reference request – Maximal order of $x^n-d$ and its dependence on $d$

It’s well known that the structure of the maximal order of $mathbb{Q}(sqrt{d})$ depends on $d$ modulo $4$: (assuming $d$ is squarefree), the maximal order is $mathbb{Z}left(frac{1+sqrt{d}}{2}right)$ if $d equiv 1 mod{4}$ and $mathbb{Z}(sqrt{d})$ otherwise. The structure of the biquadratic extension $mathbb{Q}(sqrt{d},sqrt{e})$ has a similar but more complicated description.

Now let’s consider $K=mathbb{Q}(sqrt(n){d})$. For simplicity, let’s consider only $d$ prime. Based on some SAGE computation, it seems that there is a degree $n$ extension $M$ of $mathbb{Q}(mu_{n^2})$ such that $mathcal{O}_K$ contains more than just $mathbb{Z}(sqrt(n){d})$ iff $d$ splits in $M$. (Note that if $n$ is prime, then $M$ is unique.) For $n=2$, one sees that $M=mathbb{Q}(i)$.

Is this known? It doesn’t seem too hard to prove that if $d$ is split, then there are more algebraic integers in $K$. But does this appear anywhere?

ag.algebraic geometry – In $mathbb{C}[x,y]$: If $langle u,v rangle$ is a maximal ideal, then $langle u-lambda,v-mu rangle$ is a maximal ideal?

I have asked the following question at MSE and got one answer. It would be nice to have more elaboration on it, please:

Let $u=u(x,y), v=v(x,y) in mathbb{C}(x,y)$, with $deg(u) geq 2$ and $deg(v) geq 2$.
Let $lambda, mu in mathbb{C}$.

Assume that the ideal generated by $u$ and $v$, $langle u,v rangle$, is a maximal ideal of $mathbb{C}(x,y)$.

Is it true that $langle u-lambda, v-mu rangle$ is a maximal ideal of $mathbb{C}(x,y)$?

My attempts to answer my question are:

(1) By Hilbert’s Nullstellensatz, $langle u,v rangle= langle x-a,y-b rangle$, for some $a,b in mathbb{C}$, so
$x-a=F_1u+G_1v$ and $y-b=F_2u+G_2v$, for some $F_1,G_1,F_2,G_2 in mathbb{C}(x,y)$.
Then, $x=F_1u+G_1v+a$ and $y=F_2u+G_2v+b$.

(2) $frac{mathbb{C}(x,y)}{langle u,v rangle}$ is a field (since $langle u,v rangle$ is maximal); actually, $frac{mathbb{C}(x,y)}{langle u,v rangle}$ is isomorphic to $mathbb{C}$. Is it true that $frac{mathbb{C}(x,y)}{langle u,v rangle}$ is isomorphic to $frac{mathbb{C}(x,y)}{langle u-lambda,v-mu rangle}$? In other words, is it true that $frac{mathbb{C}(x,y)}{langle u-lambda,v-mu rangle}$ is isomorphic to $mathbb{C}$?
See this question.

(3) If $langle u-lambda,v-mu rangle$ is not maximal, then it is contained in some maximal ideal: $langle u-lambda,v-mu rangle subsetneq langle x-c,y-d rangle$, $c,d in mathbb{C}$.
It is not difficult to see that $(u-lambda)(c,d)=0$ and $(v-mu)(c,d)=0$,
so $u(c,d)-lambda=0$ and $v(c,d)-mu=0$, namely,
$u(c,d)=lambda$ and $v(c,d)=mu$.

Remark: Is it possible that $langle u-lambda,v-mu rangle = mathbb{C}(x,y)$. If so, then there exist $F,G in mathbb{C}(x,y)$ such that
$F(u-lambda)+G(v-mu)=1$. Then at $(a,b)$ we get:
$F(a,b)(-lambda)+G(a,b)(-mu)=1$ (since, by (1), $u(a,b)=0$ and $v(a,b)=0$).

Thank you very much!

Maximal Result

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dg.differential geometry – Reductive Lie groups and existence of maximal compact subgroup

I am reading Knapp’s book “Lee groups beyond introduction” (2nd edition). I am struggling to understand the following point. Recall that $G$ is a reductive Lie group. If the Lie algebra $mathfrak g$ of $G$ is reductive and equipped with a involution and a nondegenerate symmetric bilinear form $B$ on $mathfrak g$ which is $theta$-invariant and $Ad$-invariant and the following hold.

