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Doubt in statement of Gauss' Lemma in Fulton's Algebraic Curves

Yes, we need $F$ to be a non-constant irreducible element. This is noted in Dummit and Foote's book just after the proof of Gauss Lemma. For example $2$ is irreducible in $\mathbb Z[X]$ but a unit in $...
Steve Morris's user avatar
1 vote
Accepted

The class $[m]_n$ generates $\mathbb{Z}/n\mathbb{Z}$ iff $\text{gcd}(m,n) = 1$

Hint Well, $[1]_n$ clearly generates, right? Then use the following.
i can try's user avatar
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0 votes

Finite subgroups of the multiplicative group of a field are cyclic

Here is an alternative argument, essentially the 8th proof presented by Keith Conrad here, but organized a little differently and trying to be clear about the flow of ideas, and what is and is not ...
Qiaochu Yuan's user avatar
0 votes

The connection between determinants and eigenvalues

The characteristic polynomial $f$ of $A$ is of the form $\lambda^n + p(\lambda)$, where $p$ is a polynomial of degree $\le n-1$. Therefore $\lim_{\lambda \to\infty} f(\lambda) = \infty$. By ...
Fnacool's user avatar
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1 vote

The connection between determinants and eigenvalues

The eigenvalues of $A \in M_n(\mathbb{R})$ are precisely the roots of $$p(\lambda) = \det(\lambda I_n - A) = \lambda^n \det\left(I_n - \frac{1}{\lambda}A\right)$$ As $|\lambda| \to \infty$, we have $\...
Tob Ernack's user avatar
  • 4,525
2 votes

The connection between determinants and eigenvalues

This uses some facts from abstract algebra. The characteristic polynomial of $A$ has no real roots. A real polynomial, that has no real roots has only complex roots. Complex roots, however, always ...
Marius S.L.'s user avatar
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3 votes
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The connection between determinants and eigenvalues

Hints: The determinant is the product of eigenvalues. Do eigenvalues come in conjugate pairs in the case of a matrix with real entries? What's a complex number times its conjugate?
J. Chapman's user avatar
1 vote
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An exotic operation on repeating sequences of natural numbers such as $\overline{6,9} = 6,9,6,9,6,9, \dots$ Having to do with common subsums.

Im not sure if I understood correctly but you seem to wonder about taylor series with periodic derivatives. So maybe this is insightful. Let $$f(x) = \sum f_n x^n$$ $$g(x) = \sum g_n x^n$$ then $$T(x) ...
mick's user avatar
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2 votes

Roots of a monic polynomial $f(X) \in k[X]$ in splitting field distinct, form a field, then char $k = p$ and $f(X) = X^{p^n} - X$

Here are two generalizations: What happens if we only ask for the roots of $f(x)$ to be closed under addition? What happens if we only ask for the roots of $f(x)$ to be closed under multiplication? ...
Qiaochu Yuan's user avatar
0 votes

(p,n)=1 and if $v \in F$ and $n v \in K$ then prove that $v \in K$

We know $(p,n)=1$ implies $\exists a,b\in \Bbb{Z}$ such that $ap+bn=1$. Then $$\begin{align} v &= (ap+bn) v \\ &= (ap)v + (bn)v \\ &= a(pv) + b(nv) \\ &= a0 + b(nv) \\ &= b(nv) \in ...
user264745's user avatar
  • 4,157
2 votes
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Does the monoid of non-zero representations with the tensor product admit unique factorization?

Counterexamples are much easier to produce than this and exist already when $G = C_2$ and with two $2$-dimensional representations, see here. Abstractly the problem is that the representation ring is ...
Qiaochu Yuan's user avatar
0 votes

A question based on intermediate fields and separability

Let $f\in K[x]$ be irreducible polynomial of separable $u$ over $K$. Let $f=f_1\cdots f_n\in E[x]$ where $f_i$ are monic irreducible factors of $f$ in $E[x]$. Then $f(u)=f_1(u)\cdots f_n(u)=0$ implies ...
user264745's user avatar
  • 4,157
0 votes

A criterion for proving that an algebraic extension $E \subset F$ of fields is normal.

