page hasn't officially been created, so it's red, a broken link. Black Carrot 20:55, 13 May 2007 (UTC)

Effect of "reroll twice" on probability distribution

Objects are to be randomly generated by rolling a die and consulting a table n times. Most of the table entries will generate one object per roll, but a "nothing" entry (P=x) is also present, as is a "reroll twice" entry (P=y). Since rerolls-twice can in turn generate more rerolls-twice, there is no upper bound on the number of objects, and the expected number of objects is not simply 2y + 1-x-y. What is the probability distribution? NeonMerlin 15:25, 13 May 2007 (UTC)

Too lazy to do full analysis, but if y>1/2, then with positive probability you sit rolling dice for ever (see Branching process). If y=1/2, then unless x=1/2 also the expected number of objects is infinite, though the actual number is almost surely finite. If y<1/2 (hopefully the case!) then the expected number E is finite and satisfies E=2yE +1-x-y, i.e. E = (1-x-y)/(1-2y). No idea about the actual distributions, I'm afraid. Algebraist 16:03, 13 May 2007 (UTC)

In case you want this, the positive probability I alluded to above is ${\frac {2y-1+{\sqrt {1-4y(1-y)}}}{2y}}$. Algebraist 16:11, 13 May 2007 (UTC)

The following is not a closed formula, but a recurrence relation that allows you to calculate the probabilities. If p_n stands for the probability of observing n objects, and xy < 1/4, then

p₀ = 2x / (1 + √(1−4xy)) ;

p₁ = (1−x−y) / (1−2yp₀) ;

p_n = yS / (1−2yp₀) for n > 1, where S = Σ_{k=1,...,n−1} p_kp_n−k .

If xy = 1/4, x = y = 1/2, and then p₀ = 1, so p_n = 0 for n > 0.  --Lambiam^Talk 20:00, 13 May 2007 (UTC)

So that's when the table is initially to be consulted once? NeonMerlin 01:48, 17 May 2007 (UTC)

Yes. For k lookups (say), the probability of n items will be $\sum _{a_{1}+a_{2}+\dots +a_{k}=n}p_{a_{1}}p_{a_{2}}\dots p_{a_{k}}$ (with the sum over all ordered k-tuples of nonnegative integers summing to n). Algebraist 16:41, 17 May 2007 (UTC)

Gun barrel bore

Why is 9mm such a popular barrel size for handguns? Why not 10mm or 8mm? Thanks, WSC

The reason is historical, not mathematical. Other sizes have also been used. See 9 mm Luger Parabellum, List of handgun cartridges, and Cartridge (firearms). nadav 17:20, 13 May 2007 (UTC)

Our cartridge page is not for victorian censored children. I read : "The alteration of the military flint-lock to the percussion musket was easily accomplished by replacing the powder pan by a perforated nipple, and by replacing the cock or hammer which held the flint by a smaller hammer with a hollow to fit on the nipple when released by the trigger.".

Nothing changes. You may find an analogy with the width or railways, originating from Roman waggons. For guns, ask yourself who manufactured iron tubes and why when the gun was invented. -- DLL ^{.. T} 19:10, 15 May 2007 (UTC)

(related to projective varieties) Is a quotient ring of a graded ring a graded ring as well?

Hello,

Let $R$ be a graded ring (see [3] for a definition)

Let $I$ be some ideal in that ring $R$ Can the quotient ring $R/I$ be interpreted as a graded ring as well?

This is actually the reason why I'm asking : I am using a book where S stands for all polynomials in $n+1$ variables over a field $k$, where $Y$ is a projective variety in $\mathbb {P} ^{n}$; and where $I(Y)$ is the ideal generated by all homogeneous polynomials vanishing on all points in $Y$. They let $S(Y)$ denote the quotient ring $S/I(Y)$ Then they speak of $S(Y)_{(}x_{0})$ , which stands for the localisation with the multiplicative set$\{1,x_{0},x_{0}^{2},\dots \}$ . And then they speak of $S(Y)_{(}x_{0})$ , which stands for "the degree zero part" of $S(Y)_{(}x_{0})$ . Now I do not understand how one can speak of "degree zero part" without the ring being graded?

As you can see, I'm really confused. Could it be that my book is a bit sloppy with definitions. I hope someone with some experience in this matter will manage to get me out of this darkness. Thank you and greetings, Evilbu 18:24, 13 May 2007 (UTC)

You'll need for your ideal to be homogenous, i.e. the generators all homogenous elements of the ring. If this is the case, then the quotient ring inherits grading from the starting ring. Michiexile 18:37, 13 May 2007 (UTC)

Okay, that is good, since I(Y) is a homogeneous ideal right? But do go on, please, how would you "grade" that new quotient ring? I mean : how would you decomposition that quotient ring? The answer may seem straightforward at first, but it isn't, since there are several representants of the same quotient class.Evilbu 19:25, 13 May 2007 (UTC)

I don't know about your specific case, but we know $R/I=\sum (R^{n}+I)/I$, assuming that I is homogenous. I can include the proof that this is true if you like. nadav 01:36, 15 May 2007 (UTC)