When people talk about Galois theory they often say that the basic idea behind it is that certain numbers are "algebraicaly indistinguishable".  I never really understood what this means in a way that doesn't make any reference to the concept of field automorphism.   So is there some intuitive way of looking at this?
121046
I heard the term couple of times (in Field theory class and book), for example: The different roots of $p(x)=x^3-2$ are "algebraically indistinguishable".  I understand the meaning intuitively, but what does it really mean ?
34729
A number is an &quot;algebraic integer&quot; if it is the root to a monic polynomial with integer coefficients. Artin says (Algebra, p. 411):  The concept of algebraic integer was one of the most important discoveries of number theory. It is not easy to explain quickly why it is the right definition to use, but roughly speaking, we can think of the leading coefficient of the primitive irreducible polynomials $f(x)$ as a &quot;denominator.&quot; If $\alpha$ is the root of an integer polynomial $f(x)=dx^n+a_{n-1}x^{n-1}...$ then $d\alpha$ is an algebraic integer, because it is a root of the monic integer polynomial $x^n + a_{n-1}x^{n-1} + ... + d^{n-1}a_0$. Thus we can &quot;clear the denominator&quot; in any algebraic number by multiplying it with a suitable integer to get an algebraic integer.  When I first learned of algebraic integers, I looked online and saw some hints that maybe they were used to prove the Abel-Ruffini theorem. So I put off questioning their usage for a while; I now think I understand one proof of this theorem (the one at the end of Artin's Algebra) and it has nothing to do with algebraic integers (that I can tell). So basically: why is it important if a number is an algebraic integer? I think I understand what he's saying about the relationship between roots of integer polynomials and algebraic integers, but I fail to see why this is &quot;one of the most important discoveries of number theory.&quot;
118804
I am having difficulty with this proof. I have previously proved that for an automorphism on the extension field $K$ of $F$ that when restricted to $F$ is the identity automorphism must take roots of $f(x)\in F[x]$ in $E$ to roots of $f(x)$ in $E$. I was thinking of using the minimal polynomial $x^{3}-2\in\mathbb{Q}[x]$ and the result of previous problem I solved, but I suspect there are automorphisms that don't fix the subfield, so the result can't be used or I need to prove that any automorphism of an extension field must fix the subfield that is being extended or always sends roots in $E$ to roots in $E$.  I am not asking for a solution to this, but maybe a hint or pointer in the right direction. This problem is introduced in the text prior to Galois theory, so please keep that in consideration.
37754
Here's the problem: if $S = \mathbb{Q}[x]/(x^2)$, compute the group  of automorphisms $\mbox{Aut}(S)$.   First off, I have the solution and I know it's isomorphic to the multiplicative group $\mathbb{Q}\backslash\{0\}$, so I'm trying to understand it. It uses what is called the substitution principle and the fact that the restriction of any endomorphism $\phi: \mathbb{Q}[x]\to \mathbb{Q}[x]$ to $\mathbb{Q}$ is the identity. It says that these two facts implies that the endomorphisms of $\mathbb{Q}[x]$ are in 1-1 correspondence with $h \in \mathbb{Q}[x]$, where the endomorphism corresponding to $h$ is $\phi_h : \mathbb{Q}[x] \to \mathbb{Q}[x]$ given by $\phi_h(f) = f(h)$ for $f \in \mathbb{Q}[x]$. I guess I just don't understand how it applied the substitution principle? It's probably obvious, but for some reason I am not getting it.  An explanation would be greatly appreciated! Or if you have a different solution, that'd be cool too. Thanks!!
