Should I Become a Mathematician?

[mattmns]

I have a question. Is it acceptable to get a letter of recommendation from faculty that are not (associate, full) professors? Maybe I should not say acceptable but will it hurt me that much when I am applying to grad schools? For example, next semester I need to get my third letter of recommendation, and I have 2 professors I am looking at: this will be the second class I take with whichever one I choose, and I did very well in the first class with them, so if I do well again I would think I could easily ask for a good letter. The problem is that one is a better instructor than the other (the part-time professor is better) and the class the tenured professor is teaching is not very important. Should I go with the part-time instructor in that he will be teaching a more important class (topology) and that he is a better instructor, or should I go with the tenured professors that is not as good at teaching and teaching something not very important (graph theory)? I asked my adviser, but she is a little biased (married to the part-time instructor). Thanks.

 

[JasonRox]

I have a question. Is it acceptable to get a letter of recommendation from faculty that are not (associate, full) professors? Maybe I should not say acceptable but will it hurt me that much when I am applying to grad schools? For example, next semester I need to get my third letter of recommendation, and I have 2 professors I am looking at: this will be the second class I take with whichever one I choose, and I did very well in the first class with them, so if I do well again I would think I could easily ask for a good letter. The problem is that one is a better instructor than the other (the part-time professor is better) and the class the tenured professor is teaching is not very important. Should I go with the part-time instructor in that he will be teaching a more important class (topology) and that he is a better instructor, or should I go with the tenured professors that is not as good at teaching and teaching something not very important (graph theory)? I asked my adviser, but she is a little biased (married to the part-time instructor). Thanks.

I'd say go with the part-time instructor. They too know what it takes to get through graduate school. Considering it's your third letter, I would think it wouldn't affect you.

It's what I'm planning on doing. My part-time professor is my favourite professor, as well as we had our conversations together. I go to him for advice usually.

 

[mathwonk]

jason, really anything is possible, but most things unusual are exceptional.

 

[mathwonk]

good letters from lesser known people who actually know you well, are often more helpful than so - so letters from famous or high ranking people, who do not.

 

[mathwonk]

day 5 notes, field and galois theory

8000 Day 5 Field extensions and homomorphisms.
Introduction: Galois theory is concerned with the self mappings, i.e. automorphisms of a field E, that are specified on some subfield, e.g. that equal the identity on some subfield k. We want to see how to construct such automorphisms, to count how many there are, and to compute their exact fixed field. If E has finite vector dimension n over k, we will see there are at most n automorphisms of E, that fix k pointwise. The reason is simple. It will turn out that such maps must send roots in E of polynomials with coefficients in k, to other roots in E of these polynomials. It follows that the number of such automorphisms will be determined by the number of distinct roots such polynomials have in E, which is always bounded by the degree of the polynomials. Galois theory is most useful when the number of distinct roots of an irreducible polynomial equals its degree. This is the "separable" case.
Since the vector dimension of a field extension can be computed in terms of the degree of the polynomials satisfied by generators for the extension, it will follow that the number of automorphisms is also bounded by the vector dimension of the extension, and that equality holds in the separable case. The proofs proceed by carrying out the "simple" case first, then using induction to deduce the result for any sequence of simple extensions, i.e. any finite extension.

First we review a few elementary facts about extensions that are technically essential, and probably familiar from math 6000.
Def: If k is a subfield of E, then E is a vector space over k, and we write [E:k] for the k dimension of E, also called the degree of E over k.

Definition: If k is a subfield of a field E, and a belongs to E, we say a is algebraic over E, iff a is a root of some non zero polynomial f in k[X].

Lemma: If k is a subfield of a field E, a belongs to E, and a is algebraic over k, there is a unique monic irreducible polynomial in k[X] satisfied by a, namely the unique monic polynomial in k[X] of lowest degree with a as root. This polynomial is called the minimal polynomial of a over k.
proof: If a is algebraic over k, the evaluation map k[X]-->E, sending g to g(a) has non trivial kernel of form (f), inducing an injection k[X]/(f)-->E. Hence (f) is a maximal and prime ideal generated by a unique monic irreducible polynomial, namely the monic polynomial of lowest degree in the kernel. QED.

Definitions:
The ring generated by a over k, =k[a], = k-algebra generated by a.
If k is a subfield of E, and a is an element of E, then k[a] denotes the intersection of all subrings of E that contain both a and k. Concretely, k[a] = {f(a): f is any polynomial in k[X]}.

