abstract algebra – Let $ phi: mathbb {Z} _7 times mathbb {Z} _7 $ be a homomorphism such that $ phi ^ 5 = text {id} $. Show that $ phi $ is the identity.

Let $ phi: mathbb {Z} _7 times mathbb {Z} _7 to mathbb {Z} _7 times mathbb {Z} _7 $ to be a homomorphism such as $ phi ^ 5 = text {id} $. CA watch $ phi $ is the identity.

My attempt: Since $ ker ( phi) subset ker ( phi ^ 2) subset cdots subset ker ( phi ^ 5) = 0 $it follows that $ phi $ is injective. Since $ | mathbb {Z} times mathbb {Z} | $ is finished, $ phi $ is automatically an isomorphism.

then $ phi $ should map a subgroup to a subgroup. Consider $ phi ( mathbb {Z} _7 times 0) $, so he should be associated with $ mathbb {Z} _7 times 0 $ or $ 0 times mathbb {Z} _7 $. Without loss of generality, we consider the first case. In particular, $ phi ($ 1.0) should be a generator for $ mathbb {Z} _7 times 0 $. assume $ phi (1,0) = (a, 0) $. then $ (1,0) = phi ^ 5 (1,0) = phi ^ 4 (a, 0) = phi ^ 3 (a ^ 2,0) = cdots = (a ^ 5,0) $. So, $ a ^ 5 = 7k + 1 $ for some people $ k in mathbb {Z} $. By testing $ a in mathbb {Z} _7 $, we see that the only possibility is that $ a = $ 1. The discussion for $ phi (0,1) $ Is similar.

Could someone help me take a look at my attempt? Is my approach reasonable and is there a better way to solve this problem?

Abstract algebra – How to apply the theory of "homology" to the Goldbach conjecture using the result of Helfgott (ternary Goldbach true)?

The result of Helfgott: every odd integer $ n geq $ 7 is the sum of $ 3 $ (odd) bonuses. I saw somewhere in the newspaper the mention of odd prime numbers. It means that every single integer $ n geq $ 10 is the sum of 4 odd prime numbers (you just added a $ 3 $ to each odd number).

Let's first assume something bold, to simplify the discussion, that for each integer $ x geq $ 13, the even number $ 2x $ can be expressed as the sum of $ 4 $ distinct prime numbers: $ 2x = p + q + r + s $. To handle all cases, I'm sure it's possible, because it would take either an enumeration of the small set of cases, or a combinatorial argument. Reason is $ x geq $ 13 is because: $ 3 + 5 + 7 + 11 = 26 = 2 cdot $ 13 is the smallest even number which is the sum of $ 4 $ distinct odd primes.

If you are not familiar with the basics of simplicial homology but you know what the free $ Bbb {Z} $-module on a set of symbols is, then google "introduction to simplicial homology". You basically need limit cards: $ partial_n: C_n to C_ {n-1} $ who take each element of the module $ C_n $ at its "limit" in $ C_ {n-1} $, its border being represented as a formal element in the free module, whatever the border map, it must be a $ Bbb {Z} $homomorphism -module and satisfy also: $ partial_ {n-1} circ partial_n = 0 $.

So, according to Helfgott's result (and the net distinction hypothesis), each integer $ x geq $ 13 is such that the even number $ 2x = p + q + r + s $, or $ {p, q, r, s } $ are distinct odd prime numbers. As a result, there is $ {4 choose 2} = $ 6 ways of forming even smaller numbers (by adding two of the prime numbers involved in the sum of $ x $) $ 2x_i, i in {1, dots, 6 } $ such as $ 2x_i + 2x_j = 2x $ or equivalent $ x_i + x_j = x $. Now, there is another problem with duplicates as $ x_i = x_j $ is a possibility. But for now, suppose that $ x_i neq x_j $, for reasons of discussion. So, to recap, everyone $ x_i = u + v $ or $ u, v in {p, q, r, s } $.

So, if we associate each $ x_i $ with the official symbol $ bar {x_i} $ which might not be an integer (for the moment) then we will call $ bar {x_i} $ the faces of the simplex $ bar {x} $. But $ bar {x} $ could also be a simplex face of another integer! It is therefore possible to obtain a long-chain complex eventually.

