Galois Theory Reading Group

[OP No.11530059]

Galois Theory Reading Group

Book: "Field and Galois Theory" by Patrick Morandi

Download link:
>https://www.medellin.unal.edu.co/~mmtoro/doc/campos/Morandi.pdf
>http://gen.lib.rus.ec/search.php?req=morandi+galois&lg_topic=libgen&open=0&view=simple&res=25&phrase=1&column=def

Errata:
>https://wordpress.nmsu.edu/pamorand/files/2018/10/Errata.pdf

This thread is dedicated to the study of Galois theory as a group. The idea is to discuss the subject with other anons as you work through the book, whether it be doing the exercises, helping each other through tough theorems and proofs, or asking interesting questions about the material itself.

I've opted for this book since it assumes very little, and has an appendix with most of the results one needs that are perhaps more complicated. It is also crystal clear in its explanations, and the exercises are fair.

>What you will get out of this
Good knowledge of the basics of field and Galois theory that are pretty much a prerequisite for any sort of algebraic number theory, arithmetic or algebraic geometry, and perhaps some use for other fields, such as algebraic topology.

>Minimal prerequisites
1. Basic ring theory: first isomorphism theorem, ideal correspondence, prime and maximal ideals, basics of UFDs, PIDs and polynomial rings.
2. Basic group theory: cosets, normal subgroups and quotients, Lagrange's theorem, actions.
3. Basic linear algebra / vector spaces over a field: bases, dimension, rank-nullity, determinants.

Most of the above prerequisites are briefly looked at in the appendix, so might be worth to read over it first.

>Minimal objectives
Chapter 1, sections 1-5 cover all the basics of field theory and the fundamental theorem of Galois theory ~ 60 pages. Can skip the sections on purely inseparable extensions and the fundamental theorem of algebra. You should at least try to complete section 5 to get the most out of the preceding 4 chapters.

[OP No.11530060]

>Optional objectives
1. Fundamental theorem of algebra: the only proof you will see that is almost purely algebraic (you cannot escape analysis, but this only requires the intermediate value theorem). Requires some knowledge of the Sylow theorems (covered in appendix). Covered in chapter 1 section 5.

2. Important examples: Chapter 2 contains several examples of fields, with the most important and interesting being finite fields and cyclotomic extensions. Covered in chapter 2 section 6/7.

3. Norms, traces, and discriminants: Pretty important concepts in number theory. Covered in chapter 2 section 8 and chapter 3 section 1.

4. Ruler and compass constructions and transcendental numbers. Probably some of the most interesting classical problems that galois theory solves. Chapter 3, section 14/15.

5. Solvability by radicals: the crux that Galois set out to solve. Requires a stronger knowledge of finite groups, specifically the permutation and alternating groups, and the Jordan-Holder theorem. Chapter 3, section 16.

6. Transcendental extensions and algebraic geometry. The first application of field theory to a different subject, one studies here the relation between these special extensions and varieties, the shapes cut out by polynomial in affine space. Chapter 5.


I will be posting solutions to the exercises as I solve them, and perhaps make notes of things I see in the book.

[Anonymous No.11530062]

Just read Milne. His notes are way better than anything else out there

[Troy(200 IQ) No.11530077]

>>11530059
> Galoy theory
not computer science

[Anonymous No.11530090]

>>11530059
Thanks OP

[Anonymous No.11530278]

>>11530060
>Ruler and compass constructions and transcendental numbers. Probably some of the most interesting classical problems that galois theory solves.
I could never figure out why people care so much about straightedge constructions. The whole concept is a relic from 2500 years ago before they had invented the ruler.

[Anonymous No.11530444]

>>11530278
Get over yourself.

