New answers tagged

3 votes
Accepted

Problem with understanding why $\mathbb{C}[x, y] / (x^2 + y^2 - 1)$ is UFD

There is an isomorphism $\mathbb{C}[X,Y]/(X^2+Y^2-1)\cong\mathbb{C}[T,T^{-1}]$ given by $T^{\pm1}=X\pm iY$. With this identification in mind, we see $X=(T+T^{-1})/2=\frac{1}{2}T^{-1}(T+i)(T-i)$ is not ...
  • 17.1k
2 votes

Finding when a relator can be removed from group presentation

A relator $R_n$ can be removed if it can be derived from the other ones $R_1...R_{n-1}$. In order to check if such derivation is possible you need to check if $R_n=e$ in group $\langle A| R_1...R_{n-1}...
  • 181
1 vote

Coordinate ring is a product of rings iff the variety is disconnected

Write $A$ the coordinate ring of $V$. If $V$ is disconnected we can write $V = X \sqcup Y$ where $X = Z(I)$ and $Y = Z(J)$. Then, $Z(J + I) = \emptyset$ so that $J + I = A$. Moreover, $Z(IJ) = V$ so ...
  • 2,112
1 vote

Possible orders of elements of the symmetric group $S_{10}$

Yes, from a standpoint of what has to be done, you simply need the set of all $\rm{lcm}'$s of partitions of $10$, in this case. So there's in a sense a nice neat solution. Tedious is the ...
  • 4,221
0 votes

Proving equivalence of two definitions of "field"

Your questions remains unclear but they seem to boil down to: let $(K,+,\cdot,0,1)$ be a (commutative) field, i.e. $(K,+,0)$ and $(K\setminus\{0\},\cdot,1)$ are abelian groups and $\cdot:K\times K\to ...
  • 3,728
1 vote
Accepted

Proving equivalence of two definitions of "field"

I'm going to try to clarify your question before i give the answer to my interpretation of it... A field (F,+,x) is often either defined using 9 axioms or by simply saying: $(F,+)$ and $(F\setminus\{...
  • 100
1 vote

Let $\alpha = 2\cos(\frac{2\pi}{7})$. What is the minimal polynomial of $\alpha$ over $\mathbb{Q}$?

Use $\cos 4\theta = \cos 3\theta$. This is satisfied by $4\theta = 2n \pi \pm \cos 3\theta$, so by $\cos\big(\frac{2n \pi}{7} \big)$, but only four of these are distinct (say for $n$ = $0$, $1$, $2$ ...
  • 586
4 votes
Accepted

Let $\alpha = 2\cos(\frac{2\pi}{7})$. What is the minimal polynomial of $\alpha$ over $\mathbb{Q}$?

Yes, you can directly show this. This is a classic trick which you can generalise to many similar examples. Let $\zeta=\exp\left(2\pi i\cdot\frac{1}{7}\right)$. So $\alpha=\zeta+\zeta^{-1}$. Consider ...
  • 14.3k
6 votes
Accepted

Ring of polynomials has zero divisors is equivalent to it being a direct product of rings

No. Let $R=k[y]/(y^2)$ for $k$ a field. Then $R[x]\cong k[x,y]/(y^2)$ which has zero divisors but is not the direct product of rings since it has a unique minimal prime $(y)$.
  • 52.8k
1 vote

Justify that |x| + |y| = 1 is contained within the unit circle

if $|x|+|y|\le 1$, both $|x|\le1 $ and $|y|\le 1$ are true, which implies $|x|^2\le |x|$ and $|y|^2\le |y|$. So $$|x|^2 + |y|^2\le |x|+|y|\le 1$$ (Edit: This proof has the disadvantage that it does ...
  • 20.6k
3 votes
Accepted

Justify that |x| + |y| = 1 is contained within the unit circle

Here is a correct proof of the whole matter: let $(x,y)\in B$ be arbitrary, then $(|x|+|y|)^2\le1$ by squaring both (nonnegative) sides of the defining equation, and $$1\ge(|x|+|y|)^2\ge|x|^2+|y|^2=x^...
  • 94.5k
2 votes
Accepted

Is this algebra simple?

The conditions on $\phi$ stated are NOT enough to show that the algebra is simple. We can pick, see below, our own $\phi$ (within the constraints) for which $A$ has an ideal. Now I get the impression ...
  • 9,555
5 votes
Accepted

Suppose $|G| < \infty$ and $H \leq G$ such that the product of any two left cosets has the same cardinality as $H$. Is $H$ is normal?

