New answers tagged

1

I am assuming that you want to characterize the unital subrings of $\mathbb{Z}[\omega]$ otherwise the statement is not true. For example $5 \mathbb{Z}[\omega]$ is a subring of $\mathbb{Z}$ not containing an identity. Assume $R$ is a unital subring of $\mathbb{Z}[\omega]$. Ofcourse $\mathbb{Z} \subseteq R$. If there is no non-integer element in $R$, then $R = ...


2

It's not hard to do it with Sage. Simply factorize the polynomial $f$ over the field $K$: K.<a> = QuadraticField(-15) R.<x> = K[] f = x^16 - 5*x^15 + 14*x^14 - 30*x^13 + 57*x^12 - 100*x^11 + 157*x^10 - 215*x^9 + 250*x^8 - 240*x^7 + 183*x^6 - 110*x^5 + 57*x^4 - 30*x^3 + 16*x^2 - 5*x + 1 print(f.factor()) Output: (x^8 + (-1/2*a - 5/2)*x^7 + (a + 2)...


1

Potential purity implies purity. Fix a prime $\mathfrak p$ of $K$ and let $\mathfrak q$ be a prime of $L$ above it. Assume that $\mathfrak q/\mathfrak p$ is unramified, with residue degree $f$. Then $\mathrm{Frob}_{\mathfrak q}\in G_L$ is exactly $\mathrm{Frob}_{\mathfrak p}^f$. Indeed, by definition, $\mathrm{Frob}_{\mathfrak p}$ is a pre-image of $x\mapsto ...


5

I'm not sure I would agree that "filling the gap" is a very reasonable way to put it (I don't really think there is a gap to fill), but the class of local principal rings (so rings that are either fields or discrete valuation rings) is between fields and euclidean rings.


1

Following @reuns' hint, I shall write up a "solution" to Question 1: For $\mathbb{Q}_2(\sqrt{3}, \sqrt{7})$, as @reuns hinted, $\zeta_3 \in \mathbb{Q}_2(\sqrt{3}, \sqrt{7})$ and hence we can consider its subextension $\mathbb{Q}_2(\zeta_3)$. As $\gcd(2,3)=1$ and $2$ is of order $2$ in $(\mathbb{Z}/3\mathbb{Z})^{\times}$, $\mathbb{Q}_2(\zeta_3)$ is ...


2

For $\Bbb{Q}_2(\sqrt3,\sqrt7)$ the trick is to say that $-7\equiv 1\bmod 8$ is a square in $\Bbb{Z}_2$ so it is $\Bbb{Q}_2(\sqrt3,i)$ which contains $\zeta_3$. For $\Bbb{Q}_2(\sqrt3,\sqrt2)$ note that $$(X-\frac{\sqrt3-1}{\sqrt2}-1)(X-\frac{-\sqrt3-1}{\sqrt2}-1)= X^2+X(\sqrt{2}-2)-\sqrt{2}\in \Bbb{Z}_2[\sqrt{2}][X]$$ is Eisenstein.


2

Sorry for the mess - the problem was that the result is "obvious" (see below) but that I wanted to avoid using the Artin isomorphisms. A normal extension $L/K$ of a quadratic number field is a ring class field if and only if $G = $Gal$(L/{\mathbb Q})$ is a group extension of an abelian group by a group of order $2$ and if the nontrivial ...


2

Observe that not only 23 but also $7=\alpha-3-(\alpha-10)=7$ is a element of $\mathcal{O}$. Note also that $23$ and $7$ are coprime. Thus there are integers $u,v$ such that $1=23u+7v$ implying that $1\in \mathcal{O}$.


2

$P(X)=X^n-a$ is a monic polynomial which is separable mod the maximal ideal. The residue field of $K^{ur}$ is algebraically closed and thus $P$ has a root in the residue field. Taking a lift of this root, we find some $x \in O_{L}$ such that $P(x)$ is not invertible and $P’(x)$ is (invertible), where $L/K$ is a finite unramified extension. Then Hensel’s ...


1

We will use the fact that $R/I$ is a domain if and only if $I$ is a prime ideal of the ring $R$. Let $f:\mathbb Z\to \mathbb Z[\sqrt d]/(p,\sqrt d)$ be a mapping given by $f(x)=x+(p,\sqrt d)$. It is routine to check that this is an onto ring homomorphism. Moreover, $\ker(f)=(p,d)\subset \mathbb Z$. By the First Isomorphism Theorem, $$\mathbb Z/(p,d)\cong\...


