ag.algebraic geometry – Cubic function $mathbb{Z}^2 to mathbb{Z}$ cannot be injective

It is easy to show, with an explicit construction, that a homogeneous cubic function $f: mathbb{Z}^2 to mathbb{Z}$ is not injective. I am seeking a proof of the same result without the condition that $f$ is homogeneous.

For homogeneous $f$, if we take a generic point $(x_0,y_0) in mathbb{Z}^2$, the tangent to the curve:

$$f(x,y)-f(x_0,y_0)=0$$

will intersect the curve at a second point, $(x_1, y_1)$, with rational coordinates. To see this, note that we can parameterise the tangent line as:

$$L(lambda) = (x_0,y_0) + lambdaspace (partial_y f(x,y), -partial_x f(x,y))|_{x=x_0,y=y_0}$$

Then the cubic in $lambda$:

$$F(lambda) = f(L(lambda)) – f(x_0,y_0)$$

will have rational coefficients and a double root at $lambda=0$, so its third root $lambda_3$ must be real and rational. So:

$$(x_1,y_1) = L(lambda_3)$$

will have rational coordinates. If we choose any integer $D$ such that:

$$(x_2,y_2) = D space (x_1,y_1) in mathbb{Z}^2$$

we can exploit the homogeneity of $f$ to obtain:

$$(x_3,y_3) = Dspace (x_0,y_0)$$
$$f(x_3,y_3) = D^3 f(x_0,y_0) = D^3 f(x_1,y1) = f(x_2,y_2)$$

My question is: when $f$ is not homogeneous, can we construct explicit pairs of distinct points in $mathbb{Z}^2$ that demonstrate $f$ is not injective?

elementary set theory – The set of all finite subsets of $mathbb{Z}_+$ is countable. Is my proof correct?

I assume the following results:

  • A countable union of countable sets is countable.
  • A finite cartesian product of countable sets is countable.
  • If for any set $A$ there is a bijective correspondence with some countable set $B$, then $A$ is countable.

This is my proof of the statement in the title:

Proof. Let $J$ denote the set of all finite subsets of $mathbb{Z}_+$. Then $J = bigcup_{n in mathbb{Z}_+} J_n$, where $J_n$ is the set of all subsets of $mathbb{Z}_+$ with $n$ elements. We prove that $J_n$ is countable by showing that for any $n in mathbb{Z}_+$ there exists a bijective function $f : mathbb{Z}_+^n rightarrow J_n$ such that for any tuple $(k_1,k_2,…,k_n) in mathbb{Z}_+^n$, the following conditions hold:

  1. The set $f(k_1, k_2,…,k_n)$ contains $n$ elements
  2. The largest element in $f(k_1,k_2,…,k_n)$ is $sum_{i=1}^nk_i$.

We use induction. Base case $n = 1$ is trivial, since we can define a function $f(n) = {n}$, which is clearly bijective and satisfies the above conditions. Thus, assume that the result holds for some $k in mathbb{Z}_+$. Hence, there exists $f_k: mathbb{Z}^k_+ rightarrow J_k$ which is a bijection satisfying the above properties. To show that the result holds for $k+1$, define $f : mathbb{Z}_+^{k+1} rightarrow J_{k+1}$ as follows:
begin{align*}
f(n_1,n_2,…,n_{k+1}) = f_k(n_1,…n_k) cup {n_1+n_2+n_3+…+n_{k+1}}
end{align*}

By the inductive hypothesis, the set $f_k(n_1,…,n_k)$ contains $k$ distinct elements and the largest element is $sum_{i=1}^{k}n_i$. Thus, since:
begin{align*}
sum_{i=1}^{k}n_i < sum_{i=1}^{k+1}n_i
end{align*}

it follows that $f$ satisfies conditions (1) and (2). Then since $f_k$ is bijective and the last element in $f(n_1,…,n_{k+1})$ is not in $f_k(n_1,…,n_k)$, the bijectivity of $f$ follows. Hence, the result holds for $k+1$. Thus, since $mathbb{Z}_+^n$ is countable and there exists a bijection $f : mathbb{Z}_+^n rightarrow J_n$, it follows that $J_n$ is countable for all $n in mathbb{Z}_+$.

Therefore, since $J_n$ is countable, we conclude that $J = bigcup_{n in mathbb{Z}_+} J_n $ is countable.

Is this proof correct? If so, how can I improve it? Thank you!

at.algebraic topology – How to show that, $ mathrm{CH}_k (X) otimes_{ mathbb{Z} } mathbb{Q} simeq Omega_k (X) otimes_{ mathbb{Z} } mathbb{Q} $?

Let $ X $ be a $ n $ – dimentional oriented closed real manifold ( compact and without boundary ).

Can you tell me how to show that,
$$ mathrm{CH}_k (X) otimes mathbb{Q} simeq Omega_k (X) otimes_{ mathbb{Z} } mathbb{Q} $$ where,
$ mathrm{CH}_k ( X ) $ is the Chow group that is the free abelian group on the set of real $ k $ – submanifolds $ Z subset X $ of $ X $, with $ 0 leq k leq n $, and, $ Omega_k (X) $ is the oriented bordism group defined as the set of all isomorphism classes of $ k $ – manifolds $ M to X $ modulo bordism, where $ M to X $ is bordant to $ N to X $ if there is a $ W $ with $ partial W = M – N $. ( Here, $ M – N $ denotes the disjoint union of $ M $ and $ N $ with the orientation of $ N $ reversed ).