  1. We have a decomposition $mathfrak g=mathfrak koplus mathfrak p$ with respect to the eigenspaces of $theta.$

  2. $G$ has a compact subgroup with Lie algebra $mathfrak k.$

  3. The map $(k,exp X)mapsto kexp X$ from $Ktimes expmathfrak p$ to $G$ is a diffeomorphism.

  4. The bilinear form $B_theta (X,Y):=-B(X,theta Y)$ is positive definite on $mathfrak g.$

  5. Every automorphism $Adg$ for $gin G$ is inner.

From this Knapp concludes that $K$ has to be a maximal compact subgroup. he argues as the following (Page 446). Let $K$ is properly contained in a compact subgroup $K_1$. Take $k_1in K_1setminus K.$ Then for some $kin K$ and $Xinmathfrak p$, we have $kexp X=k_1$ which implies that $exp Xin K_1.$ Since $(exp X)^n=exp (nX)$ is in $K_1$ the sequence $exp(nX)$ must have a convergent subsequence. I get easily to this point. Now Knapp says that this contradicts 3.! I do not understand this. Can someone help me out?

p adic number theory – conjugation of maximal algebraic tori

Accept $ G $ is a connected, reductive algebraic group over a non-Archimedean local field $ F $which is divided over a finite extent $ E / F $,

I often see a result that says "everything is maximum $ F $-Tori are conjugated over $ E $", by which I understand the following: Let $ G (E) $ denote the $ E $-Dots of the algebraic group $ G $;; then for each maximum $ F $-tori $ T, T $ $ of $ G $is there $ x in G (E) $ so that $ T (E) = xT & # 39; (E) x ^ {- 1} $,

In addition, the definitions show that if $ T, T $ $ are maximum $ F $-tori from $ G $then there is an isomorphism of $ T (F) $ on to $ T & # 39; (F) $ which is defined via $ E $,

My question is: Can the isomorphism be assumed to be conjugation in the second statement (as in the first statement)? That means: it follows from these results that if $ T, T $ $ are maximum $ F $-tori in $ G $then it exists $ x in G (E) $ so that $ T (F) = xT & # 39; (F) x ^ {- 1} $?

Any help (including proof of the first statement) is greatly appreciated!

Ring Theory – Let $ I: = (3,1+ sqrt {5} i) $ and $ R: = mathbb {Z}[sqrt{5}i]$. Show that $ I $ is not maximal.

To let $ I: = (3,1+ sqrt {5} i) $ and $ R: = mathbb {Z} ( sqrt {5} i) $, Show that $ I $ is not maximal. My goal is to show that $ R / I $ is a field. So far I have the following:

begin {align *}
mathbb {Z} ( sqrt {5} i) / (3,1+ sqrt {5} i) & cong ( mathbb {Z} (X) / (x ^ 2 + 5)) / (3 , 1 + X, X ^ 2 + 5) / (X ^ 2 + 5)) \
& = mathbb {Z} (X) / (3,1 + X, X ^ 2 + 5) \
& = mathbb {Z} / 3 mathbb {Z} (X) / (1 + X, X ^ 2 + 2).
end {align *}

I can not see how you reduce $ mathbb {Z} / 3 mathbb {Z} (X) / (1 + X, X ^ 2 + 5) $ further. Ideally, if I could write $ X ^ 2 + 2 $ as a multiple of $ 1 + X $I would be ready.

abstract algebra – Be $ R = mathbb Z[x]$ is the ring of polynomials over $ mathbb Z $. Prove that the ideal $ (x, p) $ is maximal if and only if $ p $ is a prime.

To let $ R = mathbb Z (x) $ let the ring of polynomials be over $ mathbb Z $, Prove that is the ideal $ (x, p) $ generated by $ x $ and $ p $, Where $ 0 <p in mathbb Z $is a maximum ideal of $ R $ then and only if $ p $ is a prime number. For a prime number $ p $, identify the field $ R / (x, p) $,

Here is my attempt:

(1) Suppose that $ (x, p) $ is maximum.

Accept $ p $ is not great. Then $ p = nk $ for some $ n, k in mathbb Z $,

The elements of $ (x, p) $ are of the form $ p + p_ {1} x + p_ {2} x ^ 2 + … $ Where $ p_ {i} $ are a multiple of $ p $ for all $ i in mathbb N $,

$ p = kn $ So the elements also have the form $ kn + kn_ {1} x + kn_ {2} x ^ 2 + … $

But then $ (x, p) = (x, kn) subseteq (x, k) $ and $ (x, n) $,

But $ (x, p) $ is maximum. Contradiction.

In order to $ p $ is prime.

(2) Suppose now $ p $ is prime, and that $ (x, p) $ is not maximal.

So there is a maximum amount $ M $ so that $ (x, p) subsetneq M subsetneq R $,

So there is a polynomial $ P in M ​​$ Where $ P = n + Q $. $ p $ does not split $ n $, and $ Q $ is a polynomial.

In order to $ (x, p) + P $ is an ideal larger than $ (x, p) $,

But $ p $ is great and does not split $ n $, in order to $ gcd (p, n) = 1 $,

In order to $ 1 in $(x, p) + P$, and is therefore $R $. A contradiction.

In order to $ (X, p) $ must be maximum.

Therefore, $ (x, p) $ is a maximum ideal if and only if $ p $ is prime.

(3) The field $ R / (x, p) $ has elements of form $ (x, p) + P $, Where $ P in R, but P notin (x, p) $, All such $ P $ are of the form $ n + Q $. $ p $ does not split $ n $, and $ Q $ is a polynomial.

I … think that's it? I'm not sure about my evidence for (1) and (2), and I'm not sure what else I need for (3). Any advice would be appreciated!