YES. Let $a$ and $b$ be elements in $G$. Note then that $a$ is in $aH$. Thus the relation $(aH)(bH) \supseteq abH$ holds. That the equation $|(aH)(bH)| = |H|$ holds as well however, gives the equation ...
  • 16.7k
4 votes
Accepted

isomorphism $\mathfrak{m}\otimes_A (A/\mathfrak{m})\cong\mathfrak{m}_\mathfrak{m}\otimes_{A_\mathfrak{m}}(A_\mathfrak{m} / \mathfrak{m}_\mathfrak{m})$

You can do this in a couple smaller, bite-sized pieces: $\newcommand\m{\mathfrak{m}}$ $\m_\m\otimes_{A_\m} (A_\m/\m_\m)\cong (\m \otimes_A A_\m)\otimes_{A_\m} (A_\m/\m_\m)$ since $S^{-1}M\cong S^{-1}...
  • 2,209
0 votes

Let $G$ group of order $pq$ where $p,q$ primes.Show that if $G$ contains normal groups $N$ and $K$ with $|N|=p$ and $|K|=q$ then is cyclic

Firstly, if $p=q$ then $G$ is abelian but not necessarily cyclic (think of $\Bbb Z_p\times\Bbb Z_p$). So I guess you are implicitly assuming $p\ne q$. In this case, if $|G|=pq$ then $G$ is centerless ...
  • 291
0 votes

Show that $k[x,y]/(xy-1)$ is not isomorphic to a polynomial ring in one variable.

Let $k$ be a field and $R:=k[x,y]/\langle xy-1\rangle$. Easy to check that $R$ is a localization of $k[x]$ by the set $\{x^n:n\in \mathbb{Z}_{\geq 0}\}$, i.e. any element of $R$ may be written as $$\...
0 votes

Is there an effective way to decompose gaussian integers into prime factors?

Dario Alpern's Gaussian integer factorisation calculator has, like many other programs there, an explanation of the mathematics behind it. Suppose you have the prime factorisation of the norm $N(z)$; ...
  • 94.5k
5 votes
Accepted

Why is finite rank necessary here? - Dummit & Foote 10.3.13

No, with the given hypotheses you cannot use infinite sums here. No object in this construction has a topology so there's no way to define convergence of such a sum. And in fact the statement is false ...
3 votes

Why is finite rank necessary here? - Dummit & Foote 10.3.13

I think finite-dimensionality is needed, even for the simple case that $R$ is a field and $F$ is a vector space over $R$. This is because $\operatorname{hom}_{R}(F,R)$ is the dual space to $F$, and an ...
0 votes

Prove that an expression is zero for all sets of distinct $a_1, \dotsc, a_n\in\mathbb{C}$

Something related to Lagrange interpolation: The polynomial $$\omega_i(x) = \frac{\prod_{j\ne i} (x - a_j)}{\prod_{j\ne i} (a_i-a_j)}$$ has the properties $$\omega_i(a_i) = 1 \\ \omega_i(a_j) = 0 \...
  • 48.8k
1 vote
Accepted

Let $[G:H]=p$ be prime. If there is an element $g\notin H$ satisfies $gH=Hg$, then $H$ is a normal subgroup of $G$

This solution was indicated by @runway44 in the comments: Since $H\le N_G(H)\le G$, we have $[G:N_G(H)]\cdot [N_G(H):H]=[G:H]=p\implies N_G(H)=H \lor N_G(H)=G$. But by assumption $H\subsetneq N_G(H)$, ...
  • 4,221
4 votes
Accepted

What are the squares in $\mathbb{R}((X))$?

Nothing mysterious. $$\Bbb{R}((X))^\times = X^\Bbb{Z}\ \times\ \Bbb{R}^\times\times\ 1\!+\!X \Bbb{R}[[X]]$$ $$\Bbb{R}((X))^{\times 2} = X^{2\Bbb{Z}}\ \times\ \Bbb{R}_{>0}\times\ 1\!+\!X \Bbb{R}[[X]]...
  • 70.6k
2 votes

Isomorphism between finite fields with polynomials $a^2 + 1$ and $b^2 + b + 2$

All you have to do is find all elements $x\in \mathbb{F}_3(b)$ such that $x^2+1=0$. Since $\mathbb{F}_3(b)$ is of dimension $2$ over $\mathbb{F}_3$, we can write $x=pb+q$ where $ p,q\in \mathbb{F}_3$. ...
  • 181
0 votes

How do I show a certain field is ordered.