So we have at least $ Delta = { bar {x}, bar {x_i}: i in {1, points, 6 } } $ like a complex. This does not necessarily correspond to an abstract simplicial complex in which $ forall sigma in Delta, $ if $ tau subset sigma $then $ tau in Delta $. Since for one, we have not defined the $ bar {x_i} $ to be sets – these are just formal symbols.

Another approach would be?

remember that $ sum limits_ {i = 1} ^ 6 alpha_i x_i = x $ or $ sum limits_ {i = 1} ^ 6 alpha_i = 2, alpha_i in {0, 1 $. It's a bit like a "convex combination". Anway, as you can see, I took elements of simplicial homology and tried to adapt it. Do you have any ideas on how to proceed?

abstract algebra – $ I $ is a free R-module if and only if $ I = Ra $

Problem: Let $ I $ to be a non-zero ideal of a commutative ring $ R $ with identity. Prove it
$ I $ is a free R-module if and only if $ I = Ra $ for some people $ a in R $ it's not a zero
divider.

My idea is:
($ Rightarrow $) Let $ I $ to be generated by $ {a_1, … a_n } (n> 1) $, then I have $ a_1x_1 + … + a_nx_n = 0 $with $ x_1 = a_2, x_2 = -a_1, … $($ n $ is same); $ x_1 = a_2, x_2 = -a_1-a_3, x_3 = a_2, x_4 = a_5, x_5 = -a_4, … $ ($ n $ is odd).
So, $ a_1 = … = a_n = 0 $. Contradiction. So, $ n = $ 1.

Is it good? Help me! Thank you!

abstract algebra – The finite Boolean ring with $ 1 neq 0 $ is isomorphic to $ mathbb {Z} _2 times mathbb {Z} _2 times dots times mathbb {Z} _2 $

Let $ R $ to be a finite boolean ring with $ 1 neq $ 0. CA watch $ R cong mathbb {Z} _2 times mathbb {Z} _2 times dots times mathbb {Z} _2 $.

This is Exercise 2 on p267 of the book Dummit and Foote's Abstract Algebra. The author gives the idea of ​​using it if $ e $ is a idempotent of a ring $ R $ with $ 1 $then

$$ Re times R (1-e) $$

or $ Re $ is a ring of identity $ e $ and $ R (1-e) $ an identity ring $ 1-e $.

My attempt:

Yes $ R = {0,1 } $then $ R cong mathbb {Z} _2 $ is trivial.

Therefore, we can assume that $ {0,1 } $ is properly contained in $ R $. So, we can choose an element $ x notin {0,1 } $. We get because every element of a boolean ring is a idempotent:

$$ Rx times R (1-x) $$

$ Rx $ is a ring of identity $ x neq $ 0 and $ R (1-x) $ is a ring of identity $ 1-x neq $ 0. In addition, $ Rx $ and $ R (1-e) $ have a cardinality strictly inferior to the cardinality of $ R $. These rings are also Boolean. The induction gives the result. $ quad square $

Is it correct?

abstract data types – Where is the original literature for the specification of an associative array ADA?

I am looking for the formal specification of abstract data type algebra for associative array, associative list (AKA multimap) and other similar ADT to quote for use in a data specification similar to binary JSON . The Wikipedia associative table entry uses non-standard mathematical terminology and their sources are all secondary sources without original sources. There is an algebra for the dictionary. It starts with Let x be a set and defines the operations on this set. Who invented the dictionary and when? It should be quite old, possible from the Journal of the Association of Computer Machinery or a similar publication.

abstract algebra – My question is about the normal subgroup of dihedral groups

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abstract algebra – How is Galois theory applied here?

I have not studied galois theory much and it's been a long time, so I could really use a sentence in this article about an algorithm for polynomial factorization.

