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Black holes are controversial. (Just browse reader comments from partisans of various sorts of "alternative" astrophysical theories – which can be found at the end of many articles dealing with black holes that allow commenting by the general public.)

Nevertheless, very solid evidence has been accumulated over the years for the existence of two types of black holes: stellar-mass black holes with masses from 3 to several tens of solar masses (M_⊙), and supermassive black holes, which are vastly larger – generally millions to billions M_⊙. Concerning some of the evidence, see here.

Stellar-mass black holes are easy to explain as supernova remnants, while supermassive black holes seem to be an inseparable concomitant of the development of all galaxies.

Perhaps surprisingly, however, there has been very little evidence for the existence of black holes of intermediate mass. If such black holes exist at all, the processes that form them must be rather more unusual. Evidence for the existence of intermediate mass black holes has been reported in the past. (Theres some discussion here of possible black holes of mass less than a million M_⊙.)

But because black holes, by their nature, are difficult to observe directly, and so their existence must be inferred indirectly, it has been difficult to come up with relatively unambiguous evidence. Now we have announcements of better evidence in two cases.

New Class Of Black Holes Discovered (7/1/09)

A new class of black hole, more than 500 times the mass of the Sun, has been discovered by an international team of astronomers.

The finding in a distant galaxy approximately 290 million light years from Earth is reported today in the journal Nature.

Until now, identified black holes have been either super-massive (several million to several billion times the mass of the Sun) in the centre of galaxies, or about the size of a typical star (between three and 20 Solar masses).

The new discovery is the first solid evidence of a new class of medium-sized black holes.

Important discoveries often dont come by themselves. Other researchers and teams tend to report related results at the same time. And this is no exception. The above reports concern a candidate object in a galaxy (ESO 243-49) about 290 million light-years away. But theres also a report of an object much closer, in the globular cluster M54 (more here), which is only about 87,000 light-years away. Its thought to belong, actually, not to the Milky Way itself, but rather to the Saggitarius Dwarf Elliptical Galaxy, a satellite of the Milky Way.

Density and kinematic cusps in M54 at the heart of the Sagittarius dwarf galaxy: evidence for a 10⁴ M_⊙ Black Hole?

We report the detection of a stellar density cusp and a velocity dispersion increase in the center of the globular cluster M54, located at the center of the Sagittarius dwarf galaxy (Sgr). The central line of sight velocity dispersion is 20.2 +/- 0.7 km/s, decreasing to 16.4 +/- 0.4 km/s at 2.5" (0.3 pc). Modeling the kinematics and surface density profiles as the sum of a King model and a point-mass yields a black hole (BH) mass of ~ 9400 M_⊙. However, the observations can alternatively be explained if the cusp stars possess moderate radial anisotropy.

M54

Further reading (ESO 243-49 candidate object):

Finally, an Average Black Hole (7/1/09) – ScienceNOW

New Candidates for Midsize Black Holes (7/3/09) – Sky and Telescope

An intermediate-mass black hole of over 500 solar masses in the galaxy ESO – Nature research article

XMM-Newton discovers a new class of black holes (7/1/09) – ESA press release

New Observations Suggest Mid-Size Black Holes Exist (7/1/09) – Space.com

Black holes: now available in size M (7/2/09) – Cosmos magazine

New Class of Black Hole Found? (7/1/09) – National Geographic

Astronomers Discover Medium-Sized Class of Black Holes (7/1/09) – Universe Today

Intermediate-mass black hole (7/1/09) – Science Centric

Astronomers sniff intermediate mass black hole (7/2/09) – The Register

Astronomers Size Up a Candidate for Midsize Black Hole (7/1/09) – Scientific American

New Class of Black Holes Discovered (7/1/09) – Wired

X-rays are smoking gun for middleweight black holes (7/1/09) – New Scientist

A New Kind of Black Hole (7/2/09) – Smithsonian.com

Team May Have Found Intermediate Black Hole (7/6/09) – New York Times

Further reading (M54 candidate object):

Density and kinematic cusps in M54 at the heart of the Sagittarius dwarf galaxy: evidence for a 10^4 M_sun Black Hole? – Astrophysical Journal research article

Tags: black holes

Our discussion of algebraic number theory returns by popular demand. Way back last April we presented some generalities on factorization of prime ideals in extension fields. (For explanation of what that means, including other necessary concepts, youll have to review earlier installments of this series, which can be found here.)