The field generated by a over k = k(a) = the intersection of all subfields of E that contain both a and k. Concretely k(a) = {f(a)/g(a): f,g are in k[X], and g(a) != 0}.

Theorem: If k is a subfield of E, and a is an element of E, then TFAE:
1) a is algebraic over k.
2) k[a] = k(a).
3) k(a) has finite dimension over k.
4) k[a] has finite dimension over k.
5) k[a] is contained in a finite dimensional k - subspace of E.
6) k(a) ? k[X]/(f) where f is an irreducible polynomial in k[X].
In particular, if a is algebraic over k, then the k dimension of k(a) equals the degree of the minimal polynomial of a over k.
proof: [sketch]: If a is algebraic over k, then for some n, an is a linear combination of lower powers of a. This implies every power am with m>n, is also a linear combination of powers of a less than n. Hence the dimension of k[a] over k is finite. If we take a monic dependency relation c0+c1a+c2a2+...+an for the smallest possible power an, then a satisifes the monic polynomial f(X) = c0+c1X+c2X2+...+Xn, but no polynomial of lower degree, so f is irreducible, and the map k[X]/(f)-->k[a] is injective and surjective. Since k[X]/(f) is a field, so is k[a] = k(a). If k[a] is contained in some finite dimensional k subspace of E, say of dimension n, then 1,a,a2,...,an are dependent/k, and a dependency relation gives a polynomial satisfied by a, so a is algebraic. The k - dimension of k[X]/(f) is deg(f) = n, and a basis is given by [1], [X],...,[Xn-1]. If a is not algebraic over k, then the infinite sequence of powers {1,a,a2,...,an,...} is linearly independent over k. QED.

Lemma: If k is a subfield of E, and E a subfield of F, then [F:k] = [F:E][E:k].
proof: If x1,...,xn is a k basis for E, and y1,...,ym an E basis for F, then {xiyj}, 1<=i<=n, 1<=j<=m, is a k basis for F. This is trivial to check by changing the order of summation. E.g. if z lies in F, then there exist constants b1,...,bm in E such that z = b1y1+...+bmym. But each bj lies in E, so there exist constants aij such that each bj = a1jx1+...+anjxn. Hence z = b1y1+...+bjyj+...bmym = ...+(a1jx1+...+anjxn)yj+... = ...+aij(xiyj)+...

Thus the products {xiyj} span F over k. In particular if both [F:E] and [E:k] are finite, so is [F:k]. You can check simialrly that the products {xiyj} are independent over k, and you should as this is a favorite little easy prelim question. QED.

Cor: The subset of E consisting of elements which are algebraic over k, is a subfield of E.
proof: If a,b in E are algebraic over k, we must show that a+b, ab, a/b, are also algebraic over k. But k(a) has finite dimension over k, and since k(b) also has finite dimension over k, it has even smaller dimension over k(a). Thus both [k(a,b):k(a)] and [k(a):k] are finite. Hence also [k(a,b):k] is finite, and so all field combinations of a,b, such as a+b, ab, a/b, etc,.. , belong to finite dimensional subfields of E, hence are algebraic over k. QED.

Note: All finite dimensional extensions of k are algebraic, but not all algebraic extensions are finite dimensional. We have shown that simple aklgebraic extensions are finite dimensional. Hence finitely generated (as fields) algebraic extensions, are finite dimensional (as vector spaces).

Cor: A field extension of k is finite dimensional (as a vector space), if and only if it is both finitely generated (as a field) and algebraic/k.

Extending homomorphisms.
Given a (homomorphic) map, f:k-->k', of fields which are subfields of larger fields E, E', we want to say exactly when it is possible to extend f to a map f':E-->E'. Of course field maps are injective whenever they exist.
Field map extensions are always done one generator at a time, and for that "simple" case we use the fundamental isomorphism F(a) ? F[X]/(g) where g is the minimal F polynomial of a, to tell us how to extend the map.