Preliminaries are like in another question of mine to find it here:

  • $ f in Z (X) $ monic, squarefree (meaning here no multiple roots) with degree $ N $
  • A prime number $ p $ such as $ f mod p $ stay squarefree (I do not know if it's important here)
  • $ f = prod_ {i = 1} ^ n f_i $ in $ mathbb {Z} _p (X) $

We also define (for a field $ F $ and some $ i in mathbb {N} $) the
$ i $-trace of a polynomial $ g in F (X) $ as $$ Tr_i (g): =
> sum_ {l = 1} ^ { deg (g)} zeta_l ^ i $$
where the $ zeta_l $ are the
$ deg (g) $ (not necessarily different) roots of $ g $, as well as, for a
full $ l $ $$ Tr_ {1 dots l} (g): = left (Tr_1 (g), , dots, Tr_l (g) right)
$$
Since $ Tr_i (gh) = Tr_i (g) + Tr_i (h) $, all these terms can also be
defined more generally for $ g in F (X) $ if we put $$ Tr_i (q / r) =
Tr_i (q) – Tr_i (R) $$
.

Also for $ i in {1, , dots, n } $, to put
$$
V_i: = Tr_ {1 dots N} (f_i)
$$

and assume $ V_i in mathbb {Q_p} ^ N $ for everyone $ i $.

Then if $ g = prod_ {i = 1} ^ n f_i ^ {v_i} $ for arbitrary integers $ v_i $, $ g in mathbb {Q_p} (X)} and using the extended definitions of the trace implied above,

$$
Tr_ {1 , points N} (g) = V: = sum_ {i = 1} ^ n V_i
$$

Now, I already know that the $ V_i $ in the case where I look are linearly independent, so we have an isomorphism of groups

$$
Tr_ {1 dots N}: left ( left { prod_ {i = 1} ^ n f_i ^ {v_i}; ; v_1, , dots, v_n in mathbb {Z} right }, , cdot right) rightarrow
left ( mathbb {Z} V_1 + , points + mathbb {Z} V_n, , + right) subset mathbb {Q} _p ^ N
$$

Suppose, moreover, $ V in mathbb {Z} ^ N $.

Now, the statement that I do not understand is

The group of Galois on $ mathbb {Q} $ leaves $ V $ invariant, so it permutes the preforms of $ V $, so the fact that the card is 1-1 implies that $ g $ is invariant and therefore defined on $ mathbb {Q} $.

I do not really see where exactly Galois theory is applied here … are we talking about the Galois group of the Galois extension $ mathbb {Q} ( alpha_1, , points, , alpha_n) / mathbb {Q} $? How / where is it applied and what is the link with the previous images, etc.?

I do not really know where to look / what to look for, because the sentence is written for people who already understand exactly what it's talking about here, and I do not …

abstract algebra – theory of representation notation doubt

I've therefore read a little representation theory in the book "Linear representations of finite groups" and I have a question.
When the author writes: Let n be the order of G, and V is a vector space of dimension n,
with a base $ (e_t) _ {t in G} $ indexed by the elements t of G. "I do not know how to look at it, I do not think it's supposed to be $ mathbb {R} ^ n $ I guess my doubt is that I do not understand what he means with this basic notation. So any help is appreciated thanks.

abstract algebra – $ R $ commutative, $ I $ maximal w.r.t. ideals not generated, show that $ I $ is not generated.

Let $ R $ be a commutative ring with the unit, with at least one ideal that is not generated finitely. Let $ S $ to be the family of such ideals. Show $ S $ has a maximum element.

I need to show every channel $ S $ has an upper bound and uses Zorn's lemma. I can take a chain $ mathcal {C}: I_ {1} subset I_ {2} subset I_ {3} subset cdots $ in $ S $ and let $ I: = bigcup_ {n} I_ {n} $. Yes $ I = (a_ {1}, ldots, a_ {t}) $ is finished generated, so I want to come to a contradiction. From here I can say that $ I_ {n} = (b_ {n, 1}, b_ {n, 2}, ldots) subset (a_ {1}, ldots, a_ {t}) $ for each $ n in mathbb {N} $, from where $ b_ {n, i} = r_ {n, 1} ^ {i} a_ {1} + cdots + r_ {n, t} ^ {i} a_ {t} $ (a linear combination). However, I do not know where to go from there, or if it's even the right direction. I guess I should use the inclusion $ I_ {1} subset I_ {2} subset cdots $ In a way, seeing as subidels of ideals finely generated does not need to be finely generated.

Abstract Algebra – Proof of Sylow's Theorem

I read Introduction to Alexander Paulin's Abstract Algebra, and I could find a line in the proof of Sylow's Theorem.

It's on the 24th and 25th page of the notes, quoted below, where the line I do not get is highlighted

Now remember that $ S $ is the disjoint union of the orbits of our action
of $ G $ sure $ S $.