3

I think the center $Z(\mathrm{GL}_{2}(\mathbb{R}))$ in your bijection should be replaced by its identity component $Z(\mathrm{GL}_{2}(\mathbb{R}))^{\circ}$. Otherwise you will get the bijection: $$\pm\mathrm{Id}_{2}\mathrm{GL}_{2}(\mathbb{Q})\backslash\mathrm{GL}_{2}(\mathbb{A}_{\mathbb{Q}})^{1}\simeq\mathrm{GL}_{2}(\mathbb{Q})Z(\mathrm{GL}_{2}(\mathbb{R}))\...


1

One important ingredient is the following theorem of Swan (Theorem 8.1. in Induced representations and projective modules): If $P$ is any projective f.g. $RG$-module, then $P \otimes_R K$ is free. With that understood, we can do that following: We actually have a splitting $K_0 (\mathcal P_R ) \cong \mathbb{Z} \oplus \widetilde{K_0 RG}$, where $\mathbb Z$ ...


1

Are you really being asked if something is prime in $\mathbf Z[\sqrt{-7}]$ or did you just make up that example? I ask because $\mathbf Z[\sqrt{-7}]$ is not a UFD, so there is a distinction between primes and irreducibles. By the way, how are you checking if something is prime in a quadratic ring? You wrote that you definitely know how to do this, but you ...


1

Your phrase "is not unramifed at $p$" should be "is not ramified at $p$" or "is unramified at $p$", but not what you wrote. As for the question itself, here is a general result that is quite useful: if $K$ is a number field and we can write $K = \mathbf Q(\alpha)$ where $\alpha$ is an algebraic integer that is the root of a ...


1

Splitting of primes in quadratic extensions is controlled by the quadratic character (the Legendre symbol). More concretely an odd prime $p\nmid d$ splits in $\mathbb{Z}[\lambda_d]$ if and ony if $d$ is a square modulo $p$. You can always find a prime $p$ such that $d$ is not a square modulo $p$, so $p$ will be inert (so irreducible in $\mathbb{Z}[\lambda_d]$...


2

The unique unramified quadratic extension of $\Bbb Q_2$ is $\Bbb Q_2(\sqrt{5})$. Can you find two ramified quadratic extensions of $\Bbb Q_2$ whose compositum contain $\Bbb Q_2(\sqrt5)$?


0

This is analogous to the Picard group argument you laid out, but anyway: Let $$A_S = \{a \in O_K: v(a) > 0 \text{ for some } v \in S\}.$$ Then $O_{K,S} = A_S^{-1}O_K$ and is still a Dedekind domain. Let $Cl(K) = Frac(O_K) / P(O_K)$, the fractional ideals mod principal ideals. We want to see how fractional ideals/principal ideals change from $O_K$ to $O_{...


2

For example, $I_1(2)/I_0(2)$ is known to be transcendental (Siegel 1929). Therefore $I_1(2)$ and $I_0(2)$ can't both be algebraic numbers.


1

The construction is as follows: first, choose an embedding $i$ of the algebraic closure $\overline{K}$ of $K$ into the algebraic closure $\overline{K_v}$ of $K_v$. Then, for any finite abelian extension $L/K$ (with $L \subset \overline{K}$), the extension $LK_v/K_v$ is finite abelian, so $i(L) \subset (K_v)^{ab}$. Now, note that $K^{ab}$ is the (filtered) ...


2

First of all, $W/IW$ is just $W_G$, the coinvariants of $W$. The modules $W^G$ and $W_G$ are not that related unless you are in a very special situation, say when $G$ is a cyclic group. If you don't impose any restriction on $\mathrm{dim}(W)$, you can first replace $G$ by taking $G \times P$ for a random $p$-group $P$ which acts trivially on $W$ (thus making ...


1

Your Q1 is answered by the next line: Observe here that $d_K : G_k \to \widehat{\Bbb Z}$ factorizes through $G(\tilde{L} | K)$ because $G_{\tilde{L}} = I_L \subseteq I_K$ Recall that $I_L$ is the kernel of $d|_{G_L}: G_L \to \widehat{\Bbb Z}$, and $G_L \subseteq G_K$, so $I_L \subseteq I_K$. Also recall from the definition of "maximal unramified ...


2

$[B:K] < \infty$, so $B$ is the union of a finite number of translations of $K$, which is closed, so $B$ is closed.


1

As suggested in the comment, we only need to construct one extension where $p$ ramifies and another extension where $p$ splits. For example, $3$ ramifies in $\Bbb Q(\sqrt{3})$ and splits in $\Bbb Q(\sqrt{7})$, so we know that $e_3 = g_3 = 2$ in $K = \Bbb Q(\sqrt{3}, \sqrt{7})$. According to PARI/GP, $\mathcal O_K = \Bbb Z[\alpha]$ where $\alpha = \sqrt{\...