If this is all Greek to you, I apologize, but thats unavoidable at this stage of a rather technical subject. You may want to go back to the earliest parts of the series to see how the subject got its start and why it may be interesting.

The discussion in the previous installment probably seems rather dry and abstract, but when we look at simple examples, such as quadratic extensions, why its interesting becomes clearer.

Because of how the ramification indexes and inertial degrees are related, for any prime ideal (p) of ℤ there are only three different possibilities for how the ideal factors in the ring of integers of a quadratic extension:

1. (p)=P₁⋅P₂, so (p) splits completely. (e=f=1, g=2)
   
2. (p)=P is a prime ideal in ℚ(√d), so p is inert. (e=g=1, f=2)
   
3. (p)=P² where P is prime, and p is ramified. (f=g=1, e=2)
   

It turns out that there are simple criteria for each of these cases. But figuring out what the criteria are is tricky.

Recall that in ℚ(√3) we found (13)=(4+√3)⋅(4-√3) and (-11)=(8+5√3)⋅(8-5√3), so both (11) and (13) split completely. Clearly, (3)=(√3)², so (3) is an example of a prime ideal if ℤ that is ramified. How about an example of a prime ideal that is inert in the extension? This is a little harder for a couple of reasons. (p) will be inert just in case it neither splits nor is ramified, but we dont yet have simple criteria to rule out those cases.

So lets back up a little and look at the details. We found examples where (p) splits in the integers of ℚ(√3) by solving the equation ±p=a²-3b², because that gave elements a±b√3 whose norm was ±p. Being able to find such elements guaranteed that the prime split. But ℚ(√d) with d=3 is a special case, since here d≡3 (mod 4). In that case, and also if d≡2 (mod 4), the integers of the extension have the form a+b√d with a,b∈ℤ.

If d≡1 (mod 4), integers can also have the form (a+b√d)/2, with a,b∈ℤ, and we might have a factorization like (p) = ((a+b√d)/2)⋅((a-b√d)/2), so we would have also to consider solvability of ±4p=a²-3b². If we were to look at solvability of the approriate equation, according as to whether or not d≡1 (mod 4), the solvability would be a sufficient condition for (p) to split (or ramify if a=0). Notice that this sufficient condition for (p) to split holds regardless of whether or not ℚ(√d) is a PID.

Now we need to find a convenient necessary condition for (p) to split. Unfortunately, solvability of one simple equation is not a necessary condition in general. It would be, as well see in a minute, if the ring of integers of ℚ(√d) happens to be a PID, as is true when d=3. However, in quadratic extensions where the ring of integers isnt a PID, being unable to solve the applicable equation doesnt guarantee (p) cannot split, because there might be non-principal ideals that are factors of (p).

So lets ignore that problem for a moment and just focus on the case where the ring of integers of ℚ(√d) is a PID. Can we then find a necessary condition on p for (p) to split or ramify, i. e. for (p) to not be a prime ideal of the integers of ℚ(√d)? That is, what must be true about p if (p) splits or ramifies?

If (p) splits or ramifies, then (p)=P₁⋅P₂ for nontrivial ideals P_i. (The ideals are the same or distinct according as (p) ramifies or splits.) Assuming ℚ(√d) is a PID, then P₁ is generated by a+b√d where both a,b∈ℤ, if d≡2 or 3 (mod 4), or else by (a+b√d)/2 with a,b∈ℤ, if d≡1 (mod 4). Likewise, the conjugate ideal P₂ is generated by a-b√d or (a-b√d)/2. Since p∈P₁⋅P₂, by definition of a product of ideals, p is of the form p = ε(a+b√d)(a-b√d) = ε(a²-db²) or p = ε(a+b√d)(a-b√d)/4 = ε(a²-db²)/4 for some integer ε of ℚ(√d).