You define $F = \{a + b\sqrt2 \mid a,b \in F\}$. This does not make sense, you probably mean $$F = \{a + b\sqrt2 \mid a,b \in \mathbb Q \} .$$ You have to show that $F$ is a subfield of $\mathbb R$ ...
  • 61.3k
0 votes

$M^{H^*}= M^{coH}$, $M^H=M^{coH^o}$ for a Hopf algebra $H$

I have solved this question on my own. For the one hand, $$ <h^*,1_H>=<h^*,u(1_k)>=<u^*(h^*),1_k>=u^*(h^*)=\epsilon_{H^*}(h^*). $$ For the other hand, we can choose a basis for $M$ ...
  • 395
2 votes
Accepted

Proving Proposition 1.1 in Atiyah and MacDonald

Hint Use the "canonical" projection $\rho:A\twoheadrightarrow A/\mathfrak a$, which sends each $x\in A$ to the coset $x+\mathfrak a$ in $A/\mathfrak a$. The bijection is between the sets $\...
  • 4,221
1 vote

Let $[G:H]=p$ be prime. If there is an element $g\notin H$ satisfies $gH=Hg$, then $H$ is a normal subgroup of $G$

HINT: Note that $[G:H]$ prime; $g \not \in H$ $\implies$ $g, H$ together generate the whole of $G$, i.e., $G =\langle g,H \rangle$. So now let $y \in G$. Then from 1., $y$ is in $\langle g, H \...
  • 16.7k
1 vote
Accepted

A smart way of proving that $\langle A/\mathfrak{a}, +, . \rangle$ is a ring.

How about using the first isomorphism theorem for rings? And the fact that $(A/\mathfrak a,+)$ is the abelian group of the quotient ring $(A/\mathfrak a,+,\cdot) $? The answer to $2)$ is clearly yes....
  • 4,221
1 vote
Accepted

Simple computation on the identity on the definition of Donaldson-Futaki invariant

It seems there is a mistake in the degrees of the polynomials, in your computation. Notice that $w_k$ is of degree $n+1$, while $d_k$ is of degree $d$. So, the two polynomials in the numerator and ...
5 votes
Accepted

Is it possible that an infinite group has exactly one infinite nontrivial proper subgroup that has a certain order?

There does indeed exist a group $G$ with precisely one proper subgroup $H$ of infinite order. Preamble Before we proceed, let's discuss the Prüfer $p$-group. Let $p$ be a prime. We know that for every ...
  • 834
4 votes

Is it possible that an infinite group has exactly one infinite nontrivial proper subgroup that has a certain order?

(Edit: this addresses the question as stated in the body, where $K = G$ is allowed.) For infinite $H$ this is impossible. If $H$ is infinite then its cardinality is unaffected by adjoining an ...
3 votes

centraliser of an element in $\mathrm{GL}(n,\mathbf Z)$

This is a difficult problem, and the answer involves algebraic number fields. There are algorithms to compute the centralizer by Eick, Hoffman and O'Brien, based on work by Grunewald and Segal. See ...
  • 81.2k
1 vote
Accepted

If $G$ has both chain conditions, then there is no proper $H \leq G$ with $G \cong H$, and no normal nontrivial $K \leq G$ with $G/K \cong G$.

The first part of the statement is actually false as given (maybe Rotman meant for $H$ to be characteristic in $G$, or to be a direct factor of $G$?) A counterexample is given by the infinite simple ...
0 votes

Show that no group of order 48 is simple

Let $G$ be the group of order $48$. $|G| = 48 = 2^4 \cdot 3$ Then number of Sylow $2$-subgroups are $1$ or $3$. If it is one then we are done. So assume there are $3$ Sylow $2$-subgroups. Let $H_1$ ...
2 votes

Weil restriction of quasi-projective varieties

You can find explicit equations in the affine case in Section 4.6 of Poonen's Rational Points on Varieties (which you can read for incidental use for free online on his webpage, google it, it's ...
  • 26.4k
1 vote

Determining All Groups of Order 308

Here are two approaches, both through the Sylow $11$-subgroup. Let $|G|=308$. Since $308=2^2\cdot 7\cdot 11$, we see that the number of Sylow $11$-subgroups is $1$ (it's one of $1$, $2$, $4$, $7$, $...
2 votes
Accepted

In any ring $R$:product of non-unit and any element is non unit.

