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Wednesday, December 29, 2010

Back Tattoos...

    Here are some recent tattoos that I've done, all of them on backs The St. Michael flying down from Heaven was this young man's first tattoo
    ...as was this young woman's flowers on her shoulder.
    This lil lucky Buddah went on the back of this kid's neck. I enjoyed doing this one...fat people are fun to shade.
    And finally, this symbolic family tree went on a young girl who's always a great person to tattoo and hang out with while she's at the shop. She always gets cool and colorful tattoos and gives me the artistic freedom I need to create great work. Everything involved in this tattoo holds meaning and represents members of her family. It was a challenge to design, but the end result is something that both of us are very proud of!Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Fire Truck in the Night...

    I captured these images of a fire truck a few nights ago in front of the tattoo shop. I enjoy experimenting with the camera to achieve these types of beautiful and unexpected results
    Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Pinstriped Toilet Seat...

    Paul, the owner of Studio One Tattoo, who's also a hot rodder and avid lover of all things custom, had me stripe this toilet seat as a Christmas gift for his friend. I hate painting on white.Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Molotov Brand...

Tuesday, December 28, 2010

Recent research findings on M87 (NGC 4486)

    M87 (Messier 87), also known as NGC 4486, is a giant elliptical galaxy, located about 53.5 million light-years away. It is noteworthy for several reasons, including the presence of an unusually large supermassive black hole (SMBH) in its active galactic nucleus, with an estimated mass of about 6.4×109 times the mass of the Sun (M), two plasma jets that emit strongly at radio frequencies and extend at least 5000 light-years from the SMBH (although only the jet pointed more towards us is readily detectable), and a population of about 15,000 globular clusters.

    The total mass of M87 is difficult to estimate, because elliptical galaxies like M87, and unlike spiral galaxies, do not tend to follow the Tully-Fisher relation between intrinsic luminosity and total mass calculated from rotation curves – which therefore includes dark matter. Estimates of the total mass of M87, including dark matter, come in around 6×1012 M within a radius of 150,000 light-years from the center. This compares with about 7×1011 M for the Milky Way, but M87 could be more than 10 times as massive.

    In other comparisons, the Milky Way has only about 160 globular clusters, and a central black hole (Sagittarius A*) with a mass of about 4.2×106 M. So M87's central black hole is about 1500 times as massive as the Milky Way's. Pretty impressive difference.



    M87 – click for 640×480 image


    Besides the recent research listed below, I've written about earlier research on M87 in these articles: Galactic black holes may be more massive than thought, Stellar birth control by supermassive black holes, Black holes in the news.

    You might also be interested in some articles from the past year on the general subject of active galaxies: Active galaxies and supermassive black hole jets, Where the action is in black hole jets, Quasars in the very early universe.


    Feedback under the microscope: thermodynamic structure and AGN driven shocks in M87 (6/29/10) – arXiv paper

    Feedback under the microscope II: heating, gas uplift, and mixing in the nearest cluster core (3/28/10) – arXiv paper

    Activity of the SMBH in M87 has a significant effect not only on the host galaxy, but also on the Virgo cluster of galaxies in which M87 is near the center. Energetic outflows of matter from near the black hole force plumes of gas out of the galaxy into the hotter intergalactic medium. The mass transported in this way represents about as much gas as is contained within 12,000 light-years of M87's center. (However, that's only about 2.5% of M87's 500,000 light-year radius.) If it had not been expelled, the gas could have formed hundreds of millions of stars.

    The first paper reports on studies using the Chandra X-ray Observatory to measure gas temperatures around M87's center. The findings include detection of 2 distinct shock wave fronts about 46 thousand light-years and 10 thousand light years from the center. This indicates that explosive events occurred about 150 million and 11 million years ago, respectively.

    The second paper uses observations from Chandra, XMM-Newton, and optical spectra to distinguish different phases of the hot gas surrounding M87's SMBH.

    Refs:
    Galactic 'Super-Volcano' in Action (8/20/10) – Science Daily (press release)
    Galactic Supervolcano Erupts From Black Hole (8/20/10) – Wired.com
    Galactic 'Supervolcano' Seen Erupting With X-Rays (9/6/10) – Space.com

    A correlation between central supermassive black holes and the globular cluster systems of early-type galaxies (8/13/10) – arXiv paper

    A study of 13 galaxies, including M87, has found a correlation between the size of a galaxy's SMBH and the number of the galaxy's globular clusters. The types of galaxies studied included nine giant ellipticals (like M87), a tight spiral, and 3 galaxies intermediate in type between spiral and elliptical. The smallness of the sample is due to the exclusion of open spiral galaxies and the further limitation to cases where good estimates of the number of globular clusters and mass of the central black hole existed.

