Skip to main content

New answers tagged

1 vote
Accepted

Check when a positive square Root $\sqrt{d}$ is contained in Cyclotomic Field

The Galois group of $\mathbb Q(\zeta_n)/\mathbb Q$ is $G_n=(\mathbb Z/n\mathbb Z)^\times$. If $n=\prod_{i=1}^k p_i^{m_i}$ is the factorization of $n$, this gives us $G_n\cong \prod_{i=1}^k G_{p_i^{m_i}...
user8268's user avatar
  • 21.6k
3 votes
Accepted

Let $F \supset K \supset L$ be fields with orders less than 100 and do not include an element $x\neq 1$ that $x^5=1$. Find the order of $F$.

Let $p$ be the characteristic of the field. Then $|F|=p^n$ for some $n$. An argument similar to yours shows that $n\geq4$. Note that we only have the two possibilities $2$ or $3$ for $p$, as already $...
spinosarus123's user avatar
3 votes
Accepted

Why is the subfield of $\mathbb{Q}(\zeta_p)$ of index $2$ expressible in terms of the sum of $\zeta_p$ to the power of all quadratic residues mod $p$?

It is called a subfield of degree $2$ (over $\mathbf Q$), or more commonly a quadratic subfield. To see a number $\alpha$ in $\mathbf Q(\zeta_p)$ is quadratic over $\mathbf Q$, we can do two things: (...
KCd's user avatar
  • 46.9k
3 votes
Accepted

Clarification on the proof that $[\mathbb F (\alpha) : \mathbb F] = \deg_{\mathbb F} \alpha$

Yes, your answer is correct. The minimal polynomial $m_{\alpha}$ divides any polynomial which has $\alpha$ as a root. Since $P(\alpha)=0$ by assumption, $m_{\alpha}$ divides $P$ which means that $P \...
wiishopwednesday's user avatar
3 votes
Accepted

Can the composite of $\mathbb Q_p$ and $\mathbb Q_\ell$ in $\mathbb C$ be $\mathbb C$?

Fix an enumeration of $\mathbb{C}$ as $(a_\alpha)_{\alpha<\mathfrak{c}}$ and fix algebraically independent elements $(x_\alpha)_{\alpha<\mathfrak{c}}$ in $\mathbb{Q}_p$ and $(y_\alpha)_{\alpha&...
Eric Wofsey's user avatar
1 vote

Let $L = \mathbb Q(\sqrt{2+\sqrt{3}})$ What is its degree over $\mathbb Q$

Notice that the polynomial $p(x)$ has got no rational root. Hence if it is reducible, it must split as product of two quadratic polynomials over $\mathbb{Q}$. Suppose $$x^4-4x^2+1 = (x^2+bx+c)(x^2+dx+...
Yathi's user avatar
  • 2,350
5 votes

Let $L = \mathbb Q(\sqrt{2+\sqrt{3}})$ What is its degree over $\mathbb Q$

As $[L:\mathbb Q]=[L:\mathbb Q(\sqrt 3)]\cdot[\mathbb Q(\sqrt 3): \mathbb Q]=2[L:\mathbb Q(\sqrt 3)]$, so $2\mid[L:\mathbb Q]$. And $[L:\mathbb Q]\le 4$, $[L:\mathbb Q]$ is either $2$ or $4$. If $[L:\...
Just a user's user avatar
  • 16.2k
1 vote

Let $L = \mathbb Q(\sqrt{2+\sqrt{3}})$ What is its degree over $\mathbb Q$

Try to show that $[L:\mathbb Q(\sqrt{3})]=[\mathbb Q(\sqrt{3}):\mathbb Q]=2$. Then using the fact that $[L:\mathbb Q]=[L:\mathbb Q(\sqrt{3})]\cdot [\mathbb Q(\sqrt{3}):\mathbb Q]$, the rest follows.
ShyamalSayak's user avatar
4 votes
Accepted

Is $\Bbb Q[a^2b+b^2c+c^2d+d^2a]$ the fixed field of the subgroup $\langle(1234)\rangle$ of Galois group $S_4$?

Yes it is always true. The following argument works for any base field assuming that it doesn't have characteristic $2$, so certainly for the field $\mathbf{Q}$. Determining the field is the same as ...
user297024's user avatar
  • 1,297
0 votes
Accepted

Subgroups and corresponding subfields of Galois group of $x^5-5x^2-3$

The fixed field of $⟨σ⟩$ is $\Bbb Q[\sum_{j=1}^5\sigma^j(r_1^2r_2)]$. ...
hbghlyj's user avatar
  • 2,842
2 votes

Irreducibility over Field extensions.