0

As explained in the comments, it might not always be abelian. However, it will always be solvable, thanks to the theory of higher ramification groups (see e.g. Local Fields by Serre): The Galois group $G$ of any (finite) Galois extension $L/K$ of local fields admits a filtration $G = G_{-1} \supseteq G_0 \supseteq G_1 \supseteq G_2 \supseteq \cdots$ of ...


0

The key point is that "completion" is the same as "closure" (because we are in a complete metric space, namely $\Bbb C$), so: If $\operatorname{im}(\tau) \subset \Bbb R$ then $\overline{\operatorname{im}(\tau)} \subseteq \Bbb R$, but $\Bbb Q$ is already dense in $\Bbb R$, so $\overline{\operatorname{im}(\tau)} = \Bbb R$. Otherwise, take ...


1

It ramifies at the infinite prime because $\Bbb Q$ is totally real but $i+\sqrt 2$ is totally imaginary. (Recall that a real place $v$ in $K$ is said to ramify over $L$ if $v$ extends to a complex embedding that is not real.)


3

Your two polynomials are both biquadratic, so you can determine their Galois groups simply by the classification of biquadratic extensions, without any local field theory: For $h = (x^2 - 1)^2 + 3$, we have $(a,b,c)=(1,-3,4)$. Here $b$ is not a square in $\Bbb Q_2$, and $c$ is a square, and $2(a \pm \sqrt c) = \{6, -2\}$ are not squares in $\Bbb Q_2$, so ...


1

For the second question, see that $L = K\mathbb{Q}(\zeta_n)$ (composite field), and a prime $p$ ramifies in $L$ if and only if it ramifies in either $K$ or $\mathbb{Q}(\zeta_n)$. Then the result Franz mentioned, along with how we defined $n$, proves that only the primes ramifying in $L$ are exactly the primes ramified in $K$.


1

Choose $\mathfrak P$ to be $\mathcal O_K \cap \mathfrak P_{\Bbb C_p}$, so the extension of $\Bbb Q_p$ generated by the image of $K$ is $K_\mathfrak P$, the completion of $K$ at $\mathfrak P$. Then clearly $v_p(x) = v_{\mathfrak P}(x)$ for any $x \in K$. Now for $\sigma \in G$, if $\sigma(\mathfrak P) = \mathfrak P$, then $v_p(\sigma(x)) = v_p(x)$; but ...


3

Since $f(X)$ is monic, with (algebraic) integral coefficients, all of its roots will be algebraic integers, so the splitting over $L$ will give a splitting over $O$.


0

Example, my little program showing class group and Gauss duplicate of each form. This shows us the principal genus, which is the squares in the group. There are numerous imprimitive forms for this discriminant. jagy@phobeusjunior:~/Desktop/Cplusplus$ ./classGroup Absolute value of discriminant? 3600 Impr 2 0 450 Impr 3 0 300 Impr 5 0 180 Impr 6 0 150 Impr ...


-1

Yes that’s exactly correct (although it’s a common thing that happens in duality theorems in other settings as well)


3

Let $2=p_{1}<p_{2}<...$ be the sequence of consecutive primes. Let $k+1<\beta<r_{1}<r_{2}<...<r_{k}$ be some chosen primes ($\beta$ is some variable, this will be used later.) We solve the following system; $w \equiv p_{1}^{r_{1}-1} \mod p_{1}^{r_{1}}$ $w+1 \equiv p_{2}^{r_{2}-1} \mod p_{2}^{r_{1}}$ . . . $w+k-1 \equiv p_{k}^{r_{k}-1} \...


7

Here's why I find it confusing: if we write $\mathbf{Q}_p/\mathbf{Z}_p = \lim_{\rightarrow n }\mathbf{Z}/p^n \mathbf{Z}$ and use the fact that tensor products commute with direct limits, we should get that $$T \otimes_{\mathbf{Z}} \mathbf{Q}_p/\mathbf{Z}_p \cong p\text{-Sylow subgroup of }T,$$ which is certainly not always zero. It's not true that the ...


2

Why not just plug in the general case like you did when n=m=1. Let $$ p(x)=x^m+\alpha_{m-1} x^{m-1} + \cdots + \alpha_1 x + \alpha_0, \\q(x)=x^n+\beta_{n-1} x^{n-1} + \cdots + \beta_1 x + \beta_0, $$ then $$ p(q(x))-q(p(x))\\ =(x^n+\beta_{n-1} x^{n-1} + \cdots + \beta_0)^m +\alpha_{m-1}(x^n+\beta_{n-1} x^{n-1} + \cdots + \beta_0)^{m-1}+\cdots+\alpha_1(x^n+\...


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