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!

turing machines – A reduction from $HP$ to ${(langle M rangle, langle k rangle) : text{M visits in at list $k$ states for any input}}$

I tried to define the next reduction from $HP$ to ${(langle M rangle, langle k rangle) : text{M visits in at list $k$ states for any input}}$.

Given a couple $(langle Mrangle , langle xrangle)$ we define $M_x$ such that for any input $y$, $M_x$ simulates $M$ on the input $x$. We denote $Q_M +c$ the number of states needed for the simulation of $M$ on $x$, and define more special states for $M_x$ $q_1′,q_2′,…,q_{Q_M + c+ 1}’$ when $q’_{Q_M +c+1}$ is defined as the only final state of $M_x$. Now, in case $M_x$ simulation of $M$ on $x$ halts (i.e $M$ reach one of its finite state) $M_x$ move to $q_1’$ and then continue to walk through all the special states till it reaches $q_{Q_M + c + 1}$.

We define the reduction $(langle M rangle , langle x rangle) longrightarrow (langle M_x rangle , langle Q_M +c+1 rangle)$

In case $((langle M rangle , langle x rangle) in HP$ then for any input $y$ , $M_x$ walks through all the special states and thus visits in at least $Q_m + c+ 1$ steps. Otherwise, $M$ doesn’t stop on $x$ so $M_x$ doesn’t visit any special state, thus visits at most $Q_M +c$ states (the states needed for the simulation).

Is it correct? If you have other ideas or suggestions please let me know.

abstract algebra – $ g ^ l in langle f_1, …, f_k rangle $, if the ideal generated by $ f_1, ~ cdots, f_k, gy -1 $ in $ Bbb C[x_1,~cdots,x_n, y]$ contains $ 1 $

To let $ f_1, ~ cdots, f_k $ Be polynomials in $ Bbb C [x_1, ~ cdots, x_n] $. I want to show that for a polynomial $ g in Bbb C [x_1, ~ cdots, x_n] $, $ g ^ l $ is contained in the ideal that is generated by $ f_1, ~ cdots, f_k $ for some $ l $ if the ideal is generated by $ f_1, ~ cdots, f_k, gy -1 $ in the $ Bbb C [x_1, ~ cdots, x_n, y] $ contains $ 1 $.

Because of the term $ gy-1 $I thought this might be similar to Hilbert's zero-digit proof, so I tried to imitate the proof, but I can't see anything. Any clues? Thank you in advance.

Probability – what does this notation mean? "$ langle M rangle_t $" or "$ langle M rangle_ infty $", where $ M_t $ is a continuous martingale

Suppose that $ M_t $ is a continuous martingale. I saw this notation in a newspaper: $ langle M rangle_t $ or $ langle M rangle_ infty $. What does this notation mean?

The paper does not explain the notation, so I suspect that it is known. I am aware that angle brackets are sometimes used to indicate expectation, but paper is used $ E M_t $ denote the expectation. Likewise, $ t $ is outside the parenthesis, which indicates that it is not an expectation notation. Does it refer to a particular notation used in martingales?

ag.algebraic geometry – intuition for the Adic Unit Disc $ operatorname {Spa} (C langle T rangle, C ^ { circ} langle T rangle) $

To let $ C $ be an algebraically closed field that is related to a non-Archimedean evaluation (e.g. $ C = mathbb {C} _p $). The points in the adic spectrum $ operatorname {Spa} (C langle T rangle, C ^ { circ} langle T rangle) $ have been fully classified and consist of five point families. Everything that follows comes from Scholze Perfectoid spaces.

  1. The classic points: take $ x in C ^ { circ} $and consider the rating $ f mapsto | f (x) | $.

Type 2 and 3 points are the "rays" of the tree and follow the recipe below. To let $ 0 leq r <1 $and fix it $ x in C ^ circ $. Expand a certain power series $ f $ how $ sum_n a_n (T-x) ^ n $and send it to $ sup | a_n | r ^ n $.

  1. The ratings as above, but with the limitation that & # 39;$ r in | C ^ { circ} | $& # 39 ;. The branching occurs exactly at the points of this type. I don't know what that notation means and would appreciate it if someone could tell me.
  2. The remaining ratings, i.e. H. These rays with $ r notin | C ^ { circ} | $.
  3. The "dead ends" of the tree. To let $ D_1 supseteq D_2 supseteq cdots , $ be a sequence of closed hard drives $ bigcap_i D_i = varnothing $ (The (non-) existence of such sequences is called spherical completeness.) Take the assessment into account $ f mapsto inf_i sup_ {x in D_i} | f (x) | $.
  4. Certain rank$ 2 $ Reviews. fix $ x in C ^ { circ} $ and $ 0 <r <1 $. Choose a character $ { lessgtr} in { < , >} $. To let $ Gamma _ { lessgtr} $ be the ordered Abelian group $ mathbb {R} _ {> 0} times gamma ^ { mathbb {Z}} $ so that $ r & # 39; lessgtr gamma lessgtr r $ for all $ r & # 39; lessgtr r $. We now define our rating. Take any power series $ f $, expand it as $ sum_n a_n (T-x) ^ n $and send it to $ sup | a_n | gamma ^ n $.

The following illustration is included with the list.

exam

I struggle with the intuition behind this picture. I can collect the following. The & # 39; arch & # 39; at the bottom represents the classic unit disc, and you can see the vertical direction as & # 39; arch & # 39; imagine.$ r $-Axis & # 39 ;. So the fact that there are rays certainly makes sense. They reflect the type 5 points that are close to type 2, which are in the closure of the nearby type 2 point.