Recall that the norm of an element of a Galois extension field is the product of all conjugates of the element. So for an element that is also in the base field, the norm (with respect to a quadratic extension, which is always Galois) is the square of the element. Taking norms of both sides of the possible equations, then either p² = N(&epsilon)(a²-db²)² or 16p² = N(&epsilon)(a²-db²)². For simplicity, consider just the first case. Then N(ε) is a positive integer that has to be 1, p, or p². If N(ε)≠1 then N(a±b√d) = a²-db² must be ±1, so a±b√d must be a unit, and both P_i must be non-proper ideals (i. e. equal to the whole ring). Hence N(ε)=1. This will be true also in the other case (when d≡1 (mod 4)), so ±p=a²-db² or ±4p=a²-db². Consequently, solvability of the appropriate equation (depending on d mod 4), is a necessary condition for (p) to split or ramify.

So we have a necessary and sufficient condition for (p) to split or ramify in ℚ(√d), in terms of solvability of Diophantine equations, provided O_ℚ(√d) is a PID. Since the only other possibility is for (p) to be inert, we also have a necessary and sufficient condition for that.

However, still assuming that O_ℚ(√d) is a PID, we can find a further necessary condition for (p) to split or ramify. Take those equations we just found and reduce them modulo p. Then both equations become a²≡db² (mod p). Since p is prime, ℤ/pℤ is a field. Assume first that b≢0 (mod p). Then b has an inverse in the finite field. So we have d≡(a/b)² (mod p). In other words, d is a square mod p. This is the additional necessary condition we were looking for on p in order for (p) to split or ramify. Since its a necessary condition, if d is not a square mod p, then (p) must not split or ramify, and thus p is inert. And so, for d to be a non-square mod p is a sufficient condition for p to be inert.

(What if b≡0 (mod p)? Then b=b₁p. So ±p = a² - (b₁p)²d or else ±4p = a² - (b₁p)²d. Either way, p∣a, hence p² divides the right side of either equation, and hence the left side also. But thats not possible unless p=2 – which is always a special case.)

To summarize, then, let p≠2 be prime and d square-free and not 0 or 1. Then the solvability of ±tp=a²-db² (where t is 4 or 1 according as d≡1 (mod 4) or not), is sufficient for (p) to split or ramify. And if the integers of ℚ(√d) are a PID, then solvability of the appropriate equation provides a necessary and sufficient condition for (p) to split or ramify. Further, in that case, d being a square mod p is necessary for (p) to split or ramify.

Remarkably, d being a square mod p is a necessary and sufficient condition for (p) to split or ramify, even if the integers of ℚ(√d) arent a PID, but thats harder to prove. Since solvability of Diophantine equations is generally not obvious by inspection, its very convenient to have a necessary and sufficient conditions for (p) to split or ramify simply in terms of the properties of d mod p.

In the next installment, which hopefully will not be as long in coming as this one, well show a much cleaner way to state necessary and sufficient conditions for (p) to split, ramify, or remain inert, in the case of any quadratic extension of ℚ, whether or not the ring of integers is a PID. This will be done in terms of what has long been called a "reciprocity law".

However, that will be only the beginning. It turns out that there are far more general kinds of reciprocity laws for many other types of field extensions. Thats what "class field theory" is all about, and why the whole subject is so appealing, once you get the basic ideas.

Tags: algebraic number theory

Disk technology takes Nobel Prize

French scientist Albert Fert and Peter Grunberg of Germany have won the 2007 Nobel Prize for physics.

They discovered the phenomenon of "giant magnetoresistance", in which weak magnetic changes give rise to big differences in electrical resistance.