Theorem 1: Assume f:k-->k' is a map of fields which are subfields of E, E'.
1) Then f extends uniquely to a ring map k[X]-->k'[X], by applying f to the coefficients of each polynomial.
Let a be any element of E with minimal polynomial g in k[X], mapping to g' in k'[X].
2) Then f extends to a map k(a)-->E' with f(a) = a', if and only if a' is a root of g'.
3) Hence the number of extensions of f to f':k(a)-->E' is equal to the number of distinct roots in E' of g'. This number is at most equal to the degree of g = dimk(k(a)).
proof: 1) we must check that (g+h)' = g'+h' and (gh)' = g'h'. The coefficient of Xn in g+h is the sum an+bn where*a,b, are the coefficients in g,h. But the coefficient of Xn in (g+h)' = f(an+bn) = (an+bn)' = (an)'+(bn)' = the sum of the coefficents of Xn in g' and h'. Hence (gh)' = g'+h'.
The coefficient of Xn in gh is the sum of the products aibj over all i,j, with i+j = n. But f applied to this sum is the sum of the corresponding products ai'bj', which is the corresponding coefficient in g'h'. Hence (gh)' = g'h'.
2) Applying f to g(a) = 0, gives g'(a') = 0, so the condition is necessary. If indeed g'(a')=0 for some a' in E', then we get a map k'[X]-->E' sending X to a', which induces a map k'[X]/(g')-->E' sending [X] to a', hence by composition a map
k(a)-->k[X]/(g)-->k'[X]/(g')-->E', sending a to a'.
3) Each different choice for f(a) gives a different map k(a)-->E', so the number of maps equals the number of roots a' of g' in E'. The number of rooits of the polynomial g' in the field E' never exceeds the degree of g', which equals the degree of g, which equals the k dimension of k(a). QED.

Cor: In the setting above, the number of extensions of f:k-->k', to f':k(a)-->E' is at most equal to dimk(k(a)), and equals this number if g' has deg(g') distinct roots in E', equivalently if g' factors into distinct linear factors in E'[X].

Note: Even if g' has d = deg(g') distinct roots in E', and there exist d distinct maps k'-->E' extending f, all of these maps may have the same field as image in E'. I.e. if {a1', a2',...,ad'} are the distinct roots of g' in E', and k-->k' is an isomorphism, then all the fields k'(ai') are isomorphic, but it may or may not happen that they are actually equal subfields of E'. We call these fields "conjugate" subfields of E', and when they are all equal we say they are all normal. We will say more about normal extensions later, and show how in this situation they define normal subgroups of the group of automorphisms of E'.

The argument for extending maps to simple extensions k(a), let's us extend maps to any algebraic extensions at all, provided the target field has enough roots.
We have to be a little careful about the hypotheses, since if k has more than one generator, it does not suffice to require just one root in E' for each of their minimal polynomials. For instance, suppose g is an irreducible polynomial in k[X] with distinct roots a,b and E = k(a,b), while E' = k(a) != k(a,b).
Then a,b, have the same minimal polynomial over k, namely g, so in both cases, the minimal polynomial of a and b does have a root in E'. Still there is no extension of the identity map k-->k to a map k(a)-->k(a,b). So we use a little stronger hypothesis on E' in the next result. We do the finite dimensional case first.

Theorem 2: Assume E = k(a1,...an) is finitely generated and algebraic, i.e. a finite (vector) dimensional field over k, f:k-->k' is a field map and E' a field containing k'.
Assume further for every generator ai of E, that gi' factors completely into linear factors in E'[X], where gi' in k'[X] is the image under f of the minimal polynomial gi of ai. Equivalently assume E' contains a splitting field for the polynomial ?gi'.
Then there exist extensions of f to f':E-->E'. The number of such extensions is at most equal to the dimension [E:k], and equals this dimension if the polynomial ?gi' factors into distinct linear factors in E'[X].
proof: (existence of extensions): From the argument in the simple case we get an extension to k(a1)-->E'. Now we want to extend further to k(a1, a2)-->E'. There is only a tiny difference from the simple case. We regard this as a simple extension k1(a2), of k(a1) = k1. But our hypothesis only says the minimal polynomial g2 of a2 over k corresponds to a polynomial g2' in E'[X] that splits completely into linear factors. We need to know about the minimal polynomial h2 of a2 over the field k1.
The point is that h2 is always a factor of g2. I.e. both h2 and g2 have coefficients in k1, and h2 is irreducible there. Since a2 is a root of both polynomials, h2 must divide g2 in k1[X]. Thus the corresponding polynomial h2' in E'[X] is a factor of g2', and since g2' factors in E'[X] completely into linear factors, some of those same factors give a factorization of h2' into linear factors.
Extending successively over each simple extension, we eventually get an extension to E-->E', by induction on the dimension [E:k].
(number of extensions): By the argument in the simple case, i.e. by the corollary above, at each stage the number of extensions from ki-->E' to ki+1-->E' is at most equal to the dimension [ki+1:ki], and equals that dimension if hi' factors in E'[X] into distinct linear factors. But the number of extensions of f from k-->k' to f'E-->E' is the product of the number at each stage, and the dimension of successive field extensions is also multiplicative, i.e. [E:k] = {E:kn-1][...][k1:k]. Thus the number of extensions is at most equal to [E:k], and equals this dimension when every polynomial hi' factors into distinct linear factors in E'[X]. But since the linear factors of hi' are a subset of the linear factors of gi', if ?gi' factors into distinct linear factors in E'[X], then so does every hi'. QED.