Your third paragraph don’t establish anything. Normal extension is by definition algebraic extension. So each $u\in F$ is algebraic over $K$, since $u\in E$ is algebraic over $K$. I would write proof ...
user264745's user avatar
  • 4,157
0 votes

(Intermediate) normal extension is stable

I’m bit sceptical of B.Swan’s answer, mainly because $L\subseteq \overline{N}$ don’t hold in general. In my proof, I have used “usual” definition of normal extension. Proof: Let $n\in N$ and $f\in F[x]...
user264745's user avatar
  • 4,157
-1 votes

Is the localization of a zero-dimensional ring a quotient?

Here is an argument when $R_m$ is of finite length, not just zero dimensional. In this case, we may assume $m^nR_m=0$ for all large $n$. Let $x\in R_m$ which is not in the image of $R$. Let $s\in R-m$ ...
Mohan's user avatar
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0 votes

Given F is Galois over E and E is Galois over K and a splitting field condition is given

Fundamental theorem stated in Henry Davii’s answer is true when $F$ is finite dimensional extension of $K$. You have to use generalized fundamental theorem instead, which is theorem V.3.12 in ...
user264745's user avatar
  • 4,157
2 votes

Does the monoid of non-zero representations with the tensor product admit unique factorization?

A counterexample is given by Nate at https://math.stackexchange.com/a/4436073/491450 : Taking $G = A_5$, we have $$ V_4 \otimes V_5 \otimes V_3 \cong V_4 \otimes V_5 \otimes {V_3}' $$ where the ...
Smiley1000's user avatar
  • 1,605
2 votes
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How to determine $\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_4,\Gamma_5$?

For $m=3$, as you mentioned, the group of characters is isomorphic to $(\mathbb{Z}/3\mathbb{Z})^\times$, which is a group of order $\varphi(3)=2$. We know that $(\mathbb{Z}/3\mathbb{Z})^\times=\{1,2\} ...
ljfirth's user avatar
  • 490
2 votes

Direct sum of free abelian group and quotient of abelian group by subgroup

Since $C$ is free abelian, $C = \mathbb{Z}^{(I)}$ for some index set $I$. For each $i \in I$, let $e_i \in \mathbb{Z}^{(I)} = C$ be the element that’s $1$ at $i \in I$ and $0$ everywhere else. Since $...
David Gao's user avatar
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0 votes

Confusion with application of butterfly lemma in Lang's Algebra

Using Lang's notation, we have that $H_{j+1} \unlhd H_j$ and $G_{i+1} \unlhd G_i $. Therefore using the butterfly lemma we have $\frac{H_{j+1}(H_j \cap G_i)}{H_{j+1}(H_j \cap G_{i+1})} \cong \frac{(...
baslerbuenzli's user avatar
1 vote

Factoring polynomials over rings and fields

Over $\mathbb{Z}$ the answer is that $x^n - 1$ decomposes into a product of cyclotomic polynomials $$x^n - 1 = \prod_{d \mid n} \Phi_d(x)$$ where $\Phi_d(x)$ is the monic polynomial whose roots are ...
Qiaochu Yuan's user avatar
1 vote

Irreducible factors of $X^n -1$ in $\mathbb{F}_q[X]$

If $f$ is an irreducible factor of $t^n-1$ over $\mathsf k=\mathbb F_q$, then if $K$ is the splitting field of $g(t)=t^n-1$, $f$ splits in $K$, and hence if we take the subfield generated by one of ...
krm2233's user avatar
  • 4,924
7 votes

Adjoining a new element to a ring $R$ that is not a field.

The notation $R[\pi]$ is unclear because you haven't defined the meaning of this symbol $\pi$; this could be interpreted as referring to the polynomial ring. The unambiguous way to adjoin an element ...
Qiaochu Yuan's user avatar
4 votes
Accepted

Galois group of $\mathbb{C}(t)$ over $\mathbb{C}(t+t^{-1})$

This is one of the rare cases in which it's eaiser to compute Galois group than the degree first. Without considering the Galois group, it's not easy to show $x^{2n}-(t^n+t^{-n})x+1$ is irreducible. ...
Just a user's user avatar
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3 votes
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$Out(F_n)$ has a free abelian subgroup of rank $2n-3$

Let me use the notation $$L_i : \begin{cases} y_i &\mapsto y_1 y_i \\ y_j &\mapsto y_j \quad\text{if $j \ne i$} \end{cases} $$ $$R_i : \begin{cases} y_i &\mapsto y_i y_1 \\ y_j &\...
Lee Mosher's user avatar
  • 123k
0 votes

Subgroup of the braid group $\mathbb{B}_n$ generated by $\sigma_1,\dots,\sigma_{n-2},\sigma_{n-1}^2$

Here is an outline of one approach, although it's more visual than group-theoretic. The statement amounts to the claim that any Artin braid with the property that the $n$th strand starts and ends in ...
DrM's user avatar
  • 81
1 vote

Localization of a module at a prime ideal and local behavior.