60440
In my notes, I am given the definition of the Galois group of a polynomial only in the case when the polynomial is separable (if $f$ is a separable polynomial over $K$ with splitting field $L$, then $\mathrm{Gal}(f) = \mathrm{Gal}(L/K)$). This makes sense, since $L$ is a splitting field implies normality, and an extension generated by separable elements is separable, so $L$ is necessarily a Galois extension of $K$.   I should probably tell you that my definition of separable is the one which involves the irreducible factors of the polynomial having distinct roots in a splitting field (I'm aware there's another definition, and that they coincide when defining a separable extension).   When I look around online, I see that the Galois group is defined for any polynomial as the Galois group of its splitting field. Why is the splitting field necessarily Galois?  Thanks
25222
On the top of my head, I cannot think of any fields of cardinality more than that of the reals. (It is known that the process of algebraic closure does not increase the cardinality of an infinite field.)  What is the simplest way to give an example of a field (and an ordered field) of a specific cardinality $\alpha$?  I see there is the "Field" of surreal numbers, but it is a proper class rather than a set (and hence do not have a cardinality as such). However, there seems to be some modified construction which gives proper fields with the cardinality of some strongly inaccessible cardinal.
118128
Suppose you're trying to teach analysis to a stubborn algebraist who refuses to acknowledge the existence of any characteristic $0$ field other than $\mathbb{Q}$.  How ugly are things going to get for him?  The algebraist argues that the real numbers are a silly construction because any real number can be approximated to arbitrarily high precision by the rational numbers - i.e., given any real number $r$ and any $\epsilon&gt;0$, the set $\left\{x\in\mathbb{Q}:\left|x-r\right|&lt;\epsilon\right\}$ is nonempty, thus sating the mad gods of algebra.  As @J.M. and @75064 pointed out to me in chat, we do start having some topology problems, for example that $f(x)=x^2$ and $g(x)=2$ are nonintersecting functions in $\mathbb{Q}$.  They do, however, come arbitrarily close to intersecting, i.e. given any $\epsilon&gt;0$ there exist rational solutions to $\left|2-x^2\right|&lt;\epsilon$.  The algebraist doesn't find this totally unsatisfying.     Where is this guy really going to start running into trouble?  Are there definitions in analysis which simply can't be reasonably formulated without leaving the rational numbers?  Which concepts would be particularly difficult to understand without the rest of the reals?
62075
I know perfectly well the Peano axioms, but if they were sufficient for defining $\mathbb N$, there would be no controversy whether $0$ is a member of $\mathbb N$ or not because $\mathbb N$ is isomorphic to a host of sub- and supersets of $\mathbb Z$.  So it seems that mathematicians, when specifying $\mathbb N$, rarely mean the set of natural numbers according to its definition but rather some particular embedding into sets defined by different axiomatic systems.  Doesn't this sort of misstate then the role of the Peano axioms as the defining characteristics of $\mathbb N$?  It seems like most of the time the Peano axioms are not more than a retrofit to a subset of $\mathbb Z$ or other sets defined by a different axiomatic system with one or several axioms incompatible with the Peano axioms.
98612
Let $F$ be a field. When happens that the additive group of $F$ is isomorphic to the multiplicative group?  It is easily to work out that $F$ must have characteristic $0$, but then what?
56213
This material is covered in detail in Dickson's "Linear Groups with an exposition of the Galois Field Theory", chapter XXII and Huppert's "Endliche Gruppen", chapter II, paragraph 8. Since I don't speak german and Dickson's treatment often requires deciphering, I was wondering if there is a "modern" account of this somewhere.
910
I stumbled across this relationship while I was messing around. What's the proof, and how do I understand it intuitively? It doesn't really make sense to me that the sum of odd numbers up to $2x + 1$ should equal $x^2$.
24913
Suppose one doesn't know any Galois theory, but they are familiar with the basic theory of field extensions. Then how would one justify the following picture?    I'm pretty sure they're just saying these are the known subfields, and there could be more but they just write down those ones (which seems weird, which is why I'm asking). Or have they deduced that this is a comprehensive list of subfields just by using basic theory? This occurs on page 573 of Dummit and Foote's Abstract Algebra.