[If you are getting lost in the theory, this is a good time to go read some examples. DF looks good in chapter 14.2, and I recommend my webnotes 844.2 where I compute very thoroughly the Galois group of X4-2 over Q.]

It may be getting tiring to repeat this same argument, but we do it anyway, to practice using Zorn's lemma in the infinite dimensional case. The moral is that there are only three steps to the argument: i) the simple case; ii) the observation that the minimal polynomial over a larger extension is a factor of the minimal polynomial for a smaller one, which allows you to repeat the simple case more than once; iii) using Zorn, i.e. "transfinite induction", to find a maximal extension. It is instructive also to note that now that E is not finitely generated over k, there will be an infinite number of hypotheses to check about E', which poses the question of how do we verify that E' satisfies them.

Def: A field F is algebraically closed if every polynomial in F[X] has a root in F, hence every polynomial in F[X] factors completely into linear factors in F[X].

Theorem 3: Assume E is an algebraic extension of k, f:k-->k' is a field map and E' a field containing k'. Assume further for every element a of E, that g' factors completely into linear factors in E'[X], where g' in k'[X] is the image under f of the minimal polynomial g of a. This is true for example if E' is algebraically closed.
Then there exist extensions of f to f':E-->E'.
proof: Consider the set of all partial extensions of f to F-->E' where F is a field intermediate between k and E. These form a partially ordered set where g > h if g extends h. (Each such partial extension is a map from a subset of E to E', hence its graph is a subset of ExE'. The collection of all such partial extensions is a subset of the set of all subsets of ExE', in particular it is a set, to which we can try to apply Zorn's lemma.)
Given any totally ordered collection or "chain" of partial extensions, they define a partial extension on the union of their domains, and this is an upper bound for the chain. Since the hypothesis of Zorn is thus satisfied, there exist maximal partial extensions. We claim any maximal partial extension is actually defined on all of E.
If g:F-->E' is an extension of f that is not defined on all of E, there is some element a of E that is not in F. Then a is algebraic over k hence also over F, and satisfies an irreducible polynomial over F which is a factor of its irreducible polynomial over k. The corresponding polynomial over E' thus factors completely into linear factors as argued above, and we get an extension of f to F(a)-->E'. This shows any extension of f whose domain is not all of E cannot be maximal, hence any maximal extension is defined on all of E. QED.

Notice theorem 2 is a corollary of theorem 3, but I thought it clearer (certainly for me) to explain these cases separately. Also the use of a finite set of generators in theorem 2, allows the splitting hypothesis to be given only for the generators.

 

[mathwonk]

day 5 part 2, separable extensions

The phenomenon of polynomials with multiple roots also bears examination, since it affects the number of extensions of a homomorphism.
Def/Ex: A polynomial g in k[X] is separable if gcd(g, dg/dx) = 1 in k[X], if and only if g, dg/dx have no common root in any extension of k, if and only if f has no multiple linear factor in any extension of k, if and only if there is an extension of k where g has deg(g) distinct roots, and where g factors into distinct linear factors.

Def: An algebraic element a of a field E containing k, is separable over k if its minimal polynomial in k[X] is separable. Notice that an element separable over k is also separable over any larger field, since its minimal poynomial there is a factor of the one over the smaller field.