The knowledge has allowed industry to develop sensitive reading tools to pull data off hard drives in computers, iPods and other digital devices.

It has made it possible to radically miniaturise hard disks in recent years.

Well, thats good timing. I just wrote about the field of disk technology here. That article isnt about giant magnetoresistance per se, but that part of physics is certainly an important part of the big picture.

Here are other reports on the news:

Nobel prize recognizes GMR pioneers

Giant magnetoresistance, or GMR, is the sudden change in electrical resistance that occurs when a material consisting of alternating ferromagnetic and non-magnetic metal layers is exposed to a sufficiently high magnetic field. In particular, the resistance becomes much lower if the magnetization in neighbouring layers is parallel and much higher if it is antiparallel. This change in resistance is due to "spin up" and "spin down" electrons scattering differently in the individual layers.

GMR has since been used to develop extremely small and sensitive read heads for magnetic hard-disk drives. These have allowed an individual data bit to be stored in a much smaller area on a disk, boosting the storage capacity greatly. The first commercial read heads based on GMR were launched by IBM in 1997 and GMR is now a standard technology found in nearly all computers worldwide and is also used in some digital cameras and MP3 players.

More:

Discoverers Of Giant Magnetoresistance Used In Hard Drives Win 2007 Nobel Prize In Physics

Magnetic Effect Nets a Nobel (Sub. rqd.)

Effect that Revolutionized Hard Drives Nets a Nobel (Sub. rqd.)

The physics prize inside the iPod (Sub. rqd.)

Nobel Focus: Sensitive Magnetic Sandwich

Small is beautiful: Incredible shrinking memory drives new IT (also here)

2007 Nobel Prize in Physics

Hard-disk breakthrough wins Nobel Prize

Hard Drive Pioneers Win Nobel Prize

Physics of Hard Drives Wins Nobel

Physics Nobel Goes to German, Frenchman (AFP)

A Nobel Nod for Giant Discovery

Hard Drive Pioneers Win Nobel Prize

Nobel research means you can read this

Hard disk pioneers win physics Nobel

Disk technology takes Nobel Prize

Heres the next installment of our series on algebraic number theory. In the last installment we had a quick look at groups and rings. Now its time to look at field theory, with special emphasis on what is known as Galois theory. The latter is all about developing a concise description of the relations among the roots of an irreducible polynomial equation using group theory. Some of this theory was famously sketched out in 1832 by Évariste Galois on the night before a duel in which he died.

Galois theory makes it possible to prove several well-known results, such as the impossibility of expressing the solution of some fifth degree polynomial equations in terms of radicals and the impossibility of trisecting some angles with straightedge and compass. We wont go into that, but instead we will eventually see Galois theory used frequently in algebraic number theory.

A field is simply a ring whose multiplication is commutative, has an identity element, and has multiplicative inverses for all elements except the additive identity element. Weve already mentioned several examples of fields, specifically number fields, which are algebraic extensions of finite degree of the rationals Q. (I. e., each element of a such a field is an algebraic number in some finite extension of Q.) More exotic examples of fields certainly exist, though, such as finite fields, fields of functions of various kinds, p-adic number fields, and certain other types of local fields. If you go far enough in algebraic number theory, youll encounter all of these.

The most important set of facts about fields for our purposes lie in what is known as Galois theory. This is the theory developed originally by Évariste Galois to deal (among other things) with the solvability or non-solvability, using radicals, of algebraic equations. It tells us a lot about the structure of field extensions in terms of certain groups – called Galois groups – which are constructed using permutations of roots of a polynomial which determines the extension. (Permutations are 1-to-1 mappings of a set to itself that interchange elements.) A little more precisely, a Galois group consists of automorphisms of a field – i. e. maps (functions) of the field to itself which preserve the field structure. All such automorphisms, it turns out, can be derived from permutations of the roots of a polynomial – under the right conditions.