$\def\p{\mathfrak{p}}$We need $M$ finitely presented for the result to hold [ref]. Contrary to what Alex Youcis says, $M$ finite is not enough. Let's show Mohan's comment in OP's question is indeed a ...
Elías Guisado Villalgordo's user avatar
0 votes

Is $\sqrt{3}$ in the field of 7-adic numbers $\mathbb{Q}_7$?

First, this is the same problem with finding the solution of the equation $x^2 = 3$ on $\Bbb Q_7$. Now $x^{2} = 3$ implies $v_7(x^2) = v_7(3) = 0 \iff v_7(x) = 0$ so $x \in \Bbb Z_{7}$. So we can ...
Afntu's user avatar
  • 1,494
0 votes

Determine the degree of a root of $X^p-X-\alpha$ (Serge Lang Algebra exercise VI.29)

I'll give you a new proof for this question. Proof. Observe that $F$($θ$) is the splitting field of a separable polynomial, hence a Galois extension of $F$. We also know that [$F$($θ$) : $F$] = $p$ so ...
Hyperplane lover's user avatar
0 votes

Example of a non-Noetherian complete local ring

As another example, you can consider the commutative ring of series with shuffle product. Let $\mathbb F$ be a field (say the rational numbers $\mathbb F = \mathbb Q$), $\Sigma$ a finite set of ...
Lorenzo's user avatar
  • 39
1 vote

Irreducible factors of $X^n -1$ in $\mathbb{F}_q[X]$

I came up with a proof that uses less terminology but that may be different from what you brought up in your post [my eyes still glaze over at the words 'congruence class', I am sorry!] Proof #1: Let $...
Mike's user avatar
  • 21k
2 votes
Accepted

$\operatorname{End}_S(M \otimes_R S) \cong \operatorname{End}_R(M) \otimes_R S$?

No, not necessarily. What you'd like to do is argue that the universal property of the tensor product allows you to identify $$\text{End}_S(M \otimes_R S) \cong \text{Hom}_R(M, M \otimes_R S)$$ and ...
Qiaochu Yuan's user avatar
1 vote

$\gcd(\text{multiples}(a,b))= \text{lcm}(a, b)$?

Define $a\leq b$ iff $a | b$. If your ring is nice enough, then $\leq$ is a partial order on the equivalence classes $x\sim y$ iff $x = c\cdot y$ for some invertible $c$. I claim your question has not ...
Jakobian's user avatar
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1 vote
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...if $\mathfrak{a}\subset\cup_{i=1}^{s}\mathfrak{p_i}$, then $\mathfrak{a_1}\subset \mathfrak{p_i}$ for some $i$

(1) I'm unsure - there doesn't seem to be much induction going on here, though I could be incorrect. (2) We chose $a_1$ and $a_2$ so that $a_1 \notin \mathfrak{p}_1$ and $a_2 \notin \mathfrak{p}_2$. ...
ljfirth's user avatar
  • 490
0 votes

$\gcd(\text{multiples}(a,b))= \text{lcm}(a, b)$?

Clearer: $ $ if a set $M = \{m_1,m_2,\ldots\}$ contains a $ $ $\rm\color{#c00}{common}$ $ $ divisor $\,\color{#c00}{m_1\mid m_2,\,m_3,\,\ldots}\,$ then $\,\gcd\,\{m_i\} \sim m_1,\ $ since $\,m_1\,$ ...
Bill Dubuque's user avatar
3 votes
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Relationship Between Subgroups of Abelian Groups & Ideals/Rings.

When the multiplication exists, it need not be unique. Fix a prime $p$, and let $A$ be the abelian group $$A = \frac{\mathbb{Z}}{p\mathbb{Z}}\times\frac{\mathbb{Z}}{p\mathbb{Z}}\times \frac{\mathbb{Z}}...
Arturo Magidin's user avatar
2 votes

Relationship Between Subgroups of Abelian Groups & Ideals/Rings.