    The correlation, in which the number of globular clusters is proportional to the black hole mass, is actually stronger than correlations between black hole mass and other galaxy properties previously studied for correlation, such as stellar velocity dispersion (an indicator of total mass), and luminosity of the galaxy's central bulge or whole galaxy (for ellipticals).

    In some cases the correlation of black hole mass with total luminosity was especially weak, but better with number of globular clusters. For instance, Fornax A (NGC 1316) is a giant lenticular galaxy with luminosity comparable to that of M87. Yet its central black hole has a mass of 1.5×108 M, 2.3% that of M87's black hole. It has 1200 globular clusters, 8% of M87's count. Clearly this is not a linear relation. Rather, the study found that the best fit was a power law with M ≈ (1.7×105)×N1.08±0.04, where M is black hole mass in units of M and N is number of globular clusters. This relation predicts a SMBH mass of 5.5×109 M for M87, which is very close, and 3.6×108 M for the SMBH mass of NGC 1316, which is high – but the SMBH mass of NGC 1316 is also unusually low in comparison with its luminosity and velocity dispersion.

    By contrast, the relation predicts that the Milky Way with a SMBH mass of 4.2×106 M should have only about 20 globular clusters, while the actual number is about 160. However, the Milky Way is a loose spiral, not one of the types that was studied, which may account for the much worse correlation. The fit is much better if only globular clusters associated with the central bulge (about 30) are considered.

    The obvious question is about why this relation between SMBH mass and number of globular clusters exists. Presumably it has much to do with the typical history of a large galaxy, which is expected to include frequent mergers with other galaxies. The existence of the relationship should provide clues to galactic history, and especially how this may be different for loose spirals like the Milky Way, in comparison with more compact galaxies.

    Refs:
    A correlation between central supermassive black holes and the globular cluster systems of early type galaxies (8/11/10) – The Astrophysical Journal
    Supermassive black holes reveal a surprising clue (5/25/10) – Physicsworld.com

    A Displaced Supermassive Black Hole in M87 (6/16/10) – arXiv paper

    It has generally been assumed that a galaxy's central SMBH is very close to the actual center of mass of the galaxy, because that is (by definition) the gravitational equilibrium point. This central point should be essentially the same as the photometric center of the galaxy, since the galaxy's stars should be distributed symmetrically around the center. Consequently, astronomers have not carefully searched for cases where a SMBH is not very near the galactic center. This lack of extensive investigation is also a result of the fact that the SMBH is often hidden inside a dense cloud of dust, so its exact position is difficult to determine. M87's SMBH (more precisely, the accretion disk around the SMBH), however, is clearly visible, and the research reported in this paper finds it is actually located about 22 light-years from the apparent galactic center.

    There are various possible reasons for this much displacement from the center, and not a lot of evidence to identify the most likely reason. Possible reasons include: (1) The SMBH is part of a binary system in which the other member is not detected. (2) The SMBH could have been gravitationally perturbed by a massive object such as a globular cluster. (3) There is a significant asymmetry of the jets. (4) The SMBH has relatively recently merged with another SMBH, subsequent to an earlier merger of another galaxy with M87.

    The displacement of the SMBH is in the direction opposite the visible jet, so the last two possibilities are more likely than the others. However, possibility (3) depends on the jet structure having existed at least 100 million years and the density of matter at the center of M87 being low enough to provide insufficient restoring force. Possibility (4) is viable if the SMBH is still oscillating around the center following a galactic merger within the past billion years.

    Refs:
    A Displaced Supermassive Black Hole in M87 (6/9/10) – The Astrophysical Journal Letters
    Black Hole Shoved Aside, Along With 'central' Dogma (5/25/10) – Science News
    Black Hole Found in Unexpected Place (5/25/10) – Wired.com
    Supermassive black holes may frequently roam galaxy centers (5/25/10) – Physog.com (press release)
    Bizarre Behavior of Two Giant Black Holes Surprises Scientists (5/25/10) – Space.com
    Galactic Black Holes Can Migrate or Quickly Awaken from Quiescence (5/26/10) – Scientific American




    M87 jet


    Radio Imaging of the Very-High-Energy γ-Ray Emission Region in the Central Engine of a Radio Galaxy (7/24/09) – Science

    Energetic plasma jets, in which matter is accelerated close to the speed of light, combined with intense electromagnetic emissions, especially at radio frequencies, are prominent in about 10% of active galaxies, including M87. However, little has been well established about what processes are responsible for the emissions, or more generally how the jets are powered, accelerated, and focused into narrow beams. Because of the relative proximity of M87 and the fact that the jet we observe is angled from 15° to 25° to our line of sight, M87 is one of the best objects to study in order to learn more about how jets work.