Question. An open question might be: How should I think about all of this? Regarding the picture, a more specific question is: what does the branching mean? Additional questions I want to understand: Can I see the open CD (for example) in this picture? How does my classic intuition for schemes relate to this? What changes when we say $ C = mathbb {Q} _p $ (i.e. something that is not algebraically closed)? Do permissible openings correspond to something in this picture?

All knowledge is valued.

Differential Geometry – Show $ langle exp ^ {- 1} _ {x} (y), exp ^ {- 1} _ {x} (y) rangle_ {g_ {x}} = langle exp ^ {- 1} _ {y} (x), exp ^ {- 1} _ {y} (x) rangle_ {g_ {y}} $

To let $ M $ a riemannian manifold n-dimensional without boundary. Show that $$ langle exp ^ {- 1} _ {x} (y), exp ^ {- 1} _ {x} (y) rangle_ {g_ {x}} = langle exp ^ {- 1 } _ {y} (x), exp ^ {- 1} _ {y} (x) rangle_ {g_ {y}} qquad, forall y in mathcal {U} $$
Where $ mathcal {U} $ is a normal neighborhood of $ x in M ​​$

My approach: Let $ mathcal {U} $ a normal neighborhood of $ x in M ​​$that contain one $ y in M ​​$, If $ c (t) $ is the geodesic curve connecting both points ($ x, y $), i. $ c (0) = x $ and $ c (1) = y $, then $ c ^ { prime} (0) = exp_ {x} ^ {- 1} (y) $,

Now consider the parallel transport of $ c ^ { prime} (0) $ along $ c $ (Note that $ c $ is the geodesic curve), then we have $$ langle c ^ { prime} (0), c ^ { prime} (0) rangle_ {g_ {x}} = langle c ^ { prime} (1), c ^ { prime} (1) rangle_ {g_ {y}} $$

I do not know if my proof is correct *. However, why, if we take the parallel transport, $ P_ {c} rvert_ {0} ^ {1} $ along the geodesic line, the end point coincides with the point $ c (1) = y $? Is this point unique?

Every idea is appreciated. THANK YOU SO MUCH!

linear algebra – If $ v $ and $ w $ are orthogonal to each other with respect to the scalar product $ langle cdot, cdot rangle $, they are also linearly independent.

To let $ V $ an Euclidean $ mathbb {R} $Vector room with a generic scalar product $ langle cdot,
cdot rangle $
, To let $ v, w $ With $ v neq 0 $ and $ w neq 0 $, Show: When $ v $ and $ w $ are orthogonal to each other with respect to the dot product $ langle cdot,
cdot rangle $
They are also linearly independent. (It should be a general scalar product)

I'm not sure how to start. I would assume that they are linearly dependent, but that did not lead to a good solution.

To let $ v = lambda w, lambda in mathbb {R} $ and $ langle v, w rangle = 0 dots $

Is $ L = { langle M rangle mid L (M) subseteq HP } in coRE $?

My intuition is that $ L notin coRE $but I did not manage to prove it $ HP le L $, as before, I have seen only from discounts $ HP $ or from $ overline {HP} $ With $ f $ so that $ f (( langle M rangle, x)) = langle M_x rangle $while $ M_x $ performs a simulation of $ M $ on $ x $,

(The answer was hidden from me for a while, so I started to write a question here.) After finding the surprisingly simple answer, I decided to post it anyway (Q & A-style).)

Turing Machines – Checking $ L = { langle M, w, n rangle $: $ M $ accepts $ w $ within $ n $ steps $ } $ is decidable

Show the following problem is decidable: Given $ w in Sigma ^ {*} $, $ n in mathbb {N} $and a Turing machine $ M $does $ M $ on $ w $ Stop inside $ n $ Steps.

My thoughts:

I am new to testing such results with Turing machines, so often I'm not sure what is legal or not legal with a Turing machine. For example,
My textbook often contains "high-level" descriptions of Turing machines that perform a calculation rather than formally step by step. That's how we're likely to describe Turing machines in the classroom, but I'm often not sure I'm strict enough in my description.


My attempt:

Construct a Turing machine $ M & # 39; $ in the following way:

$ M & # 39; $ = "When entering $ langle M, w, n rangle $:

(1) Representation $ n $ on the tape of $ n $ consecutive $ 1 $& # 39; s

(2) simulate $ M $ on $ w $

(3) Each "execution" of the transition function is followed by a mark on one of the $ 1 $& # 39; s

(4) If $ M $ enters an accepting state, accepting; Otherwise, refuse if $ M $ enters a negative state or after all $ 1 $'S are marked. "


My main concern is step (1) of the algorithm. I'm really not sure if that's possible. I'm not sure how $ n $ is displayed as a string for input $ langle M, w, n rangle $, Can I choose how it is encrypted?

Thank you for your time and feedback.

Linear Algebra – Find the first $ 3 $ orthogonal polynomials of Laguerre (in terms of $ langle f, g rangle = int_0 ^ { infty} f (x) g (x) e ^ {- x} dx $)

Find the first one $ 3 $ orthogonal polynomials of Laguerre (in relation to
$ langle f, g rangle = int_0 ^ { infty} f (x) g (x) e ^ {- x} dx $)

I do not know what definition to use for the Laguerre polynomials, and whether my definition produces orthogonal polynomials.

I found this recursive definition:

$$ L_0 = 1, L_1 = 1-x \ $$

$$ L_ {k + 1} (x) = frac {(2k + 1-x) L_ {k} (x) -kL_ {k-1} (x)} {k + 1} $$

So I can easily get the third. However, I do not know if these polynomials are orthogonal because the Wikipedia article mentions this generalized Laguerre polynomials are orthogonal and I do not think they are generalized. Besides, I did not use the dot product from the exercise. I think my way is not what my teacher intended for me.