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January 12

Normal Closures Qual. Exam

I'm stuck on the problem: Given a non-normal separable extension [E: F] = 4, bound the degree [K: F] of the normal closure of E. [Bergman] from a grad. qual. exam http://www.math.harvard.edu/graduate/quals/topics/galois.pdf. Please help! I think that [K: F] is bounded by 8 by testing a couple of examples like F = Q, E = Q[a] where a⁴ = 2 where the minimal polynomial of a over Q has a splitting field of degree 8 and other examples. Obviously [K: F] is bounded by 24, but I'm not sure that's what the question is after (I don't even know whether 24 is a strict bound, that you can get a normal closure of degree 24 in the problem). I haven't much experience with normal closures, but I don't believe I require it for this question. The other qual. exam problems in the link are dead easy and that's why this one question is bugging me. Thanks. —Preceding unsigned comment added by 122.109.42.56 (talk) 06:25, 12 January 2010 (UTC)[reply]

I'm fairly certain that the bound 24 is optimal. — Emil J. 11:11, 12 January 2010 (UTC)[reply]

Can you show me how [E: F] = 4 can exist with [K: F] = 24? Thanks for your answer but I can't think of an explicit example of how the degree could get as high as 24. Why I'm unsure that the degree is 24, is because of the specificity of the problem ([E: F] = 4, and not [E: F] = n for any n), and that it's a qual. problem.—Preceding unsigned comment added by 122.109.42.56 (talk) 11:15, 12 January 2010 (UTC)[reply]

Let

K=\mathbb {Q} (x_{0},x_{1},x_{2},x_{3})

(the field of rational functions in 4 variables, that is, the x_i are algebraically independent). Any

\sigma \in \mathrm {Sym} _{4}

induces an automorphism of K by permuting the corresponding variables. Let F be the fixed field of all these automorphisms, and let E = F(x₀). The minimal polynomial of x₀ over F is

f(x)=\prod _{i<4}(x-x_{i})\,

, hence [E : F] = 4. The normal closure of E (i.e., the splitting field of f over F) is K, and the construction ensures Gal(K/F) = Sym₄, hence [K : F] = 24. Of course, the same argument works for any n, there's nothing special about 4. — Emil J. 11:30, 12 January 2010 (UTC)[reply]

Maybe I should also stress that this F can be easily described explicitly, namely it is the field generated by the elementary symmetric polynomials:

{F=\mathbb {Q} (x_{0}+x_{1}+x_{2}+x_{3},x_{0}x_{1}+x_{0}x_{2}+x_{0}x_{3}+x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3},x_{0}x_{1}x_{2}+x_{0}x_{1}x_{3}+x_{0}x_{2}x_{3}+x_{1}x_{2}x_{3},x_{0}x_{1}x_{2}x_{3}).}

— Emil J. 18:42, 12 January 2010 (UTC)[reply]

Thanks muchly so F can be chosen as the fixed field of that Galois group. Just to investigate furthur suppose F = Q. Is it then true that the bound of 8 is optimal? I've been basically looking at the problem for F = Q so that's probably why I wasn't thinking clearly but in the case F = Q my bound seems to work. Thanks.—Preceding unsigned comment added by 122.109.42.56 (talk) 11:58, 12 January 2010 (UTC)[reply]

Please, sign your posts, and don't remove signatures supplied by others without adequate replacement. This is a basic rule of conduct here.

No, you can get degree 24 even for F = Q. The problem is equivalent to asking for a normal extension K of Q such that Gal(K/Q) = Sym₄. This is a particular instance of the inverse Galois problem. While the general problem for arbitrary finite groups remains unsolved, many special cases are known, and in particular, one can always find such an extension for any Sym_n (the argument is described at Inverse Galois problem#Symmetric and alternating groups; your polynomial is x⁴ − t(x + 1) for a suitably chosen rational number t). — Emil J. 13:35, 12 January 2010 (UTC)[reply]

If you're concerned about your IP address being visible, don't worry (or worry more, depending on how you look at it). Your IP address is publicly available in the history page (and this is common knowledge, so don't accuse me of further compromising your privacy). -- Meni Rosenfeld (talk) 15:33, 12 January 2010 (UTC)[reply]

And if you are indeed concerned about your IP address being visible, the easy solution is to create an account. — Emil J. 15:53, 12 January 2010 (UTC)[reply]

Sorry about that, I was just worried that people may find me out. I didn't know you were adding the signature, so I thought it was alright to remove it. Thanks for the help. —Preceding unsigned comment added by 122.109.42.151 (talk) 02:15, 13 January 2010 (UTC)[reply]

solution of differential equation of motion of a quarter car

Where can i find a solution of differential equation of motion of a quarter car? —Preceding unsigned comment added by 113.199.176.179 (talk) 17:04, 12 January 2010 (UTC)[reply]