The importance of Galois theory is that it sketches out some of the "easy" background facts about a given field extension, into which some of the more difficult facts about the algebraic integers of the extension must fit.

Before we proceed, lets review some notations and definitions that will be used frequently. Suppose F is a field. For now, we will assume F is a subset of the complex numbers C, but not necessarily a subset of the real numbers R. If x is an indeterminate (an "unknown"), then F[x] is the set of polynomials in powers of x with coefficients in F. F[x] is obviously a ring. If f(x)∈F[x] is a polynomial, it has degree n if n is the highest power of x in the polynomial. f(x) is monic if the coefficient of its highest power of x is 1. If f(x) has degree n, it is said to be irreducible over F if it is not the product of two (or more) nonconstant polynomials in F[x] having degree less than n.

A complex number α, which is not in F, is algebraic over F if f(α)=0 for some f(x)∈F[x]. f(x) is said to be a minimal polynomial for α over F if f(x) is monic, f(α)=0, and no polynomial g(x) whose degree is less than that of f(x) has g(α)=0. (Note that any polynomial such that f(α)=0 can be made monic without changing its degree.) A minimal polynomial is therefore irreducible over F. F(α) is defined to be the set of all quotients g(α)/h(α) where g(x) and h(x) are in F[x] and h(α)≠0. F(α) is obviously a field, and it is referred to as the field obtained by adjoining α to F.

If E is any field that contains F, such as F(α), the degree of E over F, written [E:F], is the dimension of E as a vector space over F. (Usually this is assumed to be finite, but there are infinite dimensional extensions also.) It is relatively easily proven that if α is algebraic over F and if the minimal polynomial of α has degree n, then [F(α):F]=n. Of course, more than one element can be adjoined to form an extension. For instance, with two elements α and β we write F(α,β), which means (F(α))(β). (Or (F(β))(α) – the order doesnt matter.)

We will frequently need one more important fact. Suppose we have two successive extensions, involving three fields, say D⊇E⊇F. This is called a tower of fields. Then D is a vector space over E, as is E over F. From basic linear algebra, D is also a vector space over F, and vector space dimensions multiply. Consequently, in this situation we have the rule that degrees of field extensions multiply in towers: [D:F]=[D:E][E:F].

Now were almost ready to define a group, called the Galois group, corresponding to an extension field E⊇F. However, Galois groups cant be properly defined for all field extensions E⊇F. The extension must have a certain property. Here is the problem: The group we want should be a group of permutations on a certain set – the set of all roots of a polynomial equation. But consider this equation: x³-2=0. One root of this equation is the (real) cube root of 2, 2^1/3. The other two roots are ω2^1/3 and ω²2^1/3 where ω=(-1+√-3)/2. You can check that ω³=1 and ω satisfies the second degree equation x²+x+1=0. ω is called a root of unity, a cube root of unity in particular. (Roots of unity, as well see, are very important in algebraic number theory.) Now, the extension field E=Q(2^1/3) is contained in R, but the other roots of x³-2=0 are complex, so not in the extension E. This means that it isnt possible to find an automorphism of E which permutes the roots of the equation. Hence we cant have the Galois group we need for an extension like E.

The property of an extension E⊇F that we need to have is that for any polynomial f(x)∈F[x] which is irreducible (has no nontrivial factors) over F, if f(x) has one root in E, then all of its roots are in E, and so f(x) splits completely in E, i. e. f(x) splits into linear (first degree) factors in E. An equivalent condition (as it turns out), though seemingly weaker, is that there be even one irreducible f(x)∈F[x] such that f(x) splits completely in E but in no subfield of E. That is, E must be the smallest field containing F in which the irreducible polynomial f(x)∈F[x] splits completely. E is said to be a splitting field of f(x). The factorization can be written

f(x) = ∏_1≤i≤n (x - α_i)

with all α_i∈E, where n is the degree of f(x). (Remember that we are assuming f(x) is monic.) When this is the case, E is generated over F by adjoining all the roots of f(x) to F. In this case it can be shown that the degree [E:F] is the same as the degree of f(x).