    Gamma rays, because of their very high energies (greater than 100 keV per photon), are not continuously produced in active galaxy jets, but are occasionally observed in short bursts lasting only a few days. One such event occurred in M87 in February 2008. At the same time, the intensity of radiation at all other wavelengths increased substantially. Such flares, at lower energies, are not unusual, since the energy output of most jets is somewhat variable in time. The flare persisted for much longer at energies below the gamma-ray band, indicating that the disturbance continued to propagate along the jet even after the gamma-ray flare subsided. However, although we don't know what the cause was, the coincidence in time of the gamma-ray emissions and the beginning of the extended flare makes it very likely that the events had the same source.

    This is significant information, because our technology for detecting gamma-ray events has very poor angular resolution (~0.1°), since gamma rays can be detected on the ground only by secondary effects that a gamma ray produces in our upper atmosphere. More than 6 orders of magnitude finer resolution can be achieved at radio frequencies, using very long baseline interferometry. With that technology, it was possible to locate the origin of the disturbance that caused both gamma ray and lower energy flaring to a region within about 100 Schwarzschild radii (Rs) of the SMBH. Since Rs = 2G×M/c2, Rs for the M87 SMBH is about 1.9×1010 km, or more than twice the radius of the solar system. So 100Rs is about 70 light-days – which is pretty small compared to the 53.5 million light-year distance to M87.

    It's also significant that the gamma-ray event occurred so close to the SMBH, because the cause must be unlike whatever is responsible for the flaring described in the following research.

    Refs:
    VLBA locates superenergetic bursts near giant black hole (7/2/09) – Physorg.com (press release)
    Mysterious Light Originates Near A Galaxy's Black Hole (7/2/09) – Space.com
    A Flare for Acceleration (7/24/09) – Science
    High Energy Galactic Particle Accelerator Located (9/14/09) – Science Daily (press release)

    Hubble Space Telescope observations of an extraordinary flare in the M87 jet (4/22/09) – arXiv paper

    Electromagnetic radiation from SMBH jets is fairly variable in both time and location along the jet. In the case of M87, high-resolution images at various wavelengths have shown the existence of many regions of enhanced emissions within the jet. One of the most prominent of these even has a name: HST-1, so-named because it was discovered by the Hubble Space Telescope. It occupies a stationary position on the jet, about a million Schwarzschild radii from the center, i. e. about 2000 light-years from the SMBH.

    HST-1 has been observable for some time, but until February 2000 it was relatively dormant. After that it began to flare more brightly across the electromagnetic spectrum up to X-rays. In 2003 it became more variable, and it reached its greatest brightness in May 2005, when the flux in near ultraviolet was 4 times as great as that of M87's central energy source, the SMBH accretion disk. This represents a brightness increase at that wavelength of a factor of 90. The X-ray flux increased by a factor of 50, and similar, synchronized changes occurred at other wavelengths. The synchronization indicates that one mechanism is responsible for the variability at all wavelengths.

    What the actual cause of the disturbance may be is not clear. Because of the great distance of HST-1 from the SMBH, its basic energy source must not be the central accretion disk itself. More likely HST-1 is a result of constriction of magnetic field lines, resulting in further acceleration of the particles making up the jet. Acceleration of charged particles causes radiation by the synchrotron process, and is evidenced by polarization of the emitted photons. Constriction of the jet may be a result of passage through a region of higher density of stars. The increased variability could mean that the jet has encountered a region of higher but varying stellar density. Alternatively, the jet may be passing through a patch of thick gas or dust, with excess radiation produced by the resulting particle collisions.

    These results could explain the variability of light from other, more distant active galaxies, at least those which have strong jets, given that it's possible for a small region of the jet far from the SMBH to outshine the central source. However, another source of variability occurs when a jet is viewed at a very low angle to our line of sight, in which case any slight change of direction could cause an apparent change of brightness.