[Anonymous No.11530447]

>>11530278
It's a matter of proportions, not of precise measurements. Measurements are irrelevant to mathematicians - implicit to it is fixing a reference frame, or rather, it is an extrinsic property. What are interesting are the intrinsic properties; things that don't depend on the ambient space, but rather is information tied to the object itself. A triangle being equilateral is a more interesting property than its sidelength being 10cm. You could stretch the plane its drawn on and it would still be equilateral, but the length will have changed - what we would care about is the proportion that it changed by.

Also tied to measurements is inevitably the real numbers, departing from the integers and algebraic numbers. Straightedge and compass constructions are to differential geometry what number theory is to real analysis. Then you could ask the number theorists the equivalent question. And the answer would be because it's cool as fuck.

[Anonymous No.11530501]

>>11530059
would participate but i'm a filthy undergrad that has lacking knowledge in ring theory.
pls point finger to rings

[Anonymous No.11530636]

>>11530060
>I will be posting solutions to the exercises as I solve them, and perhaps make notes of things I see in the book.
pls do, tag the posts properly with what exercise or part of the book the concern

>>11530077
I remember that there's some server math making use of algebra employing it

https://en.wikipedia.org/wiki/Standard_RAID_levels#General_parity_system

[Anonymous No.11531669]

Too lazy to type this out in latex
Solution for Chapter 1 Problem 10

[p 17 ex 10 No.11531826]

>>11531669
Are you OP?

Can we put posts concerning different exercises apart and put the page and exercise in the name field like so?

[ch1 ex10 p60 No.11531831]

>>11531669
Are you OP?

Can we put posts concerning different exercises apart and put the page and exercise in the name field like so?

Also, don't put a
>
before a link.

[Anonymous No.11531839]

i took galois theory last semester though so I won't be participating
good luck op

[Anonymous No.11531895]

>>11531831
Not OP. Took a course in galois theory last semester and will probably be doing some excercises to refresh some of the material.

[Anonymous No.11531901]

Why should I care about Galois theory? Why is it so spectacular?

[Anonymous No.11531924]

>>11531669
I can barely read this shit. How about light or line free paper?

[Anonymous No.11531930]

>>11531924
>line free paper

[Anonymous No.11531964]

>>11531924
It's really not that hard to read.
But I agree, he should have texted it. It's a good habit of taking notes anyway.

>>11531901
You can also considering it learning field theory. Fields are basic and relatively nice algebraic structures.

[Anonymous No.11532154]

I already took a course in Galois theory but I will be revising it soon so I might join these threads if they're at an appropriate level to me (although looking at the problems in the first chapter this book might be too easy).

>>11531669
>>11531964
Your proof reads like shit. I normally wouldn't even bother with someone as bad at maths as you are but I'll do it just this once for the sake of this reading group:
-you're deriving a contradiction from nothing and pretend this finishes the proof. This is wrong and you are retarded.
-the minimal polynomial of a over F(a^2) has to divide x^2-a^2 but doesn't have to be equal to it. In fact, the whole point of the problem is showing that it's just x-a.
The correct way to do it is by saying the min poly has to divide x^2-a^2. Suppose it is x^2-a^2, then derive the contradiction. Therefore it must have degree 1, so it must be x-a i.e. a is in F(a^2), so done.
To reiterate, this is basic fundamental stuff and you having studied Galois theory already but being this shit at presenting your proofs in a readable way is sad. inb4
>muh semantics
yes, your proof has all the correct ideas in it but this is worthless if you can't write it down in a comprehensible way.

[Anonymous No.11532158]

>>11532154
>This is wrong and you are retarded.
Thanks prof.

[Anonymous No.11532162]

>>11532154
>yes, your proof has all the correct ideas in it but this is worthless if you can't write it down in a comprehensible way.
My PhD supervisor Montezuma said otherwise

[Anonymous No.11532169]

>>11532162

[Anonymous No.11532200]

>>11532154
>I normally wouldn't even bother with someone as bad at maths as you
do people really talk/act like this?