57490
This question is motivated from an exercise from Rudin. The exercise says that prove that set of all algebraic numbers is countable.   Proof: We know that a number $z$ is called algebraic if it is the root of a polynomial $a_0z^n+a_1z^{n-1}+\cdots +a_n=0$, where $a_i$'s $\in \Bbb{Z}$. Since there are only finitely many equations of the form $n+|a_0|+|a_1|+\cdots+|a_n|=N$ for a positive integer $N$ this implies there are only finitely many such polynomials. Now since each of these polynomials have finitely many roots therefore there are countably many algebraic numbers.   Now the next exercise in the process asks us to prove that there is a real number which is not algebraic which is very trivial. So this proves that there exists transcendental numbers. Also this shows that there are uncountably many of them. So that means there are fairly many of them.   But where are they? Now if there were uncountably many of them then it should be fairly easy to find them but it is not . Because we can get all sorts of wonky numbers being the root of polynomials.    mentions all the transcendental numbers and the numbers which might be transcendental. These do not seem to be many. So what about the rest of the transcendental numbers. Where are they? And why is it so difficult to find them and prove that they are transcendental?   A side question: Is there some nice property that the class of all these transcendental numbers might satisfy . By a nice property I mean that if I take all transcendental numbers and form a group then what about the isomorphisms on this group. Similarly what about if I take all transcendental numbers and form a topological space?  Note: Now if this is a question that has already been asked then please do notify me in the comments I will delete this question
100917
I was reading about this in a cryptography book last night.  I have a hunch about this, but I can't quite put my finger on it.  I think this is a similar situation to an affine cipher, where the multiplication factor has to be relatively prime with the size of the alphabet in order for the function to be surjective.  Obviously for practical purposes, it would have to be $p^n$ where $p$ is the arity of the number system and $n$ is the bit width of a memory unit.  But is there also a theoretical basis for requiring that a Galois field be of size $p^n?$
16562
I think I may have caught a mistake in my professor's lecture notes, and just want to be sure I'm not mistaken.   We want to construct $\mathbb{Q}$ from $\mathbb{Z}$, and we consider the mapping $$f: \mathbb{Z} \to \mathbb{Q}, \; n \mapsto \frac{n}{1}.$$ He writes that because $f$ is surjective, $\mathbb{Z} \subset \mathbb{Q}$.   However, it is surely not surjective because there are surely rational numbers whose  (unique) simplest form is $\frac{a}{b}$ for $b \neq 1$. The function is injective, surely, because the simplest representative in $\mathbb{Q}$ is unique, but that alone would not imply that $\mathbb{Z} \subset \mathbb{Q}$.   I think he meant to say that because $f$ is well-defined over the entirety of $\mathbb{Z}$, meaning we actually can identify each integer with a single rational number via the above inclusion, $\mathbb{Z} \subset \mathbb{Q}$.   Am I correct or glossing over something?
96664
How symmetric is the hierarchy of indecomposable ordinals?   The process of ordinal arithmetic follows a repetitive, inductive pattern from addition to multiplication to exponentiation etc., in which the morphism from one operation to the next looks pretty much the same - in the respect that every step up is an induction beyond the transinfinite limit of the previous operation.  I'm calling this a symmetry.  We have the concepts of additively indecomposable ordinals, and beyond that, multiplicatively indecomposable ordinals etc.  Does this hierarchy enjoy the same inductive symmetry infinitely far in the same way - i.e. to exponentiation, tetration and beyond?  $1,\omega^\beta$ are additively indecomposable (for all ordinals $\beta$).  $2,\omega^{\omega^\beta}$ are multiplicatively indecomposable.  How does this list continue and what does it look like?    It looks like $0^0=1$ and $1^1=1$ but $(n-1)^{n-1}&gt;n\forall n&gt;2\implies $ so the first two exponentially indecomposable ordinals would be:  $2,\omega$  Is it $2,\omega,\omega^{\omega^{\omega^\beta}}$ for all ordinals $\beta$ and so on?  Or does the list of special cases at the start get ever more complicated, and do the power towers just get taller or do they change?  Tetration is a complicated matter (at least over regular numbers) as the orders of exponents matter and the rule to always apply powers from the right isn't necessarily well-justified, so I imagine there's the potential to get into considering Dyck words etc.  How messy does it get?
69532
What is the intuition behind the fact that if every element in a group is of order $2$, we have that the group is abelian? I can prove it, but I do not know the intuition behind it.