    Refs:
    Hubble Space Telescope observations of an extraordinary flare in the M87 jet (3/6/09) – The Astronomical Journal
    Hubble Witnesses Spectacular Flaring in Gas Jet from M87's Black Hole (4/14/09) – Physorg.com (press release)
    Black Hole Creates Spectacular Light Show (4/14/09) – Space.com
    Black hole jet brightens mysteriously (4/15/09) – New Scientist
    Black hole spews out impressive light show (4/20/09) – Cosmos Magazine
    Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Monday, December 27, 2010

Pinwheel of Star Birth

    Pinwheel of Star Birth (10/19/10)
    This face-on spiral galaxy, called NGC 3982, is striking for its rich tapestry of star birth, along with its winding arms. The arms are lined with pink star-forming regions of glowing hydrogen, newborn blue star clusters, and obscuring dust lanes that provide the raw material for future generations of stars. The bright nucleus is home to an older population of stars, which grow ever more densely packed toward the center.

    NGC 3982 is located about 68 million light-years away in the constellation Ursa Major. The galaxy spans about 30,000 light-years, one-third of the size of our Milky Way galaxy.




    NGC 3982 – click for 984×1000 image


    More: hereSource URL: https://bechikkethitek.blogspot.com/2010/12/
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Friday, December 17, 2010

Roots of unity and cyclotomic fields

    In preparation for many good things that are to come, we need to have a talk about another important class of field extensions of ℚ – the cyclotomic extensions. (Check here for a list of previous articles on algebraic number theory.)

    A cyclotomic field in general is a field that is an extension of some base field formed by adjoining all the roots of the polynomial f(x) = xn-1=0 for some specific positive n∈ℤ to the base field. Usually, though not always, this will mean roots that lie in some large field in which f(x) splits completely and that contains ℚ as the base field, such as ℂ, the complex numbers. f(x) is known as the nth cyclotomic polynomial. Mostly the same theory applies if the base field is a finite algebraic extension of ℚ, but we'll use ℚ as the base field for simplicity.

    Since f(1)=0, x-1 is one factor of f(x), and f(x)/(x-1) = xn-1 + … + x + 1 ∈ ℤ[x], with all coefficients equal to 1. If n is even, -1 is also a root of f(x). However, all other roots of f(x) in ℂ are complex numbers that are not in ℝ. Some of these roots, known as the "primitive" nth roots of unity – denoted by ζn (or just ζ if the context is clear) – have the property that all other roots are a power ζk for some integer k, 1≤k<n. So the smallest subfield of ℂ that contains ℚ and all roots of f(x) is ℚ(&zeta), known as the nth cyclotomic field.

    It is possible to express all the roots of f(x) in the form e2πi/n, where ez is the complex-valued exponential function, which can be defined in various ways. The most straightforward way is in terms of an infinite series, ez = Σ0≤n<∞zn/n!. The exponential function ez can also be defined as the solution of the differential equation dF(z)/dz = F(z) with initial value F(1)=e, the base of the natural logarithms. So there is the rather unusual circumstance that the roots of an algebraic equation can be expressed as special values of a transcendental function. Mathematicians long hoped that other important examples like this could be found (a problem sometimes referred to as "Kronecker's Jugendtraum", a special case of Hilbert's twelfth problem), but that hope has mostly not been fulfilled.

    The most well-known nontrivial root of unity is the fourth root, i=&radic(-1), which satisfies x4-1 = (x2+1)(x2-1) = 0.

    All complex roots of unity have absolute value 1, i. e. |ζ|=1, since |ζ| is a positive real number such that |ζ|n=1. The set of all complex numbers with |z|=1 is simply the unit circle in the complex plane, since if z=x+iy, then |z|2 = x2+y2 = 1. (Note that the linguistic root of words like "circle", "cyclic", and "cyclotomic" is the Greek κύκλος (kuklos).) Since e = sin(θ) + i⋅cos(θ) for any θ, with θ=2πk/n the real and imaginary parts of a general nth root of unity ζ=e2πi⋅k/n are just Re(ζ)=sin(2πk/n) and Im(ζ)=i⋅cos(2πk/n).

    There are many reasons why cyclotomic fields are important, and we'll eventually discuss a number of them. One simple reason is that roots of algebraic equations can sometimes be expressed in terms of real-valued roots (such as cube roots, d1/3 for some d), and roots of unity. See, for example, this article, where we discussed the Galois group of the splitting field of f(x)=x3-2.

    The set of all complex nth roots of unity forms a group under multiplication, denoted by μn. This group is cyclic, of order n, generated by any primitive nth roots of unity. (Any finite subgroup of the multiplicative group of a field is cyclic.) As such, it is isomorphic to the additive group ℤn = ℤ/nℤ, the group of integers modulo n. Because of this, many of the group properties of μn are just restatements of number theoretic properties of ℤn. For instance, each element of order n in μn is a generator of the whole group – one of the primitive nth roots of unity. Since μn⊆ℚ(ζn), adjoining all of μn gives the same extension ℚ(ζn) = ℚ(μn).

