## 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.

## complex analysis – evidence of the existence of a maximal analytical continuation of a holomorphic germ

The following happens Lectures on Riemann's surfaces by O. Forster:

I do not understand the second paragraph of the proof. Why is the card? $$F: Z to Y$$ well defined if we choose another curve $$a$$ to $$x$$, we could get another one $$eta$$

For the relevant definitions:

## Proofing Techniques – Prove that every complete prefix-free language is maximal

I am practicing a problem where I have to prove that every prefix-free language is maximal.

I know that a prefix-free language A is a maximum if it is not a proper subset of a prefix-free language, where a prefix-free language is full if

$$sum_ {x in A} 2 ^ {- | x |} = 1$$

Also, I know that a language A 0 {0, 1} * is a prefix-free if no element of A is a prefix of another element of A and that the force inequality says that for each prefix-free language A .

$$sum_ {x in A} 2 ^ {- | x |} leqslant 1$$

I'm pretty sure that a full free prefix language is maximal because it belongs to the maximum prefix-free set. But I do not know how to prove it formally. Should I deal with Kraft's inequality and what I know about the relationship between maximum and prefix-free quantities?

Any help would be appreciated!

## ag.algebraic geometry – Are maximal tori conjugated locally localizable?

To let $$S$$ to be and leave a scheme $$G to S$$ to be a reductive group scheme. Then $$G$$ admits that there is a maximal torus etale-local, and two maximal tori are conjugate etale-local, according to Theorem 3.2.6 and Corollary 3.2.7 in [B. Conrad, Reductuve group schemes] http://math.stanford.edu/~conrad/papers/luminysga3.pdf.

As stated in the book under Proposition 3.1.9 and at the beginning of Chapter 4, the presence of maximum Tori Zariski locally can be achieved (cf. [SGA3, XIV, 3.20]). My question is the following

To let $$T$$ and $$T #$$ be maximum tori $$G$$ over defined $$S$$are they conjugated Zariski-local?

In my case, $$T$$ is split $$S$$, and $$S$$ is affine with $$Pic (S) = 0$$,

## Determine the probability that X + Y is maximal for a given X

Consider some random variables $$X_1$$ , , , $$X_n$$ they are independent and distributed identically to CDF $$F (x)$$and some random variables $$Y_1. , , Y_n$$ which are independent and identical to CDF distributed $$G (y)$$, Suppose that everyone $$X_i$$ is independent of everyone $$Y_i$$,

These pairs of random variables give random sums. $$X_1 + U_1$$ , , , $$X_n + U_n$$, What is the probability that the first random sum is the highest? That's clear $$1 / n$$ through symmetry. Equally clear is the likelihood that the first random sum is the highest Fewer when $$1 / n$$ if I assume that the first random sum falls below a certain limit. This is,

$$P (X_1 + U_1 geq X_j + U_j hspace {0,1cm} for all i hspace {0,1cm} | hspace {0,1cm} U_i leq u) leq 1 / n$$

Presumably this is true with strict inequality, though $$n> 1$$, Is that true? I've spent a lot of time showing this, and would appreciate any help, ideas, or hints.

Let $$K$$ be an algebraically closed field. By a variety $$V$$ definable over $$K$$, I mean a quasi-projective or an algebraic variety in sense of Weil. It is the set of points in an affine or a projective space over some bigger algebraically closed field $$L$$.
Now consider the Zariski topology $$tau$$ on $$V$$ together with Krull dimension on closed sets. Is it possible to enrich this topology in the naive sense by adding new closed sets, so that the enriched topology $$tau^prime$$ is also a Noetherian topology?
In other words, is the Zariski topology on $$V$$ a maximal noetherian topology?