## 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!

## inequality – Maximal value of \$a_1x_1+…+a_nx_n\$

Given $$a_1,…,a_n$$ real numbers such that $$a_1^2+…+a_n^2 = 1$$, prove that the maximal possible value of $$a_1x_1+…+a_nx_n$$ such that $$x_1^2+…+x_n^2=1$$ is achieved when $$x_1=a_1,…,x_n=a_n$$. I would like to hear a proof of it that does not use lagrange multipliers.

## Maximal Result

All trader actually want to get maximum trading result, but in forex trading, this is not a simple way, loss also as part in trading which as a trader must considering with this risk, learn how to determine best entry point and exit point is a crucial part to make better skill, forex trading will have more fun if we can make money regularly

## 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?

## prime and maximal ideals and associates of Integers (44)?

(i) Find all prime and maximal ideals in Z44.
(ii) Find all associates of 2 in Z44.

Can anyone show those ? Thanks

## 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 $$0 is a maximum ideal of $$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!

## \$ I = {a + ib in mathbb Z[i]: 5 | a, 5 | b } \$ forms a maximal ideal of \$ mathbb Z[i], \$

How to show it $$I = {a + ib in mathbb Z [i]: 5 | a, 5 | b }$$ makes a maximum ideal of $$mathbb Z [i].$$

I tried to prove $$mathbb Z [i] Backslash I$$ is a field could not prove it.