    Now, ℤ/nℤ is a ring, and its elements that are not divisors of zero are invertible, i. e they are units of the ring. They form a group under ring multiplication, which in this case is written as (ℤ/nℤ)× (sometimes Un for short). An integer m is invertible in ℤ/nℤ if and only if it is prime to n, i. e. (m,n)=1 (because of the Euclidean algorithm). The number of such distinct integers modulo n is a function of n, written φ(n). This number is important enough to have its own symbol, because it was studied by Euler as fundamental to the arithmetic of ℤ/nℤ. Thus φ(n) is also the order of the group (ℤ/nℤ)×.

    Let ζ=e(2πi)m/n, for 0≤m<n, be an element of μn. The correspondence m↔e(2πi)m/n establishes a group isomorphism between the additive cyclic group ℤ/nℤ and the multiplicative group μn. Modulo n, m generates ℤ/nℤ additively if and only if (m,n)=1, which is if and only if the corresponding ζ generates μn. So the number of generators of μn – which is the number of primitive nth roots of unity – is the same as the order of (ℤ/nℤ)×, i. e. φ(n).

    One has to be careful, because the multiplicative structure of μn parallels the additive structure of ℤ/nℤ, not the multiplicative structure of (ℤ/nℤ)×. (Because if ζM and ζN are typical elements of μn then ζM×ζNM+N.) Hence even though there are φ(n) generators of μn, these generators do not form a group by themselves (a product of generators isn't in general a generator), so the set of them isn't isomorphic to (ℤ/nℤ)×, even though the latter also has φ(n) elements. Give this a little thought if it seems confusing.

    Moreover, the group (ℤ/nℤ)× is not necessarily cyclic. It is cyclic if n is 1, 2, 4, pe, or 2pe for odd prime p, but not otherwise. Confusingly, if the group does happen to be cyclic then integers modulo n that generate the whole group are called "primitive roots" for the integer n. If (ℤ/nℤ)× happens to be cyclic, then only those m∈(ℤ/nℤ)× having order φ(n) are "primitive roots" that generate the group, while all m∈(ℤ/nℤ)× have the property that if ζ∈μn has order n and generates the latter group, then so does ζm, as we showed above. Got that straight, now? This needs to be understood when working in detail with roots of unity.

    Another reason for the importance of cyclotomic fields is that the Galois group of the extension [ℚ(ζn):ℚ] is especially easy to describe. Indeed, it is isomorphic to the group of order φ(n) we've just discussed: (ℤ/nℤ)×. There's a little work in proving this isomorphism, but let's first note what it implies. Let G=G(ℚ(ζ)/ℚ) be the Galois group. It is an abelian group of order φ(n) since it's isomorphic to (ℤ/nℤ)×. Further, any subgroup of G′ of G is abelian and by Galois theory determines an abelian extension (i. e., an extension that is Galois with an abelian Galois group) of ℚ as the fixed field of G′. Conversely, it can be shown (not easily) that every abelian extension of ℚ is contained in some cyclotomic field. (This is the Kronecker-Weber theorem.)

    Half of the proof of the isomorphism is easy. Pick one generator ζ of μn, i. e. a primitive nth root of unity. We'll see that it doesn't matter which of the φ(n) possibilities we use. Suppose σ∈G is an automorphism in the Galois group. Since σ is an automorphism and ζ generates the field extension, all we need to know is how σ acts on ζ. Since σ is an automorphism, σ(ζ) has the same order as ζ, so it's also a primitive nth root of unity. Therefore &sigma(ζ) = ζm for some m, 1≤m<n. As we saw above, m is uniquely determined and has to be a unit of ℤ/nℤ, with (m,n)=1, in order for ζm to be, like ζ, a generator of the cyclic multiplicative group μn. Hence m∈(ℤ/nℤ)×. Call this map from G to (ℤ/nℤ)× j, so that σ(ζ)=ζj(σ). To see that it's a group homomorphism, suppose σ12∈G, with j(σ1)=r, j(σ2)=s. Then σ21(ζ)) = σ2r) = (ζs)r = ζsr, hence j(σ2σ1) = j(σ2)j(σ1). j is clearly injective since j(σ)=1 means σ(ζ)=ζ, so σ is the identity element of G. Finally, to see that j doesn't depend on the choice of primitive nth root of unity, suppose ζm with m∈(ℤ/nℤ)× is another one. Then σ(ζm) = σ(ζ)m = (ζj(σ))m = (ζm)j(σ).