Theorem 4: If E is any field containing k, the elements of E that are separable over k form a subfield of the field of algebraic elements.
proof: Let a,b, be elements of E that are separable over k. We claim every element of k(a,b) is separable over k. Let c be any element of k(a,b), and let f,g,h, be the minimal polynomials of a,b,c over k, with f,g separable. Let E' be a splitting field for (fgh) over k. Since f,g both factor into distinct linear factors in E', there are exactly [k(a,b):k] extensions of the inclusion map k-->E' to maps of k(a,b)-->E'.
If c is not separable over k, there are fewer than deg(h) extensions of k-->E' to k(c)-->E', hence fewer than [k(a,b,c):k] extensions of k-->E' to k(a,b,c)-->E'. But since k(a,b,c) = k(a,b) this is a contradiction. So c is separable over k. QED.

With this terminology, Theorem 2 implies the following statements.
Cor 5: Assume E = k(a1,...an) is a finite separable extension of k, g is a polynomial in kl[X] satisfied by the generators ai, and E' is a field containing a splitting field for g. Then there exist exactly [E:k] maps E-->E', which are the identity on k.

Cor 6: If k is any field, f a separable polynomial over k, and E a splitting field for f, there are exactly [E:k] automorphisms of E which equal the identity on k.

Remark: All algebraic extensions are separable in characteristic zero. We will see later that all algebraic extensions of finite fields are also separable.

Def: The "fixed field" of a set S of automorphisms of a field E is the subfield of E of elements left identically fixed by every element of S. Automorphisms of E leaving a subfield k fixed are called k automorphisms.

Cor 7: If k is any field, f a separable polynomial over k, and E a splitting field for f, the fixed field of G = Galk(E) equals k.
proof: If G fixed a larger subfield F of E containing k, there would be more than [E:F] automorphisms of E fixing F, contradicting Theorem 2. QED.

Cor 8: If E is a splitting field for a separable polynomial f over k, and F is an intermediate field, the fixed field of the subgroup GalF(E) of Galk(E), is F.
proof: E is also a splitting field for f over F. QED.

If E is a splitting field of a finite separable polynomial over k, this shows the map from subgroups of G = Galk(E) to fields intermediate between k and E, is surjective. I.e. every field between k and E is the fixed fielkd of some subgroup of Galk(E).

Theorem: If f is a separable polynomial of degree n over k, with splitting field E, the group Galk(E) is isomorphic to a subgroup of the symmetric group S(n). In particular it has finite order dividing n!
proof: Every k automorphism of E permutes the roots of f, hence G acts on the set of n roots. But the roots generate E, so this action determines the element of G, hence the map G-->S(n) is injective. QED.

Cor: If E is the splitting field of a separable polynomial over k, there are only a finite number of intermediate fields between k and E.

It remains to show in this setting that no two subgroups of Galk(E) can fix exactly the same subfield of E, so that the numer of subgroups and intermediate fields is the same. I.e. it could conceivably occur that a subgroup H fixes exactly F, but that actually also more elements of G also fix F, so the full subgroup fixing F is larger than H. This never happens, as the following cute little argument apparently stemming from work of Dedekind and Artin shows.

Theorem: If G is a finite group of automorphisms of a field E, and k the subfield fixed by G, then [E:k] = #G.
proof: We know #G <= [E:k], so we want to show any subset x1,...,xn of E with n > #G is dependent over k. This a matter of solving linear equations. If f1,...,fm are the elements of G, the m by n matrix [fj(xi)] over E has a non zero solution vector c = (c1,...,cn) in En, since n > m. Thus for all i, (*) <sum>j cjfi(xj) = 0. Choose c with as few non zero entries as possible. There must be at least two != 0 entries, since we may assume all xj != 0, else dependency is obvious, and the fi are automorphisms of E. By reordering the x's, we may assume c1 != 0, and multiplying through by c1-1 we may assume c1 = 1 belongs to k.
We claim actually all cj belong to k, yielding a k - linear relation among the elements (fi(x1),...fi(xj),...,fi(xn)) for every i. Since one of the f's is the identity, this will give a k linear relation among the (xj) as claimed.
If some cr != c1 is not in k, then since k is the fixed field of G, some fs != id does not fix cr. We can renumber so that fr(cr) != cr. Then apply fr to the system of equations getting <sum>j fr(cjfi(xj)) = <sum>j fr(cj) frfi(xj) = 0 for all i. But as fi runs over the group G, the product frfi does the same. So we can say also that
(**) <sum>j fr(cj) fi(xj) = 0 for all i. Now if we subtract the two systems of equations (*) and (**), we get <sum>j [fr(cj)-cj] fi(xj) = 0 for all i. In this system, since c1 is in k, fr(c1)-c1 = 0, but fr(cr)-cr != 0. Hence we have a non zero solution vector
(,...,fr(cj)-cj,...) in the kernel of the matrix, but with fewer non zero entries than before, a contradiction. QED.