(On the notation: $ Stab ( omega) $ is the stabilizing subgroup; $ Orb (omega) $ is the orbit; and HCF means the highest common factor):


Sylow's theorem. Let $ (G, *) $ to be a finite group such as $ p ^ n $ Split $ | G | $, or $ p $ is first.
Then there is a subgroup of order $ p ^ n $.

Evidence. Suppose that $ | G | = p ^ nm $, or $ m = p ^ ru $ with $ HCF (p, u) = $ 1. Our central strategy
is to consider an intelligently chosen group action of G and to prove one of the subgroups of stabilizers
to the size $ p ^ n $. We will need to heavily exploit the orbit stabilizer theorem.

Let $ S $ be the set of all subsets of $ G $ size $ p ^ n $. An element of $ S $ is an unordered $ n $-tuple
of distinct elements in $ G $. There is a natural action of $ G $ sure $ S $ by term term composition
on the left.

Let $ ∈ S $. If we correct an order $ ω = {ω_1, · · ·, _ {p ^ n}} S $then $ g (ω): = {g * _1, · · ·, g * _ {p ^ n}} $.

• We claim first that $ | Stab (ω) | ≤ p ^ n $. To see this define the function
$$ f: Stab (ω) → ω $$
$$ g → g * _1 $$
By the property of cancellation for groups it is an injective card. Therefore $ | Stab (ω) | ≤ | ω | = p ^ n $.

• Observe this
$$ | S | = pmatrix {p ^ nm \ p ^ n} = frac {(p ^ nm)!} {p ^ n! (p ^ mp ^ n)!} = prod_ {j = 0} ^ {p ^ n-1} frac {p ^ nm-j} {p ^ nj} = m prod_ {j = 1} ^ { p ^ n-1} frac {p ^ mj} {p ^ nj} $$

Note that if $ 1 ≤ j ≤ p ^ n – $ 1 then j is divisible by $ p $ at most $ n – $ 1 time. It means
this $ p ^ nm – j $ and $ p ^ n – j $ have the same number of $ p $ factors, namely the number of $ p $
factor of $ j. It means that
$$ m prod_ {j = 1} ^ {p ^ n-1} frac {p ^ m-j} {p ^ n-j} $$

do not have $ p $ factors. Therefore $ | S | = p ^ rv $, or $ HCF (p, v) = $ 1.

Now remember that $ S $ is the disjoint union of the orbits of our action of $ G $ sure $ S $.

Therefore
there must be a $ ∈ S $ such as $ | Orb (ω) | = p ^ st $, or $ s ≤ r $ and $ HCF (p, t) = $ 1.

By
the orbiting theorem we know that $ | Stab (ω) | = p ^ {n + r – s} frac {u} {t} $.

Because $ | Stab (ω) | ∈ N $
and $ u $ and $ t $ are coprime to $ p $, we can deduce $ frac {u} {t} N $. Therefore $ | Stab (ω) | ≥ p ^ n $.

• For this choice of S, Stab () is a subgroup of size pn.


So I do not understand this line:

Now remember that $ S $ is the disjoint union of the orbits of our action
of $ G $ sure $ S $.

What does it mean? for example, if take $ G $ as $ Sym_3 = {e, a, b, c, d, f } $, the transformers of a triangle, where $ a $, $ b $ are counterclockwise and clockwise rotation; $ c $,$ d $,and $ f $ like reflections. Let $ p = $ 3, $ n = $ 1, $ m = $ 2.

then $ S $ would be the set of all subsets of $ Sym_3 $ size $ 3 $. Picking $ 3 $ elements of $ 6 $ at $ pmatrix {6 \ 3} = $ 20 choice, so $ S $ at the size of $ 20 $to know
$$ S = {{{e, a, b }, {e, a, c }, {e, a, d }, {e, a, f }, {e, b, c }, {e, b, d }, {e, b, f }, {e, c, d }, {e, c, f }, {e, d, f ,
{a, b, c }, {a, b, d }, {a, b, f }, {a, c, d }, {a, c, f }, {a, d, f },
{b, c, d }, {b, c, f }, {b, d, f }, {c, d, f } $$

So, what disjointed orbits formed this $ S $?