An extension that satisfies these conditions is said to be a Galois extension, and it is the kind of extension we need in order to define the Galois group G(E/F). (Sometimes the type of extension just described is called a normal extension, and a further property known as separability is required for a Galois extension. As long as we are dealing with subfields of C, fields are automaticaly separable, so the concepts of Galois and normal are the same in this case.)

Suppose E⊇F isnt a Galois extension. If E is a proper extensions of F (i. e. E≠F), if α∈E but α∉F, and if f(x) is a minimal polynomial for α over F, then the degree [E:F] of the extension is greater than or equal to the degree of f(x). The degrees might not be equal, because all the roots of f(x) must be adjoined to F to obtain a Galois extension, not just a single root. If α is (any) one of the roots, [F(α):F] is equal to the degree of f(x). But this is the degree [E:F] only if α happens to be a primitive element for the extension, so that E=F(α), which isnt usually the case, and certainly isnt if E isnt a Galois extension of F.

In the example above with f(x)=x³-2, we have E = Q(ω,2^1/3) = Q(ω)(2^1/3), [Q(ω):Q]=2 and [Q(ω,2^1/3):Q(ω)]=3, so the degree of the splitting field of f(x) over Q is 6, because degrees multiply. Q(2^1/3)⊇Q is an example of a field extension that is not Galois. But Q(ω,2^1/3)⊇Q(ω) is Galois, since f(x) is irreducible over Q(ω) but splits completely in the larger field. Likewise, Q(ω)⊇Q is Galois, and in fact all extensions of degree 2 are Galois. (If f(x)∈Z[X] is a quadratic which is irreducible over Q and has one root in E, then the roots are given by the quadratic formula and involve √d for some d∈Z, so if one is in E, both are.)

Well come back to this example, but first well look at a simpler one to get some idea of how Galois groups work. Consider the two equations x²-2=0 and x²-3=0. The roots of the first are x=±√2, and the roots of the second are x=±√3. We will start from the field Q and adjoin one root of each equation. This yields two different fields: E₂=Q(√2) and E₃=Q(√3). If we adjoin a root from both equations we get a larger field that contains the others as subfields: E=Q(√2,√3).

Consider the field extension E₂⊇Q first. We use the notation G(E₂/Q) to denote the Galois group of the extension. In this example, call it G₂ for short. We will use Greek letters σ and τ to denote Galois group elements in general. G₂ consists of two elements. One of these is the identity (which we denote by "1") which acts on elements of the field E₂ but (by definition) leaves them unchanged. This can be symbolized as 1(α)=α for all α∈E₂. The action of a Galois group element can be fully determined by how it acts on a generator of the field, meaning √2 in this case. So it is enough to specify that 1(√2) = √2. This Galois group has just one other element σ₂, which is defined by σ₂(√2)=-√2. An important property that a Galois group must satisfy is that the action of all its elements leaves the base field (Q in this case) unchanged. A Galois group is an example of a group that acts on a set – a very important concept in group theory. But there is an additional requirement on Galois groups: each group element must preserve the structure of the field it acts on. In technical terms, it must be a field automorphism. Well see the importance of this condition very soon.

As you can probably anticipate, the Galois group G₃=G(E₃/Q) has elements 1 and σ₃ defined by σ₃(√3)=-√3. We can now ask: what is the Galois group of the larger extension E⊇Q? It must contain 1, σ₂ and σ₃. We have to think about how (for instance) σ₂ acts on √3. The clever thing about Galois theory is that its easy to say what this action should be: σ₂ should leave √3 unchanged: σ₂(√3)=√3. In particular, σ₂(√3) cannot be ±√2 The reason is that σ₂ leaves the coefficients of x²-3=0 unchanged, and because σ₂ is a structure-preserving field automorphism it cannot map something that is a root of that equation (such as √3) to something that is not a root of that equation (±√2).