So unless I am misunderstanding your question, I believe such a multiplication is not uniquely defined. Let $G =\{0,a,b,c\}$ be the Klein-4 group (with $0$ being the identity). Then $(c) = \{0,c\}$ is ...
NaturalLogZ's user avatar
0 votes
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Why $P(y/x) = 0$?

This is just a direct result of $P(\varphi)(x) = 0$.
Functor's user avatar
  • 797
0 votes

Left coset is generated by the element of G, but not H

This is your definition of a left coset: left coset of H in G is a subset of G of type aH for some element a of G. Your definition of coset does not require that $H$ is a subgroup of $G$. This is ...
Ka Wa Yip's user avatar
  • 938
0 votes

Are cosets groups?

This is your definition of a left coset: $aH=\{ah\ |\ h\in H\}$, where $H$ is a subgroup of $G$, and $a\in G$. Are cosets groups? Some of them are not. The identity $I$ is not contained in the coset ...
Ka Wa Yip's user avatar
  • 938
2 votes

Signed measure and bounded total variation on algebras

Here is a counterexample, if I did not misunderstand the notations: Let $X = \mathbb{N} \cup \{\infty\}$ be the one-point compactification of the naturals. Let $\mathcal{B}$ consists of finite subsets ...
David Gao's user avatar
  • 7,830
0 votes

Characteristic of a ring R divides the number of elements m of a ring

Set n $=$ char(R). Since R $=$ m, we know that $mx = 0$ for any x $\in$ R By a corollary to Lagrange's Theorem In particular, this tells us that $n = 0$. Hence, by the division algorithm, $\exists$ q,...
Anindya Das's user avatar
1 vote
Accepted

Example of a residuated prelinear lattice that isn't linear

You can verify that any Boolean algebra is a residuated lattice where $\star$ is the meet operation ($\wedge$) and $$x \to y = x' \vee y.$$ Then we certainly have $$(x\to y) \vee (y \to x) = (x' \vee ...
amrsa's user avatar
  • 13.1k
2 votes

Automorphism of order 2 of $D_q$

Hint: You can also use, that the automorphism group of $D_n$ for $n\ge 3$ is isomorphic to $$ {\rm Aut}(D_n)\cong C_n\rtimes U(n), $$ of order $n\phi(n)$, where $U(n)=(\Bbb Z/n\Bbb Z)^{\times}$ is the ...
Dietrich Burde's user avatar
0 votes

Idempotents in a local ring

Let $m$ be the unique maximal ideal of a local ring $A$. The Jacobson radical ($J(R)$) is defined as the intersection of all maximal ideals of $A$. Since $J(R)$ contains the nilradical $R$, which is ...
Aristarchus_'s user avatar
1 vote
Accepted

Image of submodule to a quotient

I would say this is right, but it's not clear that you understand the crux of what's happening here. Say $N, M'$ are submodules of $M$ an $A$-module. Say $\pi \colon M \to M/M'$ natural quotient. What'...
tiny's user avatar
  • 95
2 votes
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Can restriction map trivialize a element in $H^1(G_K,M)$ by finite extension?

$\DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\res}{res}$ Here is a simple explicit construction of such an $L$. Since $M$ is finite, the map $G_K\to \Aut M$ ...
Alex B.'s user avatar
  • 19.8k
9 votes

Is there an associative binary operation $\otimes:\mathbf N^2\to\mathbf N$ such that $n\otimes n=0$ and $0\otimes n=n\otimes 0=n$ other than xor?

$\otimes$ gives $\mathbb{N}$ the structure of a monoid with identity $0$ in which every element has order dividing $2$. Since in particular every element is its own inverse, every element is ...
Qiaochu Yuan's user avatar
3 votes

Confusion about rings and ideals and how ideals are always subrings.

This is an issue in the definition of "subring," which I am sorry to report can vary from author to author. Some authors don't require rings to have multiplicative identities, and some ...
Qiaochu Yuan's user avatar
1 vote

Localization commutes with arbitrary direct sums

I realize this probably will not be useful for the original poster, but I'm doing this exercise in Vakil right now and I think this explanation is worth having in mind. To make the use of universal ...
Connorlindquist's user avatar

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