What equation is that? And what is a "quarter car", anyway? — Emil J. 17:25, 12 January 2010 (UTC)[reply]

"Quarter car model" gets a lot of Google hits (but I haven't read them !). Gandalf61 (talk) 17:29, 12 January 2010 (UTC)[reply]

Something like Quarter Midget racing, may be? --CiaPan (talk) 18:16, 12 January 2010 (UTC)[reply]

The only thing you can look at by only looking at a quarter of a car is the suspension of one wheel, i.e. disregarding all effects (lateral, turning, anti-roll bar, drag etc.) which require knowledge of the other wheels or the body. The page Vehicle dynamics links to Tuned mass damper which may be what you need.--JohnBlackburne^words_deeds 18:45, 12 January 2010 (UTC)[reply]

Drag races are traditionally a quarter mile... -- Coneslayer (talk) 20:05, 13 January 2010 (UTC)[reply]

Locally invertible function from R^2 to R

Hi all,

I want to show that for $f(x,y)=(x,x^{3}+y^{3}-3xy)$ , and $C=\{(x,y):x^{3}+y^{3}-3xy=0\}$ , f is locally invertible around each point of C except (0,0) and $(2^{\frac {2}{3}},2^{\frac {1}{3}})$ , i.e. show that there are open sets U and V about any $(x_{0},y_{0})$ in C (but for the aforementioned 2 points) such that f maps bijectively: U $\to$ V. Could anyone give me a hand? I've tried saying f(a,b)=f(c,d), so a=c, and then if b is not equal to d, we get 3a= $b^{2}+bd+d^{2}$ , and I'm not sure where to go from here or whether I am indeed going in the right direction. I haven't actually used the fact that we're local to points where the second coordinate of f(x,y) is 0, and I can't really see what makes the above 2 points special either. Could anyone give me a hand finishing this off?

I also want to find the derivative of the inverse function (locally) - what would be the neatest way to find this?

Many, many thanks! Spamalert101 (talk) 20:09, 12 January 2010 (UTC)[reply]

Jacobian matrix and determinant might be a good place to start. Basically the Jacobian gives you the linear transformation that your function looks like locally. The two points you mentioned are the points on C where the Jacobian determinant is zero. The derivative of the inverse function at a point will be the inverse of the Jacobian there. Note that won't work when the determinant is zero. Rckrone (talk) 21:58, 12 January 2010 (UTC)[reply]

Just want to add, if the Jacobian determinant is zero that doesn't necessarily imply a problem, but those are the places where things can potentially go wrong. Rckrone (talk) 00:16, 13 January 2010 (UTC)[reply]

To be precise, if the Jacobian is zero at a point, the map is not locally invertible there in the C¹ sense, that is with a C¹ local inverse. So, to give a complete answer to the question as you stated it, now you just have to look more closely at the two points where the Jacobian determinant vanishes (as Rckrone sais, the Jacobian determinant being zero doesn't necessarily imply that the map is not locally bijective there). For instance, the map is not locally bijective at (0,0), because e.g. for any 0 < ε < 1 there are at least two distinct solutions (x₁,y₁) and (x₂,y₂) of f(x,y) = (ε,0) close to (0,0), namely with x₁ = x₂ = ε and 0 < y₁ < ε^1/2 < y₂ < 2ε^1/4 (just look at the sign of y³ - 3εy + ε³). Can you see how to do the other point? (It is also possible a more detailed geometric description of the local structure of C at these points, of course) --pm a 08:24, 13 January 2010 (UTC)[reply]

That's great - thanks all, i've got it now :) Spamalert101 (talk) 13:46, 13 January 2010 (UTC)[reply]

Estimating number ratios

For instance 2.68:1 ratio, if I want to round to nearst 0-5-0 should 2.68 round up to 3 or it should round to 2.5 if I'm douing halfway rounding scale. For 3.04:1 to estimating sentence is it could be cite as more than 3 times xx and vice versa, or 3.04 is usually said like about 3 times the xx vice versa since 3.04:1 is only 4 cents past 3?--209.129.85.4 (talk) 20:26, 12 January 2010 (UTC)[reply]

It's unclear to me what you are asking; maybe you will find something helpful at rounding. If I were to choose between rounding 2.68 to either 2.5 or to 3, I would likely choose to round to the nearer number. As for your second question, it is correct to say that 3.04 is more than 3 and it is also correct to say that 3.04 is about 3, so either sentence is correct. Eric. 131.215.159.171 (talk) 09:14, 13 January 2010 (UTC)[reply]

I was also unclear about what you were asking. It is not usual to round to the nearest half, but if you really need to do this, then any value more than .00 and less than .25 gets rounded down, and so does any value more than .50 and less than .75
For values from .25 up to .50, and values from .75 up to the next whole number, the convention is to round up. Dbfirs 20:01, 13 January 2010 (UTC)[reply]