Consider the abelian group $A=\Bbb Z^{\omega}$ of sequences of integers and let $R=\operatorname{End}(A)$ be its ring of endomorphisms. For $f\in A$, define $D(f)$ per $$ D(f)(n)=f(n+1)$$ and $I(f)$ ...
2 votes

Twisted graded algebras [J.J. Zhang]

Let $A=\bigoplus_{i \in \mathbb{Z}} A_i$ be a $\mathbb{Z}$-graded algebra and $\sigma$ an automorphism of $A$ that respects the grading (so $\sigma(A_i) \subset A_i$). The "Zhang twist" of $...
  • 279
0 votes

Exercise 2.3 in Serre's Linear Representations of Finite Groups (and proof verification)

Your confusion about "right" (in the comments) is a common one. Let $\rho: G \rightarrow \mathrm{GL}(V)$ be the given representation. We define $\rho': G \rightarrow \mathrm{GL}(V')$ as ...
  • 2,018
0 votes
Accepted

Ideal in $A[x]$ generated by $a,x$

This is not true. Consider the ring $A=\mathbb Z/6\mathbb Z$. We have the principal ideal $(2x+3)$ contains $x$, as $$(2x+3)(3x+2)=x$$ But it's clear that $2x+3\not\in(x)$. Note that $(2x+3)\not=(1)$. ...
  • 9,026
0 votes

Let $F$ be a flat $R$-module, and let $a \in R$ be an eleemnt which is not a zero-diisor. Show that if $ax = 0$ for some $x \in F$ then $x = 0$

Multiplication by $a$ gives a monomorphism $R\to R$. Now tensor with $F$, which is flat, so preserves monos. We see that multiplication by $a$ is a mono $F\to F$, so $ax=0$ implies $x=0$.
0 votes

Ideals generated by all nilpotent elements in noncommutative rings

I think I have a partial answer for 1. Consider the ring $$ R = \{ a + bi + cj + dk : a,b,c,d \in \mathbb{Z} / m \mathbb{Z}\},$$ where $m$ is a composite integer such that the power of some of its ...
  • 161
2 votes
Accepted

Difficulty proving the second isomorphism theorem: $\frac{X}{X\cap Y}\cong\frac{X\cup Y}{Y}$ in an Abelian category

The sum $A+B$ is the pushout of the span $A \leftarrow A\cap B \rightarrow B$. Whence you have a diagram ...
  • 136
2 votes
Accepted

Question about the ring $A=k[X,Y]/(Y^2-X^3)$, Miles Reid section 4.4 example (iii)

Given a commutative ring $A$ and an ideal $I$ of $A$, an element of the quotient ring $A/I$ is a coset of $I$, that is, a subset of $A$ of the form $a + I = \{a + b : b \in I\}$ for some $a \in A$. ...
  • 4,385
2 votes

calculate $II=${$xy \mid x,y \in I$} for $I=(2,1+\sqrt{-5}) \in \Bbb{Z}[ \sqrt{-5}]$

I suspect the OP may just be starting a course in algebraic number theory, so this may not help until the end, but the answer is that $z \in \mathbb{Z}[\sqrt{-5}]$ can be written as $xy$ for $x$, $y \...
1 vote
Accepted

exact sequence of differential

$N$ is not just any $B$-module, it’s a $B$-module where the $B$-action factorizes through $C$ – which means exactly that $I$ annihilates $N$.
  • 28.1k
0 votes

Show that $f$ is an epimorphism of groups if and only if $f$ is surjective as a map of groups.

Here is an alternative answer to this old question. The key point is to show the following: Given a subgroup $H$ of a group $G$, there is a group $K$ and homomorphisms $a, b:G\to K$ such that $a(g)=b(...
  • 1,186
1 vote
Accepted

Prove $ \bar{u}\in \operatorname{Hom}_{A} (\operatorname{Hom}(M,N), \operatorname{Hom}(M',N))$

Suppose $u : M' \to M$ is an $A$-module homomorphism. Then $\overline{u} : \text{Hom}(M, N) \to \text{Hom}(M', N)$ is defined by $$\overline{u}(f) = f \circ u.$$ First, observe that $f \circ u$ is an $...
0 votes

elements contained in the sigma algebra generated by the union of two sigma algebras

here are the two missing elements: $$(A \cup B) \cap (A^c \cup B^c), (A \cap B) \cup (A^c \cap B^c)$$ These are complementary each other. The first one is the symmetric defference of $A,B$, while the ...
  • 35.6k
2 votes

Is every ordered abelian group the additive group of an ordered ring?

For instance, the lexicographically ordered group $(\mathbb{Z}^2,+,<)$ is the underlying ordered group of no ordered ring. edit: in fact, what I show below is rather that there is no structure of ...
  • 4,831

Top 50 recent answers are included