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13 votes
Accepted

Cubic polynomials over finite fields whose roots are quadratic residues or non-residues

EDIT: Following a clever observation of user44191 in the comments: If $f(x)$ is a monic polynomial, and $c$ a number, then the polynomial $xf(x)^2+c$ has a similar property to your example (the case $...
Will Sawin's user avatar
  • 139k
13 votes

Cubic polynomials over finite fields whose roots are quadratic residues or non-residues

Yes, it does. Here is a stronger claim: Theorem 1. Let $F$ be a field of characteristic $\neq2$. Let $c\in F$. Let $r\in F$ be a nonzero square, and let $n_{1},n_{2}\in F$ be such that the ...
darij grinberg's user avatar
11 votes

What are the primes that are ramified?

All the information on the higher ramification groups can be derived from Theorem II.5.6, which TKe notes. The Galois group of the ray class field is the group of fractional ideals relatively prime ...
Will Sawin's user avatar
  • 139k
9 votes

Sum of subfields of $\mathbb{C}$

I think I can show that if the cardinality of the continuum is a regular cardinal (for example, if the continuum hypothesis is true, or more generally if $2^{\aleph_0}=\aleph_n$ for any natural number ...
Jeremy Rickard's user avatar
8 votes
Accepted

Image of the norm map for degree $3$ galois extension over $\mathbb{Q}$

There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ divides a ...
S. carmeli's user avatar
  • 4,074
7 votes
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If $G=\mathsf{Aut}_k(F)$ acts on field $F$ algebraic over $k$ then do we have: orbit $G\alpha=\text{ roots of minimal polynomial of }\alpha$?

Here is a different interpretation of the question, which is hopefully closer to OP's intent: Let $F/k$ be an algebraic field extension, and let $\alpha\in F$. Does $Aut_k(F)$ act transitively on the ...
Wojowu's user avatar
  • 27.6k
5 votes
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Concerning $k \subset L \subset k(x,y)$

No. $M\mathrel{:=}k(x,y)$ has a $k$-automorphism $\sigma:x\mapsto 1/x,\,y\mapsto 1/y$, of order 2. Let $G\mathrel{:=}\langle\sigma\rangle$, and put $L\mathrel{:=}M^{G}$, the fixed field. The elements $...
Matthé van der Lee's user avatar
4 votes

A quantity associated to a field extension

For the rest of this post, I'm talking about fields not of characteristic $2$. Under this assumption, we will show that any such $V$ is either a subfield of $E$ or is a complement of a subfield $F'$ ...
user44191's user avatar
  • 4,991
4 votes
Accepted

Is the minimal polynomial of an algebraic formal Laurent series always separable?

Yes. Let $p=char(K)$ and $\alpha\in \overline{K(x)}\cap K((x))$ assumed to be inseparable over $K(x)$. Let $L= K^{1/p^\infty}$ which is perfect. If $\alpha$ is inseparable over $L(x)$ then $\alpha$'s ...
reuns's user avatar
  • 3,425
3 votes

Artin-Schreier theorem for rings (a little different)

I am assuming that you intended for $R$ to already be integrally closed in $K$, i.e. normal. Then the answer to your question is yes. Claim: Suppose the domain $R$ is integrally closed in its field of ...
Manny Reyes's user avatar
  • 5,152
3 votes

A quantity associated to a field extension

For $F$ of characteristic $2$, a space $V$ has this form if and only if there are fields $F_1, F_2$ with $F \subseteq F_1 \subseteq F_2 \subseteq E$ and $F_2^2 \subseteq F_1$, and $V$ is an $F_1$-...
Will Sawin's user avatar
  • 139k
3 votes
Accepted

Absolutely irreducible representation and splitting field

Yes, $F$ is a splitting field of $A$. Put $A_E = A\otimes_F E$ for convenience. Let $S'$ be a simple $A$-module. Then $S'\otimes_F E$ has a simple $A_E$-submodule, which by hypothesis is of the form $...
Benjamin Steinberg's user avatar
3 votes

Fields with restrictions on their finite extensions: Given $n\in\mathbb{N}_{>1}$ which fields $F$ do not have extensions of degree $n$?

Let $F$ be the field of real numbers constructible by straightedge and compass, and let $K=F[\sqrt{-1}]=\{\,a+bi:a,b{\rm\ real\ constructible}\,\}$. Then $K$ has no extensions of degree two. For if $(...
Gerry Myerson's user avatar
3 votes

Counter example of a radical extension that is not Galois/normal over $\mathbb{Q}(\omega)$?