For any finite group G, the order of the group is the number of distinct elements. We symbolize the order of G by #(G). In Galois theory it is shown that the order of a Galois group is the same as the degree of the corresponding field extension. Symbolically: #(G(E/F))=[E:F]. Basically this is because we can always find a primitive element θ such that E=F(θ), and θ satisfies an equation f(x)=0, where the degree of f(x) is [E:F]. The other n-1 roots of that equation are said to be conjugate roots. We get n automorphisms, the elements of G(E/F), generated from mapping θ to one of its conjugates (or to itself, giving the identity automorphism). Since the degrees of field extensions in towers multiply, so too do the orders of Galois groups in field towers, as long as each extension is Galois. That is, if D⊇E⊇F, where each extension is Galois, then #(G(D/F)) = #(G(D/E))#(G(E/F)). In our example, the degree of the extension is [Q(√2,√3):Q] = [Q(√2,√3):Q(√2)][Q(√2):Q] = 4. So this is also the order of the Galois group G=G(Q(√2,√3)/Q), and therefore we need to find 4 elements.

Weve already identified three of the elements (1, σ₂ and σ₃). Its pretty clear that the remaining element must be a product of group elements: τ=σ₂σ₃. The product of Galois group elements is just the composition of the elements, which are field automorphisms (which happen to be derived from permutations on roots of equations), and hence they compose like any other function (or permutation). (Composition is just another term for the the function which is the result of applying one function after another.) Because of how σ₂ and σ₃ are defined, it must be the case that τ(√2)=-√2 and τ(√3)=-√3. Since E⊇Q is generated by √2 and √3, and τ is a field automorphism, we can figure out what τ(α) must be for any other α∈E. For instance, τ(√6)=√6, since √6=√2√3.

(Remember that we specified σ₂(√3)=√3. You may have been wondering why we didnt just define the action of σ₂ as an element of the full Galois group G=G(E/Q) by σ₂(√3)=-√3. Had we done that, σ₂ would have been what we found as τ, while the τ we got as the product of σ₂ and σ₃ would turn out to be the "old" σ₂, so the only difference would be a relabeling of group elements.)

For a slightly more complicated example, suppose f(x)=x²+x+1 and g(x)=x³-2, with roots ω and 2^1/3 respectively, as above. Then in the tower Q(ω,2^1/3) ⊇ Q(ω) ⊇ Q both the extensions are Galois. (We already saw this isnt so with the tower Q(ω,2^1/3) ⊇ Q(2^1/3) ⊇ Q – order matters.) So the full extension E=Q(ω,2^1/3) ⊇ Q is Galois. Its Galois group G=G(E/Q) has order 6, because 6 is the degree of the whole extension, since the intermediate extensions are of degree 3 and 2 and the degrees of the extensions multiply.

It turns out to be easy to determine the Galois group of this extension, although there are some tedious calculations needed to verify this. So bear with us a moment here. We can define two automorphisms of E that leave Q fixed, as follows. It suffices to specify them on generators of the field. Let one automorphism σ be defined by σ(&omega)=ω² and σ(2^1/3)=2^1/3. Let the other automorphism τ be defined by &tau(2^1/3)=ω2^1/3 and τ(ω)=ω. σ and τ are defined to leave elements of Q unchanged. For sums and products elements of E, σ and τ are defined to preserve the field structure, so they really are automorphisms (though, to be rigorous, this should be checked). So σ and τ are elements of the Galois group G=G(E/Q).

We can also see that σ²(ω) = σ(σ(ω)) = σ(ω²) = ω⁴ = ω, because ω³ = 1. So σ² is the identity automorphism. (Note that the exponents on σ and τ refer to repeated composition, not ordinary exponentiation, because composition "is" multiplication in the group G.) If we compute τ² and τ³ in the same way, applied to 2^1/3, we find that τ²(2^1/3) = ω²2^1/3, and τ³(2^1/3) = 2^1/3, again because ω³ = 1. Thus τ² isnt the identity automorphism, but τ³ is.