This argument only works for $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}, ..., \sqrt{p_{k-1}})$. It uses concepts from algebraic number theory, which you may or may not have studied. Note that $p$ is ...
Lukas Heger's user avatar
  • 21.7k
3 votes
Accepted

What is the index of $\textrm{SL}_2(\mathbb{R})$ as a subgroup of $\textrm{SL}_2(\mathbb{C})$

The index is infinite. When $z \in \mathbf C^\times$, the matrix $M_z := (\begin{smallmatrix} z&0\\0&1/z\end{smallmatrix})$ is in ${\rm SL}_2(\mathbf C)$. For $z$ and $w$ in $\mathbf C^\times$,...
KCd's user avatar
  • 46.9k
1 vote
Accepted

Compute $\operatorname{Gal}(\mathbb{Q}(\sqrt{1+2i},\sqrt{1-2i}) / \mathbb{Q})$

The primitive element theorem dictates there is an $c$ such that $\mathbb Q(a,b)=\mathbb Q(c)$, but $c=a+b$ may not work. For example, $$\mathbb Q(\sqrt 2, \sqrt 3) = \mathbb Q(\sqrt 2 + \sqrt 3, \...
Just a user's user avatar
  • 16.2k
1 vote
Accepted

When normal extensions are normal

Neither normal nor "finite and Galois" have property $\mathcal{P}$. (Note that there is such a thing as an infinite Galois extension...) Note: I misread the problem initially. What is below ...
Arturo Magidin's user avatar
2 votes
Accepted

Degree of field extension $\Bbb Q(\sum\limits_{k=1}^{\text{ord}_n(2)}\zeta_n^{2^k}):\Bbb Q$

The answer to your question is yes when $n$ is squarefree. Here is a general result. When $L/K$ is Galois with Galois group $G$, $L = K(\alpha)$, and $H$ is a subgroup of $G$, then we can ask ...
KCd's user avatar
  • 46.9k
2 votes
Accepted

Inverse problem of the field extension $K/\Bbb{Q}$ of given number of prime above $p$

It is a theorem that for every number field, infinitely many primes split completely in it. So let $K$ be an arbitrary number field with degree $g$ over $\mathbf Q$, e.g., $\mathbf Q(\sqrt[g]{2})$ and ...
KCd's user avatar
  • 46.9k
2 votes
Accepted

Do I need the assumption that $\operatorname{char}(k) = p$ here?

I'd like to add more details to your proof of $\{t^au^b\}_{0\le a,b<p}$ are linearly independent. Assume $\sum_{a,b}c_{a,b}t^au^b=0$, we want to show $c_{a,b}=0$. By multiplying the product of all ...
Just a user's user avatar
  • 16.2k
3 votes
Accepted

What is the fixed field of $\mathbb{Q}(\sqrt[3]{5}, \sqrt[3]{5}\zeta_3)$ with 3-cycle group?

The fixed field is $\mathbb{Q}(\zeta_3)$. To verify that $\zeta_3$ lies in the fixed field, note that $$\zeta_3 = \frac{\sqrt[3]{5}\zeta_3}{\sqrt[3]{5}}.$$ So the image of $\zeta_3$ under the ...
Arturo Magidin's user avatar
1 vote

Minimal polynomials and field extensions

Yes. Let $V$ be the $F[x]$-module whose underlying vector space is $F^n$ and on which $x$ acts via $A$. Then $V$ is a finitely generated module over a PID, and so it decomposes uniquely as a sum $$ F[...
hunter's user avatar
  • 30.9k
0 votes
Accepted

How to view $k'$-algebra structure on $k$?

So here is a proof via flatness. First notice that $k' \otimes k[x] = k'[x]$ according to the defintion of module of polynomials. Each ideal of $k[x]$ can be viewed as a vector space of $k$(of ...
RHspqr's user avatar
  • 358
1 vote
Accepted

Proof of Newton's Formulas.

Wikipedia states the identity as $$k e_k(x_1,\ldots,x_n) = \sum_{q=1}^k (-1)^{q-1} e_{k-q}(x_1,\ldots,x_n) p_q(x_1,\ldots,x_n).$$ We have for the RHS $$- \sum_{q=1}^k (-1)^q [w^{k-q}] \prod_{r=1}^n (1+...
Marko Riedel's user avatar
  • 61.7k
1 vote
Accepted

Annihilator basis after extension of scalars

Note that $S_a$ is the kernel of the map $a : A \to A \oplus A$ given $$ x \mapsto (ax, xa). $$ Thus, we have an exact sequence $$ 0 \to S_a \to A \xrightarrow{a} A \oplus A. $$ By exactness of the ...
Frank's user avatar
  • 2,603
2 votes

$K_1,K_2$ are isomorphic subfields of $L$ via $\sigma$, every polynomial $P(x)$ over $K_1$ have the same number of roots in $L$ as $\sigma(P(x))$?

The problem is that $K_1$, $K_2$ may be isomorphic, but not "sitting the same way" in $L$. Now, an example, $K_1= L = k(t)$, and $K_2 = k(t^2)$, with the isomorphism $\sigma \colon t\mapsto ...
orangeskid's user avatar
  • 54.4k
4 votes
Accepted

$K_1,K_2$ are isomorphic subfields of $L$ via $\sigma$, every polynomial $P(x)$ over $K_1$ have the same number of roots in $L$ as $\sigma(P(x))$?

No, not always. Let $K_1=K_2=\Bbb{Q}(\sqrt2)$, $\sigma(\sqrt2)=-\sqrt2$. Let $L=\Bbb{Q}(\root4\of2)\subseteq\Bbb{R}$. Then $P(x)=x^2-\sqrt2$ has two roots in $L$ whereas $\sigma(P(x))=x^2+\sqrt2$ has ...
Jyrki Lahtonen's user avatar

Top 50 recent answers are included