Cor: If E is a finite dimensional extension of a field k, TFAE:
i) #Galk(E) = [E:k].
ii) E is the splitting field of a separable polynomial.iii) k is the fixed field of some finite group of automorphisms of E.
proof: We proved ii) implies i) and iii), and that iii) implies i) above. So it suffices to prove i) implies ii). By the arguments used above, i) implies the minimal polynomial of every generator of E over k splits into distinct lienar factors in E, so E is a splitting field for the product of the minimal polynomials of a finite set of generators for E over k, where each of these polynomials is separable. It remains to show their product may be assumed separable. If any two of these polynomials are equal we may omit one. So we want to rule out that two distinct separable irreducible polynomials over k share a root in E. But any element of E has a unique irreducible minimal polynomial over k, so this is impossible. QED.

Def: Under any of these conditions, we say E is a finite galois extension of k.

Cor: If E is a finite galois extension of k, then different subgroups of G = Galk(E) have different fixed fields. Thus the correspondence between subgroups of G and fields intermediate between k and E is bijective.
proof: If E is galois over k, and H is a subgroup of G, let F be the fixed field of H, and K the subgroup of all elements of G fixing F. By the previous corollary, #H = [E:F]. H is a subgroup of K. But theorem 2 shows that #K <= [E:F] = #H. Thus H = K. QED.

 

[cubert]

I'm thinking of taking the real analysis course next term, but the class has a notorious reputation for its difficulty. I think only two students out of about 25 got A's in it last spring, and they're no less than academic superstars (read: putnam fellows).

Therefore, I want to have some idea of what I'm up against. The class uses Rudin's book, which I already bought. However, I have not written a proof since my sophomore year's geometry class. How do you think I should proceed with the book? Reading doesn't seem to cut it; I have a fair understanding of the underlying concepts, but I'd think a little more than would be required to solve the problems at the end of each chapter.

 

[balletomane]

That sounds a lot like the analysis classes at my school. I have to take it in the spring or in the fall and I'm quite nervous to say the least. That said, several of my friends who are certainly very bright, but not superstars passed it and learned a lot. I think sometimes the reputation is worse than the class itself. (At least that's what I'm hoping.)

Anyhow, maybe practice with proofs would be helpful?

I'm interested to see what someone who actually knows what they're talking about has to say.

 

[mathwonk]

real analysis is thorough going proofs class in a topic whose concepts are hard and precise. Rudins book moreover treats the material very briefly and succinctly, and is far less than ideal as a learning place.

You need a lot of preparation to ace this class; 1) practice in proofs, 2) preliminary study from an easier book.

you do not have much time now, so clear your schedule as much as possible to leave double free time for this one course. then get some other books, such as simmons introduction to topology and analysis, and study them.

practice writing proofs and expect to get less than an A with your weak background. try for a B.

 

[mathwonk]

heres a cheap copy of simmons:

Introduction to Topology and Modern Analysis
Simmons, George F.
Bookseller: Logos Books
(Davis, CA, U.S.A.) Price: US$ 14.00
[Convert Currency]
Quantity: 1 Shipping within U.S.A.:
US$ 3.50
[Rates & Speeds]
Book Description: McGraw-Hill, 1963. Hardcover. Book Condition: Good. Dust Jacket Condition: Good. 1st Edition. A small amount of marginalia; a few pages have indents from a paperclip. A good reading copy. Bookseller Inventory # 010260

 

[mathwonk]

spivak's calc book is also good preparation, or apostol.

 

[Beeza]

If one were to be in a differential equations class that is taught with no theory, theorems, but just methods to solving differential equations (Yes, this is the course math majors have to take too). The teacher also considers integration using anything beyond "u substitutions" too hard. How much trouble would this person be in when they move on to more advanced math courses-- let's say partial differential equations? The text is Boyce, DiPrima.