    Thus G is isomorphic to a subgroup of (ℤ/nℤ)×. That's enough to show G is abelian, so the extension ℚ(ζ)/ℚ is abelian. To complete the proof of an isomorphism G≅(ℤ/nℤ)× we would need to show that the injective homomorphism j is also surjective, i. e. every m∈(ℤ/nℤ)× determines some σ∈G such that m=j(σ). We can certainly define a function from ℚ(ζ) to ℚ(ζ) by σ(ζ)=ζm for a generator ζ of the field ℚ(ζ). One might naively think that's enough, but the problem is that one has to show that σ is a field automorphism of ℚ(ζ).

    The map σ defined that way certainly permutes the nth roots of unity in μn, the roots of the polynomial f(x)=xn-1. However, not all permutations of elements of μn, of which there are n!, yield automorphisms of ℚ(ζ). The problem here is that if z(x) is the minimal polynomial of some ζ, i. e. the irreducible polynomial of smallest degree in ℤ[x] such that z(ζ)=0, then by Galois theory the order |G| of the Galois group G is the degree of the field extension, which is the degree of z(x). Since G is isomorphic to a subgroup of the group (ℤ/nℤ)×, and the latter has order φ(n), all we know is that |G| divides φ(n). It could be that other primitive nth roots of unity have minimal polynomials in ℤ[x] that are not the same as z(x), though they have the same degree |G|. For σ to be an automorphism, σ(ζ) needs to have the same minimal polynomial as ζ, and we don't know that immediately from the relation σ(ζ)=ζm.

    We will defer discussion of the rest of the proof that G(ℚ(μn)/ℚ)≅(ℤ/nℤ)× for the next installment, since some new and important concepts will be introduced.Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Monday, December 13, 2010

Splitting of prime ideals in algebraic number fields



    Our series of articles on algebraic number theory is back again. Maybe this time it won't be so sporadic. Stranger things have happened. The previous installment, of which this is a direct continuation, is here. All previous installments are listed here.

    When we left off, we were talking about how to determine the way a prime ideal factors in the ring of integers of a quadratic extension of ℚ. Such a field is of the form ℚ(√d) for some square-free d∈ℤ. We were using very simple elementary reasoning with congruences, and we found a fairly simple rule, namely:

    If p∈ℤ is an odd prime (i. e., not 2), and K=ℚ(√d) is a quadratic extension of ℚ (where d is not divisible by a square) then
    1. p splits completely in K if and only if p∤d and d is a square modulo p.
    2. p is prime (i. e. inert) in K if and only if d is not a square modulo p.
    3. p is ramified in K if and only if p|d.
    The prime 2 behaves a little more weirdly, but the result is that 2 ramifies if and only if d≡2 or 3 (mod 4); 2 is inert if and only if d≡5 (mod 8); 2 splits if and only if d≡1 (mod 8).

    One limitation was that our simple reasoning made it necessary to assume that OK, the ring of integers of K, was a PID (principal ideal domain).

    Let's review what we were trying to do. We were investigating the factorization of a prime ideal (p)=pOℚ(√d) in Oℚ(√d). If Oℚ(√d) is a PID, then there is a simple approach to investigate how p splits. If p splits then (p)=P1⋅P2, where Pi=(αi), i=1,2. Any quadratic extension is Galois, and the Galois group permutes the prime ideal factors of (p). The factors are conjugate, so if α1=a+b√d we can assume α21*=a-b√d. Hence (p)=(α1)⋅(α1*)= (α1α1*)= (a2-db2).

    Taking norms (to eliminate possible units ε∈Oℚ(√d)) reduces the problem to a Diophantine equation of the form ±p=a2-db2. With the problem thus reduced, a necessary condition for (p) to split (or ramify) is that the equation can be solved for a,b∈ℤ. A sufficient condition to show that (p) is inert, i. e. doesn't split or ramify, is to show that the equation can't be solved.