[Anonymous No.11532211]

>>11532169

[Ch1 ex11 p14 No.11532689]

Let [math]K/F[/math] be an algebraic extension and let [math]R[/math] be a subring of [math]K[/math] with [math]F \subset R \subset K [/math]. Clearly, [math]K[/math] is commutative, so we only need to show that for every nonzero element [math]a\in R[/math], we also have [math]a^{-1}\in R[/math].
So, let [math]a\in R[/math] and let [math][F(a):F]=n[/math]. Then the elements [math]1, a, a^2,..., a^{n-1}[/math] form a basis of [math]F(a)[/math] over [math]F[/math] (Proposition 1.15). Because [math]a^{-1} \in F(a)[/math] it follows that there are
[math]\lambda _1,...,\lambda _n \in F [/math] such that [math]a^{-1}=\sum_{1}^{n} \lambda _i a^{i-1} [/math].
Now, since the [math]1, a, a^2,..., a^{n-1}[/math] are already in [math]R[/math], and [math]F \subset R[/math],
[math]a^{-1}=\sum_{1}^{n} \lambda _i a^{i-1} \in R[/math].
Since a was arbitrary, every nonzero element in [math]R[/math] has an inverse. Thus, [math]R[/math] is a Field

[Ch1 ex11 p14 No.11532692]

>>11532689
sorry for shit formatting

[OP No.11532699]

OP here, had to do some things so couldn't get started on the book. Just read the first section, and there's a thing that came up that wasn't entirely obvious in Example 1.17.

First, is that the author mistakenly uses a future result (precisely, Proposition 1.20) without mention, but I think this sort of mistake only comes up once, at least from what I remember in my past reading.

Second, is that the reasoning is a bit complicated - he uses Gauss lemma twice, once to go forward, and once to go back. Last is the fact that [math]k[x]\cong k[u][/math] doesn't appear to be used explicitly, but is actually used to transfer the irreducibility from "[math]x[/math] over [math]u[/math]", to "[math]u[/math] over [math]x[/math]" when doing the [math]k[u][x]=k[x][u][/math] maneouvre.

[Anonymous No.11532972]

>>11530278
I think it's mostly because knowing which points were constructible was an open problem for so long, and it was surprising when the answer came from field theory, which wasn't well understood at the time. Even more surprising that the structure of the field made restrictions on the constructible polygons, and that there is a formula for determining constructibility of a regular n-gon.

[wiki server No.11533131]

>>11531964

>>11532689

I was just thinking one can in addition also post a screenshot of the exercise before then responding to it!

[Anonymous No.11534147]

So here's my attempt at p13 problem 8. I am in a very sleep deprived state so this might be terribly wrong.

Suppose [math][F(a):F]=n[/math] is the degree of the field extension, and let [math]\{1,a,...,a^{n-1}\}[/math] be a basis. Let [math]p(x)=x^n+b_{n-1}x^{n-1}+...+b_0[/math] be the minimal polynomial of [math]a[/math], which is necessarily of degree [math]n[/math], and hence we have an equation [math]a^n= -b_{n-1}a^{n-1}-...-b_0[/math].

We know what the effect of [math]L_a[/math] is on the basis, so we can write down its matrix to be

[eqn]\begin{bmatrix}
0 & 1 & 0 & \dots &0 \\
0 & 0 & 1 & \dots &0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-b_0 & -b_1& -b_2 & \dots & -b_{n-1}
\end{bmatrix}[/eqn]
In particular, [math]xI-L_a[/math] has matrix [eqn]\begin{bmatrix}
x & -1 & 0 & \dots &0 \\
0 & x & -1 & \dots &0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
-b_0 & -b_1& -b_2 & \dots & x-b_{n-1}
\end{bmatrix}[/eqn]
And its determinant can be calculated to be precisely the expression of the minimal polynomial.

For the last part, we know that this determinant expression is always a degree [math]n[/math] polynomial, so if such a degree polynomial is minimal for an [math]\alpha\in K[/math], then we must have that \{1,\alpha,...,\alpha^{n-1}\} is a linearly independent set. Therefore it's all [math]\alpha\in K[/math] such that [math]F[a]=F[\alpha][/math]. I don't know if there is a more specific answer.