 

[ircdan]

If one were to be in a differential equations class that is taught with no theory, theorems, but just methods to solving differential equations (Yes, this is the course math majors have to take too). The teacher also considers integration using anything beyond "u substitutions" too hard. How much trouble would this person be in when they move on to more advanced math courses-- let's say partial differential equations? The text is Boyce, DiPrima.

When I studied PDEs, the main things from ODEs we used were the solution techniques. We needed to be able to solve ODEs right away. Solving a PDE often comes down to solving an ODE, or several ODEs. I don't think it's a problem, and if you feel it is, read your text. Anyways make sure you remember how to solve the various types of ODEs. I would suggest taking a course on PDEs right after a course on ODEs if possible. Also, being familiar with vector calculus helps. It seems that when vector calculus was used, it was mostly the concepts being applied. Anyways every course on PDEs is different, and this is just my experience. Goodluck.

 

[mathwonk]

if you read the book, can't you get a lot more out of it than the teacher is offering? boyce diprima has proofs in it right?

or read a better book, like arnol'd.

 

[quasar987]

That's what I was thinking. Boyce & DiPrima proves stuff. So even if you teacher only mentions the result without proof, you can look the proof up yourself.

 

[mathwonk]

here is a site with free notes on a wide variety of topics:

http://us.geocities.com/alex_stef/mylist.html

 

[balletomane]

Awesome. Thanks.

 

[mathwonk]

Galois' theorem on solvability of polynomials by radicals
We will prove next that in characteristic zero, a polynomial whose
galois group is not a solvable group, is not "solvable by radicals", and
give an example of a polynomial that is not solvable by radicals.

Lemma: The galois group of a polynomial is isomorphic to a subgroup of
permutations of its distinct roots. If the polynomial, is irreducible, the
subgroup of permutations is transitive on the roots.

Cor: If an irreducible polynomial over Q has prime degree p, and exactly 2
non real roots, its Galois group is isomorphic to S(p).

Def: A primitive nth root of 1 (or of "unity"), is an element w of a field
such that w^n = 1, but no smaller power of w equals 1.

Lemma: If char(k) = 0, for every n > 0, there is an extension of k which
contains a primitive nth root of 1.

Theorem 1: If char(k) = 0, the Galois group G of X^n -1 over k is
isomorphic to a subgroup of the multiplicative group (Z/n)*, hence G is
abelian.
Rmk: If k = Q, then G ≈ (Z/n)*, as we will show later.

Theorem 2: If k is a field containing a primitive nth root of 1, and c is
an element of k, the galois group G of X^n-c is isomorphic to a subgroup of
the additive group Z/n, hence G is abelian.
Rmk: It is NOT always true that G equals Z/n, even if k is the splitting
field of X^n-1 over Q.

Theorem 3: If ch(k) = 0, and w is a primitive nth root of 1, and if k0 =
k(w), ki = k(w,a1,...,ai), where for all i = 1,...,m, ai^ri = bi is some
element of ki-1, and ri divides n, then the galois group of km =
k(w,a1,...,am) over k is solvable.

Def: A radical extension E of k, is one obtained by successively adjoining
radicals of elements already obtained. I.e. E = k(a1,...,am) where for
each i, some positive integral power of ai lies in the field
k(a1,...,ai-1).

Def: A polynomial f in k[X] is "solvable by radicals" if its splitting
field lies in some radical extension of k.

Theorem 4: If k has characteristic zero, and f in k[X] is solvable by
radicals, then the galois group of f is a solvable group.
Rmk: The converse is true as well, also in characteristic zero.

Cor 5: The polynomial f = X^5-80X + 2 in Q[X], is not solvable by radicals.
proof: The derivative 5X^4 - 80, has two real roots 2,-2, so the graph has
two critical points, (-2, 130), and (2, -126). Since f is monic of odd
degree, it has thus exactly 3 real roots, and 2 non real roots. The galois
group is therefore isomorphic to S(5), which has a non solvable subgroup
A(5) ≈ Icos. QED.

 

[Ki Man]

does fractal geometry have any practical uses

 

[mathwonk]

check this out;

http://classes.yale.edu/fractals/

 

[0rthodontist]

I read once that somebody made a fractal image compression algorithm for nature scenes. I don't think it was dramatically better than other algorithms for most subjects, but apparently it worked.