Now lets compute with the composed automorphisms στ and τσ. First, στ(2^1/3) = σ(ω2^1/3) = ω²2^1/3. However, τσ(2^1/3) = τ(2^1/3) = ω2^1/3. So we have στ ≠ τσ, because ω≠ω². Instead, we will find by a similar calculation that στ(2^1/3) = ω²2^1/3 = τ²σ(2^1/3). Hence στ = τ²σ. A little more checking will show that 1 (the identity automorphism), σ, τ, τ², τσ, and στ give a complete list of distinct automorphisms that can be formed from σ and τ. Thats just right, because G must be a group of order 6.

In abstract group theory there are only two distinct groups of order 6. (That is, distinct up to an isomorphism, which is a 1-to-1 structure-preserving map between groups that shows they are essentiall the "same" group.) One is the cyclic group of order 6, denoted by C₆. This is isomorphic to the direct product of a cyclic group of order two and one of order 3, i. e. the group C₂×C₃. However, since στ ≠ τσ, G isnt abelian, it cannot be C₆, which is abelian. The only other group of order 6 is (up to isomorphism) S₃, the group of permutations of three distinct objects, also known as the symmetric group. (An isomorphic group is the dihedral group D₃, the group of symmetries of an equilateral triangle.) Since this group is the only nonabelian group of order 6, G(E/Q) must be isomorphic to it.

Theres a whole lot more that could be said about Galois theory, but that would take up quite a bit of space, and the intention here is only to give a feel for what it is about. The basic idea to take away is this: A great deal is known about abstract groups and their subgroup structure. Galois theory is a way to "map" extensions of fields to groups and their subgroups in such a way that most of the interesting details about the extension are reflected in details about the groups, and vice versa. The group structure is sensitive to relationships among elements in the subextensions of a Galois extension. In Galois theory it is proven that there is a precise correspondence between subextensions and subgroups of the Galois group.

It thus becomes possible to infer facts about field extensions easily from a knowledge of their Galois groups. One example of the power of this method is that it made possible proving facts that had remained mysterious for hundreds of years – for example, the unsolvability by radicals of general polynomial equations of degree 5 or more, and the impossibility of certain geometric constructions by straightedge and compass alone (trisecting angles, for example).

Galois theory is an absolutely indispensible tool in algebraic number theory. It will come up again and again. We will mention other results in the theory when they are needed.

In the next installment well circle back to take a deeper look at ring theory, which is the most basic tool used in algebraic number theory – because there are generalizations of "integers" in an algebraic number field, and they are rings analogous to the familiar ring Z of ordinary integers.

Tags: algebraic number theory, field theory, Galois theory, Galois group

Snap judgments about candidates are the best way to pick winners, study suggests

After watching ten-second silent video clips of competing gubernatorial candidates, participants in the study were able to pick the winning candidate at a rate significantly better than chance. When the sound was turned on and participants could hear what the candidates were saying, they were no better than chance at predicting the winner. For the study, Benjamin and Shapiro showed 264 participants, virtually all Harvard undergraduates, ten-second video clips of the major party candidates in 58 gubernatorial elections from 1988 to 2002.

Researchers found that the accuracy of predictions based solely on silent video clips was about the same as or greater than the accuracy of predictions based on knowledge of which candidate was the incumbent and information about the prevailing economic conditions at the time of the election, including the unemployment rate and any changes in personal income for the year prior to the election.

I understand that "leadership" is important for forming consensus and hence getting things done. And charisma is a large part of what makes some people seem like "leaders".

The findings also underscore the importance of charisma as distinct from policy positions or party affiliations in winning elections.

But what if the policy positions of the candidates with more charisma in fact stink out loud? Happens all too often...

Given candidates who lack principles and ethics, is there much difference between having charisma and being a good con artist?

Tags: political science, charisma, elections

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