## 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?

Posted on Categories Articles

## Are \${frac{1}{n}|n∈mathbb{N}}cup{0}\$ and \$mathbb{Z}\$ homeomorphic?

Would anyone be able to explain why not?

I feel like you can show it using sequences, but I’m not sure how.

Posted on Categories Articles

## 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!

Posted on Categories Articles

## 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 $$MO_k(X) := displaystyle lim_{longrightarrow \ k} pi_ {n + k} (MO(k) wedge X_+)$$.

Posted on Categories Articles

## 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)*?

Posted on Categories Articles

## 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?

Posted on Categories Articles

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

Consider the commutative diagram 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?

Posted on Categories Articles

## Identify quotient group on \$mathbb{Z} times mathbb{Z}_n\$

Let $$F: mathbb{Z} longrightarrow mathbb{Z} times mathbb{Z}_n$$ be the homomorphism of groups given by $$F(a)=(ma,-(a)_n)$$.

What is, up to isomorphism, the quotient group $$mathbb{Z} times mathbb{Z}_n/mathrm{Im}(F)$$

Posted on Categories Articles

## 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 $$mathbf{VERIFY}$$.

Posted on Categories Articles