The polynomial $f:=(x^2-1)^3-2$ has Galois group $G \simeq S_4 \times C_2$ over $\mathbb{Q}$, so the commutator subgroup $[G,G] \simeq A_4$. Let $L$ be the splitting field (over $\mathbb{Q}$) of $f$. ...
Sander Dahmen's user avatar
2 votes

Complete reducibility and field extension

It is worth pointing out that there are general statements that semisimplicity can be checked over an algebraic closure. The following statement is used: Let $k$ be a field and $A$ a $k$-linear ...
Thorsten Heidersdorf's user avatar
2 votes
Accepted

Algorithms for Polynomials Over a Real Algebraic Number Field, a reference

The thesis can be found here.
Carl-Fredrik Nyberg Brodda's user avatar
2 votes
Accepted

General linear group action on extensions of finite fields

Let $F$ be any field and $F<E$ a finite field extension. Fix $x\in E^*$ and consider the multiplication operator $g_x\in \text{GL}_F(E)$, the group of invertible $F$-linear transformations of $E$. ...
Uri Bader's user avatar
  • 11.5k
2 votes
Accepted

Existence of intermediate field extensions for tamely ramified p-adic extensions

The answer is no. For example the group $S_3\times C_3$ occurs (as TransitiveGroup(6,5)) as the Galois group of an (automatically tame) extension of $\mathbb{Q}_5$ (and more generally over $\mathbb{Q} ...
Joachim König's user avatar
2 votes

If $G=\mathsf{Aut}_k(F)$ acts on field $F$ algebraic over $k$ then do we have: orbit $G\alpha=\text{ roots of minimal polynomial of }\alpha$?

Yes, this can hold for non-normal extensions, for example $F=\mathbb Q(2^{1/3})$. For any zeros $\alpha,\beta\in F$ of an irreducible $f\in\mathbb Q[x]$, we can always map $\alpha\mapsto\beta$ by a $\...
Christian Remling's user avatar
1 vote
Accepted

Shrinking the base field of an affine variety

A standard reference is Chapter 6 of "Neron Models" of Bosch, Lütkebohmert and Raynaud, Springer, 1990. The original Grothendieck's exposés in the Cartan Seminar (1960--1961) are available ...
F Zaldivar's user avatar
  • 1,432
1 vote

How do I extend the $2$-adic absolute value to prove Monsky's Theorem?

You are asking how to extend $|\cdot|_2$ from $\mathbf Q$ to a finite extension $L = \mathbf Q(\gamma)$. In the ring of integers $\mathcal O_L$, let $(2) = \mathfrak p_1^{e_1}\cdots \mathfrak p_g^{...
KConrad's user avatar
  • 49.9k
1 vote

General linear group action on extensions of finite fields

Let $k=\Bbb{F}_{q},K=\Bbb{F}_{q^n}$, $x\in K$ and $g_x\in M_n(k)\cong End_k(K)$ the matrix of the multiplication by $x$. If $k(x)$ is smaller than $K$ then every $k(x)$-linear endomorphism of $K$ ...
reuns's user avatar
  • 3,425
1 vote

Counter example of a radical extension that is not Galois/normal over $\mathbb{Q}(\omega)$?

The smallest counterexample (in terms of degree) should be to take the fixed field of a non-central involution in any realization of the dihedral group $D_4$. EDIT: Here are some details, since the ...
Arno Fehm's user avatar
  • 2,041
1 vote

Counter example of a radical extension that is not Galois/normal over $\mathbb{Q}(\omega)$?

I think I have a better solution: $\sqrt{1 + \sqrt{3}}$. Since the "accident" in the comment above does not happen. $\sqrt{3} -1= \frac{ (\sqrt{3}- 1)^2 }{2} (1+ \sqrt{3})$. Edit: it does ...
Cyril's user avatar
  • 221
1 vote

The variety induced by an extension of a field

Let $n=2$ and $k=3$, and suppose by the sake of simplicity that the three lines are in general position. Then, up to projective transformations, we can assume that they are the three coordinate lines ...
Francesco Polizzi's user avatar
1 vote
Accepted

The variety induced by an extension of a field

I decided to turn my comment into an answer not because it is complete but because I think it can be of use. Let $z=(z_0:z_1:z_2)$ and $u=(u_1:\ldots:u_k)$ be homogeneous coordinates of $\mathbb{P}^2$ ...
amateur's user avatar
  • 375
1 vote

Complete reducibility and field extension

Amazingly, I cannot see an elementary solution. I believe there should be one. Otherwise, one can expand the comment of YCor with some standard Ring Theory. Let $A$ be the image of $U(L)$ in $End_F(V)...
Bugs Bunny's user avatar
  • 12.1k
1 vote

Extension of isomorphism of fields

In order to avoid having a question without answer in the forum, Eric Wofsey posted an answer here.
1 vote

Field extensions (non-algebraic)

If $X$ is fibered over $C$ and $C$ has positive genus then, as $C$ has non-zero holomorphic differentials, you can pull back those to $X$, which is already a non-trivial condition on $X$. Moreover, ...
Felipe Voloch's user avatar

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