Here, I would like to point out that, the oriented bordism group $ Omega_k (X) $ is the one which is isomorphic to $ MSO_k (X) := displaystyle lim_ {longrightarrow \ k} pi_{n + k} (MSO(k) wedge X_+) $ if I’m not mistaken, and should not be confused with the unoriented bordism group $ mathfrak {M} _k (X) $ which is isomorphic to $ MO_k(X) := displaystyle lim_{longrightarrow \ k} pi_ {n + k} (MO(k) wedge X_+) $.

Thanks in advance for your help.

Exhibit the complete table of multiplication for (mathbb{Z} / 15 * mathbb{Z})∗, and use this to give a solution to the equation 7X ≡ 3 (mod 15).

Exhibit the complete table of multiplication for (mathbb{Z} / 15 * mathbb{Z})∗, and use this to give a solution to the equation 7X ≡ 3 (mod 15).

I have no idea how to start this question? How do you build a multiplication table for (Z/15Z)*?

ag.algebraic geometry – Automorphy of mixed Tate motives over $mathbb{Z}$

Deligne, Goncharov and Levine have constructed a $mathbb{Q}$-Tannakian category of mixed Tate motives, MTM($mathcal{O}_{K,S}$), over the ring of integers of a number field $K$ unratified outside a finite set of its places $S$.

In particular there is a category MTM($mathbb{Z}$) of mixed Tate motives unramified over $mathbb{Z}$. Its objects are extensions of pure Tate motives $mathbb{Q}(n)$ where $n$ is an integer. They arise from motivic sheaves on the moduli spaces $mathcal{M}_{0,m}$ of marked genus 0 curves.

Brown has shown that MTM($mathbb{Z}$) is generated by the motivic fundamental group of $mathbb{P}^1-{0,1,infty}$, and that periods of mixed Tate motives are $mathbb{Q}((2pi i)^{-1})$-linear combinations of multi-zeta values $zeta(n_1, dots, n_r)$, $n_i in mathbb{N}, n_1 geq 2$.

The $ell$-adic realization of the pure Tate motive $mathbb{Q}(n)$ is the $G_mathbb{Q}$-module $mathbb{Q}_ell$ where the Galois action is given by the $n$-th power of the $ell$-adic cyclotomic character $chi_ell$.

By Langlands philosophy every motive should correspond to an automorphic form. For pure Tate motives $mathbb{Q}(n)$ class field theory shows how: they correspond to powers of idèle class characters.

Which automorphic form/representation does a truly mixed Tate motive correspond to?

It might be made out of idèle class characters corresponding to its constituent pure Tate motives, but I am not able to see how.

For example we know that $M^1_n := Ext^1(mathbb{Q}(0), mathbb{Q}(n)) = K_{2n-1}(mathbb{Z})otimes mathbb{Q}$ for $n > 1$ where $K_.$ are the Milnor K-groups. What makes $M^1_n$ automorphic?

homological algebra – Limit in $mathbb{Z} / P^{2} to mathbb{Z} / P $

Consider the commutative diagram

![enter image description here

in Ab, where the morphisms are the canonical ones. Let us denote the top horizontal sequence by A and the bottom one by B. Then the vertical maps induce a morphism of abelian groups $alpha : Lim A to Lim B$. I am going to state which of the following statement are correct:

  1. $alpha$ is bijective
  2. $alpha$ is injective but not surjective
  3. $alpha$ is surjective but not injective
  4. $alpha$ is neither injective nor surjective.

However, I am having some trouble seeing what the limits are of both sequences as I am new to homological algebra. I believe the limit of A is $mathbb{Z}$ but I am not sure. Is this correct? I am also having some trouble seeing what the limit of B is, but I believe it could simply be 0. Is this correct?

general topology – Is the topological space $(mathbb{Z}, tau)$ compact/connected?

$mathbf{Question:}$ Consider the topology $tau={U subseteq mathbb{Z}: mathbb{Z} setminus U$ is finite or $0notin U }$ on $mathbb{Z}$. Then, the topological space $(mathbb{Z}, tau)$ is

(A) compact but NOT connected; (B) connected but NOT compact; (C) both compact and connected; (D) neither compact nor connected.

$mathbf{Attempt:}$ Consider $A={1} in tau$ and $B={…,-2,-1,0,2,3,4,…} in tau$. Now, $A cup B= mathbb{Z}$ but $A cap B= emptyset$ where $A$ and $B$ are open sets. Therefore $(mathbb{Z},tau)$ is not connected.

Now, take an arbitrary family of open sets ${G_alpha}_{alpha in I}$ where $G_alpha in tau$ for all $alpha in I$. Further suppose that it covers $mathbb{Z}$.

Since ${G_alpha}_{alpha in I}$ covers $mathbb{Z}$, there exists an $alpha=p$ such that $0 in G_p$. Now, by hypothesis, $mathbb{Z}setminus G_p$ is finite, say $| mathbb{Z}setminus G_p|=n$

We now require “at most” $n$ members of the family such that $(mathbb{Z}setminus G_p)cap G_alpha neq emptyset$. Since it is an open cover, we can find $G_{k_1},G_{k_2},…,G_{k_m}$, $m leq n$ such that $(mathbb{Z}setminus G_p) subset G_{k_1} cup…cup G_{k_m}$.

Therefore, for every open cover ${G_alpha}_{alpha in I}$ of $mathbb{Z}$, there exists a finite subcover $G_{k_1} cup…cup G_{k_m}cup G_p$.

Hence, $(mathbb{Z}, tau)$ is compact. So, Option (A) is the right choice.

Is this correct?

Kindly $mathbf{VERIFY}$.