    Let's look at how that might work. For example, let d=3. Looking at the equations modulo 3, we have ±p≡a2 (mod 3). That is, either p or -p is a square modulo 3. Say p=5. The only nonzero square mod 3 is 1, and 5≢1 (mod 3). However -5≡1 (mod 3), so could we have -5=a2-3b2? Suppose there were some a,b∈ℤ such that -5=a2-3b2. Then instead of looking at the equation modulo 3, we could look at it modulo 5, and find that then a2≡3b2 (mod 5). If 5 divides either a or b, it divides both, and so 25 divides a2-3b2, which is impossible since 25∤5. Therefore 5∤b. ℤ/(5) is a field, so b must have an inverse c such that cb≡1 (mod 5). Therefore, (ac)2 ≡ 3(bc)2 ≡ 3 (mod 5), and so 3 is a square mod 5. But that can't be, since only 1 and 4 are squares modulo 5. The contradiction implies -5=a2-3b2 has no solution for a,b∈Z.

    All that does show 5 doesn't split or ramify in ℚ(√3), hence it must be intert, but this approach is messy and still requires knowing that the integers of ℚ(√3) form a PID. We need to find a better way. Fortunately, there is one. But first let's observe that this elementary discussion shows there is a fairly complicated interrelationship among:
    1. Factorization of (prime) ideals in extension fields,
    2. Whether a given ring of integers is a PID,
    3. Whether an integer prime can be represented as the norm of an integer in an extension field,
    4. Whether an integer can be represented by an expression of the form a2+db2 for a,b∈Z (in the case of quadratic extensions),
    5. Whether, for primes p,q∈Z, p is a square modulo q and/or q is a square modulo p.
    The problem of representing an integer by an expression like a2+db2 is a question of solving a Diophantine equation, and more specifically is of the type known as representing a number by the value of a quadratic form. This question was studied extensively by Gauss, who proved a remarkable and very important result, known as the law of quadratic reciprocity, which relates p being a square modulo q to q being a square modulo p, for primes p,q.

    We will take up quadratic reciprocity soon (and eventually much more general "reciprocity laws"), but right now, let's attack head on the issue of determining how a prime of a base field splits in the ring of integers of an extension field. We will use abstract algebra instead of simple arithmetic to deal with this question. For simplicity, we'll assume here that the base field is ℚ, even though many results can be stated, and are often valid, for more arbitrary base fields.

    Chinese Remainder Theorem

    The first piece of abstract algebra we'll need is the Chinese Remainder Theorem (CRT). Although it's been known since antiquity to hold for the ring ℤ, generalizations are actually true for any commutative ring.

    Let R be a commutative ring, and suppose you have a collection of ideals Ij, for j in some index set, j∈J. Suppose that the ideals are relatively prime in pairs. In general that means that Ii+Ij=R if i≠j, and further, the product of ideals, Ii⋅Ij, is Ii∩Ij when i≠j. If R is Dedekind, then each ideal has a unique factorization into prime ideals, and they are relatively prime if Ii and Ij have no prime ideal factors in common when i≠j. Let I be the product of all Ij for j∈J, which is also the intersection of all Ij for j∈J, since the ideals are coprime in pairs.

    The direct product of rings Ri for 1≤i≤k is defined to be the set of all ordered k-tuples (r1, ... ,rk), for ri∈Ri, with ring structure given by element-wise addition and multiplication. The direct product is written as R1×...×Rk, or &Pi1≤i≤kRi.

    Given all that, the CRT says the quotient ring R/I is isomorphic to the direct product of quotient rings &Pi1≤i≤k(R/Ii) via the ring homomorphism f(x)=(x+I1, ... ,x+Ik) for all x∈R.

    The CRT is very straightforward, since f is obviously a surjective ring homomorphism, and the kernel is I, since it's the intersection of all Ii. (It's straightforward, at least, if you're used to concepts like "surjective" and "kernel".)

    Now we'll apply the CRT in two different situations. First let R be the ring of integers OK of a finite extension K/ℚ, and Ii=Pi, 1≤i≤g, be the set of all distinct prime ideals of OK that divide (p)=pOK for some prime p∈ℤ. Then (p)=P1e1 ⋅⋅⋅ Pgeg, where ei are the ramification indices of each prime factor of (p). An application of CRT then shows that OK/(p) ≅ Π1≤i≤g(OK/Piei). Recall that for each i, OK/Pi is isomorphic to the finite field Fqi, where qi=pfi for some fi, known as the degree of inertia of Pi. (This field is the extension of degree fi of Fp=ℤ/pℤ.) Further, Σ1≤i≤geifi=[K:ℚ], the degree of the extension. Check here if you need to review these facts. Specifying how (p) splits in OK amounts to determination of the Pi and the numbers ei, fi, and g.