[Anonymous No.11534917]

>>11530059
Started reading. It's not exactly the easiest book.

[Anonymous No.11535247]

p14 ex10.

We have the inclusions as fields [math]F\subseteq F(a^2)\subseteq F(a)[/math]. In particular, we have the equation [math][F(a):F]=[F(a):F(a^2)][F(a^2):F][/math]. A vanishing polynomial for [math]a[/math] over [math]F(a^2)[/math] is [math]x^2-a^2[/math]. If it were irreducible, then this would be the minimal polynomial for [math]a[/math] in this field, giving us a field extension of degree [math]2[/math]. But this is a contradiction to [math][F(a):F][/math] being odd. Hence the polynomial splits; equivalently, [math]a\in F(a^2)[/math].

[Anonymous No.11535274]

p14 ex11

Let [math]0\neq a\in R[/math]. We need to show [math]a^{-1}\in R[/math]. Since [math]K[/math] is algebraic over [math]F[/math], there is a minimal polynomial over [math]F[/math], [math]p(x)=x^n+b_{n-1}x^{n-1}+...+b_0[/math] such that [math]p(a^{-1})=0[/math]. Rearranging and multiplying the latter equation by [math]a^{n-1}[/math] gives us [math]a^{-1}=-(b_{n-1}+...+b_0 a^{n-1})[/math], an equation that exists over [math]R[/math].

[Anonymous No.11535297]

p14 ex12

Any such field isomorphism [math]\varphi[/math] would necessarily be an isomorphism of the prime field [math]\mathbb Q[/math], so this forces a correspondence [math]\pm\sqrt2\leftrightarrow \pm\sqrt3[/math]. But [math]2=\varphi(2)=\varphi(\sqrt2\cdot \sqrt2)=\varphi(\sqrt2)^2=(\sqrt3)^2=3[/math], a contradiction.

They are however isomorphic as vector spaces by basic dimension theory - as the algebra structure is forgotten, no contradiction arises.

[Anonymous No.11535376]

>>11535297
Wrong. It doesn't follow that phi(+-sqrt(2))=+-sqrt(3).
This thread is absolute trash and all of you are literally retarded and should kill yourselves or go back to your meritard "pre-pre-algebra" kindergarten tier shit as you don't seem capable of any maths that's even marginally more difficult than basic addition.
Sage with rage

[Anonymous No.11535384]

>>11535376
Why not? there's a set theoretic bijection between every single element of the fields except for two in each of the fields, so to have a full bijection, it means that the nonrational have to be sent to the nonrational elements

[Anonymous No.11535604]

p14 ex15

[math]\Rightarrow[/math]: Any element of [math]L_1,\: L_2[/math] is also in [math]L_1L_2[/math], so it must be algebraic over [math]F[/math].

[math]\Leftarrow[/math]: Let [math]\alpha\in L_1L_2[/math]. We have that [math]\alpha[/math] is generated by a finite amount of elements of [math]L_1[/math] and [math]L_2[/math], each being algebraic over the base field, so [math]\alpha[/math] itself is contained in a finite extension of [math]F[/math], hence algebraic.

[Anonymous No.11535816]

p14 ex17

We have that [math][L_1L_2:F]=[L_1L_2:L_1][L_1:F][/math], so it is enough to show that [math][L_1L_2:L_1]\leq [L_2:F][/math]. But this is obviously true since the minimal polynomial over [math]F[/math] of any element in [math]L_2[/math] is also a polynomial over [math]L_1[/math] of same degree.

The last part follows because each of [math][L_i:F][/math] are factors of [math][L_1L_2:F][/math], so together with the inequality shows the result.

[Anonymous No.11535847]

>>11535376
What exactly did you get out of this? Not only are you wrong, but you sound like a manchild with self-esteem issues.