 

[pivoxa15]

In your very first post you mentioned

"basically 3 branches of math, or maybe 4, algebra, topology, and analysis, or also maybe geometry and complex analysis"

But what about branches like Statistics, Probability/
Stochastic Processes, Operations Research?
Do they fit into one of the 4 major fields you suggested? If so how would you put them in?

 

[jammidactyl]

The comment probably should have been, "three branches of PURE math".

Those topics you mentioned all fall under applied mathematics, with things like statistics and stochastic process borrowing heavily from analysis. The question is more, can I use some or invent some mathematical technique to solve this problem? So it doesn't really matter what branch that technique comes from.

http://www.math.niu.edu/~rusin/known-math/index/mathmap.html"

 

[mathwonk]

yes, i admitted later that i am incompetent in those other areas of applied math, and i appreciate any input on those topics anyone is willing to offer. i apologize if i gave the impresion my advice is comprehensive, as i am obviously limited by my own knowledge and experience.

i myself have studied only pure math, with courses in algebraic topology, algebraic geometry, functional analysis, riemann surfaces, homological algebra, complex manifolds, real anaylsis and representations.

then my research was entirely in riemann surfaces, singularities of theta divisors of jacobians and prym varieties, and their moduli.

so i am pretty ignorant of analysis, algebra, topology, finite math, probability, statistics, gosh almost everything.

I do know a little about theta divisors.

But I still feel free to offer advice!

i just meant to start this thread, not to dominate it. my apologies for its shortcomings.

 

[sherlockjones]

can one be both a pure and applied mathematician?

 

[pivoxa15]

Why don't you include complex analysis in analysis and geometry in topology so that there are (definitively) 3 branches of (pure) mathematics.

 

[mathwonk]

you could very reasonably do that. complex analysis , at least the homological kind I know most about, does have a rather different flavor from real analysis, but deep died analysts use a lot of real analysis to do complex analysis, via harmonic functions.

 

[pivoxa15]

you could very reasonably do that. complex analysis , at least the homological kind I know most about, does have a rather different flavor from real analysis, but deep died analysts use a lot of real analysis to do complex analysis, via harmonic functions.

That is interesting. So the complex analysis you do is more related to algebra than analysis?

Now I am going to ask, most likely a very stupid questions.

If someone wanted to solve the Riemann hypothesis, which branch of mathematics should they get into?

 

[radou]

In general, the more branches you get in, the better.

 

[JasonRox]

In general, the more branches you get in, the better.

Yeah, but you might want to do the relevant ones first.

I'd say Complex Analysis, Number Theory and Abstract Algebra would be the most relevant.

Not entirely sure if there are more important areas.

You might want to read "The Music of the Primes". That might give you an idea of what you're getting into.

 

[quasar987]

What is everyone's favorite book on the Riemann hypothesis? There's

Prime obsession
The Music of the Primes
The Riemann Hypothesis: The Greatest Unsolved Problem in Mathematics
Riemann's Zeta Function

And probably a miriad of other ones. Which one's the most interesting to read?

 

[mathwonk]

i rather liked Riemann's own paper.

 

[mathwonk]

the riemann hypothesis is clearly an application of compelx analysis to number theory. put very simply, riemanns point of view was that a complex functon is best understood by studying its zeros and poles.

the zeta function is determined by the distribution of primes among the integers, since its definition is f(s) = the sum of (forgive me if tjhis is entirely wrong, but someone will soon fix it) the terms 1/n^s, for all n, which by eulers product formula equals the product ofn the sum of the powers of 1/p^s, which equals by the geometric d]series, the product of the factors 1/[ 1 - p^(-s)].

now this function, determined by the sequence primes , is by riemanns philosophy best understood by its zeros and poles.

hence riemanns point of view requires an understanding of its zeroes, which he believed to lie entirely on the critical line.

this hypothesis the allowed him to estimate the number of primes less than a given value, to an accuracy closer than gauss' integral estimate.

even with its flaws, this brief discussion allows you to see what areas of math you might need to know at a minimum. complex analysis, number theory, integral estimates, and (excuse me that it was not visible) mobius inversion.

 

[sherlockjones]

mathwonk, what do you think of the math department of this college:

http://www.rose-hulman.edu/math/home.php

 

[mathwonk]

i never heard of it before but it looks like a really good undergraduate college department, with a deep commitment to teaching and nurturing undergraduates. i like what i see on their website.