    The second situation where we apply CRT involves the ring of polynomials in one variable over the finite field Fp=ℤ/pℤ, denoted by Fp[x]. Let f(x) be a monic irreducible polynomial with integer coefficients, i. e. an element of ℤ[x]. Let f(x) be f(x) with all coefficients reduced modulo p, an element of Fp[x]. f(x) will not, in general, be irreducible in Fp[x], so it will be a product of powers of irreducible factors: Π1≤i≤g(fi(x)ei), where fi(x)∈Fp[x]. Each quotient ring Fp[x]/(fi(x)) is a finite field that is an extension of Fp of some degree fi. In general, ei, fi, and g will be different, of course, from the same numbers in the preceding paragraph. But the CRT gives us an isomorphism Fp[x]/(f(x)) ≅ &Pi1≤i≤g(Fp[x]/(fi(x)ei)).

    Now, here's the good news. For many field extensions K/ℚ, there exists an appropriate choice of f(x)∈ℤ[x] such that for most primes (depending on K and f(x)), the numbers ei, fi, and g will be the same for both applications of the CRT. Consequently, we will have OK/(p) ≅ Fp[x]/(f(x)), because for corresponding factors of the direct product of rings, OK/PieiFp[x]/(fi(x)ei). As it happens, most primes don't ramify for given choices of K and f(x), so that things are even simpler, since all ei=1, and all factors of the direct products are fields.

    We can't go into all of the details now as to how to choose f(x) and what the limitations on this result are. However, here are the basics. Any finite algebraic extension of ℚ (and indeed of any base field that is a finite algebraic extension of ℚ) can be generated by a single algebraic number θ: K=ℚ(θ), called a "primitive element". In fact, &theta can be chosen to be an integer of K. Then the ring of integers of K, OK, is a finitely generated module over ℤ. (A module is like a vector space, except that all coefficients belong to a ring rather than a field.) The number of generators is the index [OK:ℤ[θ]]. (ℤ[θ] is just all polynomials in θ with coefficients in ℤ.) If p∈ℤ is any prime that does not divide [OK:ℤ[θ]], then the result of the preceding paragraph holds. If for some p and some choice of θ p does divide the index, then there may be another choice of θ for which p doesn't divide the index. Unfortunately, there are some fields (even of degree 3 over ℚ) where this isn't possible for some choices of p.

    The situation is especially nice in the case of quadratic fields, K=ℚ(√d), square-free d∈ℤ. If d≢1 (mod 4); we can take θ=√d and f(x)=x2-d, since OK=ℤ[√d]. If d≡1 (mod 4), then the index [OK:ℤ[√d]]=2, and there's a possible problem only for p=2. However, we still have OK/(p) ≅ Fp[x]/(x2-d) for all p≠2. From that it's obvious that, except for p=2, (p) ramifies if p|d, (p) splits if d is a square modulo p, or else (p) is inert. That is exactly the conclusion we began with at the beginning of this article, on the basis of elementary considerations. Only now we need not assume that OK is a PID.

    There are four important lessons to take away from this discussion.

    First, there is a very close relationship between the arithmetic of algebraic number fields and the arithmetic of polynomials over a finite field. Not only do we have the isomorphism discussed above, but it turns out that a number of similar powerful theorems are true for both algebraic number fields and the field of quotients of polynomial rings over a finite field.

    Second, a lot of the arithmetic of algebraic number fields can be analyzed in terms of what happens "locally" with the prime ideals of the ring of integers of the field.

    Third, many of the results of algebraic number theory are fairly simple if the rings of integers are PIDs (or, equivalently, have unique factorization). Such results often remain true when the rings aren't PIDs, though they can be a lot harder to prove. Often the path to proving such results involves considering the degree to which a given ring of integers departs from being a PID.

    Fourth, and perhaps most importantly, abstract algebra is a very powerful tool for understanding algebraic number fields – and it is much easier to work with and understand than trying to use "elementary" methods with explicit calculations involving polynomials and their roots.

    We will see these lessons validated time and again as we get deeper into the subject.

    So where do we go from here? There are a lot of directions we could take, so we'll probably jump around among a variety of topics.Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Saturday, December 11, 2010

Merry Christmas And Happy New Year


    Funnymadworld team wishing our entire new and loyal visitor’s “Merry Christmas and Happy New Year’s 2011”. Hope you all will always visit our blog and give a full support to us :P































    Source URL: https://bechikkethitek.blogspot.com/2010/12/
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Masterpiece Art by Apple product


    How about you destroy Apple product like iPad or iPhone? Then snap the picture and make more money as a contemporary art? This is done by Michel Tompert, a former Apple employee.
























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