[Anonymous No.11535874]

>>11535847
kill yourself nigger retard
I'm not wrong and you would know that if you had half a brain cell
I'm not going to explain trivial garbage like this. I'll just point it out whenever I see it, so people who are studying galois theory without a sliver of a grasp of elementary mathematics get discouraged and hopefully kill themselves and stop shitting up this board and threads I have an interest in.

[Anonymous No.11535883]

>>11530059
kek that image is so fitting

[Anonymous No.11535904]

>>11535874
wew lad, it's wrong but you can't explain why, so you resort to telling me it's trivial garbage (which it is, I wont deny it).

There's a bijection. And a known correspondence between every single element of both fields, except for two elements in each field. Therefore these two elements in each field must be taken to each other, because they could not possibly be taken to any other. No matter how you pair them up, you arrive at a contradiction. QED

[Anonymous No.11535930]

11535904
I'm not even going to give you a (You) for such obvious bait. You're wrong and you should kill yourself.

[Anonymous No.11535996]

ok, I made a dumb mistake after not doing any math for a couple years. i guess i must kill myself now

[Anonymous No.11536007]

>>11535904
>And a known correspondence between every single element of both fields, except for two elements in each field. Therefore these two elements in each field must be taken to each other
But that's not true, there are loads of elements that aren't in Q and so aren't fixed. Why can't I send root 2 to 3 + 17 root 3 ?

[Anonymous No.11536019]

>>11535996
>coping this hard
AHAHAHAHAHAHAHA
suck my dick retard

[Anonymous No.11536103]

>Minimal prerequisites
>1. Basic ring theory: first isomorphism theorem, ideal correspondence, prime and maximal ideals, basics of UFDs, PIDs and polynomial rings.
The only relevant rings are Q and extensions. You can essentially avoid all ring-theoretic language.
2. Basic group theory: cosets, normal subgroups and quotients, Lagrange's theorem, actions.
Same here. You don't need to know what abstract groups are. Only what permutations are
>Ruler and compass constructions and transcendental numbers. Probably some of the most interesting classical problems that galois theory solves
Just fyi, the proof essentially doesn't involve any Galois tgeory stuff. You just note that e.g. trisections would yield cubic root coordinates which the usual ruler and compass moves can't do.

[Anonymous No.11536115]

>>11535874
Nice bait retard.

[Anonymous No.11536141]

>>11536115
And what's your problem, bitch? Why do you call me a retard when I was right and that retarded insect I deigned to discipline even admitted his mistake?
Be grateful that I grace this thread and correct idiots when they're wrong.

[Anonymous No.11536164]

you're honestly pathetic. i wish you the best in solving your mental problems. i was there once too, but didnt realise it until a few years later when i matured

[Anonymous No.11536346]

>>11530059
didn't this guy die in a duel at age 20 or something??

Imagine if he didn't die being shot with a pistol in the stomach at such a young age, he was truly the smartest mathematician of all time.

[Anonymous No.11536422]

>>11530059
Are there any major use cases for galois theory in physics or another field of science?

[Anonymous No.11536441]

>>11536346
No actually he drove into a bunch of harlots with his brand new carriage (on purpose). And then he was guillotined right then and there.

[Anonymous No.11536483]

>>11536422
Well, most of modern physics involves the study of fields somehow, so yes, certainly

[Anonymous No.11537865]

>>11535297
Wrong, you have to assume that [math] \varphi (\sqrt{2})=a+b\sqrt{3} [/math] with a and b integers.

>>11535384
>there's a set theoretic bijection between every single element of the fields except for two in each of the fields
[math] \varphi (\mathbb{Q})=\sqrt{3} \mathbb{Q} [/math] and [math] \varphi (\sqrt{2}\mathbb{Q})= \mathbb{Q} [/math] is also a set theoretic bijection between every single element of the fiels, you are just dumb.