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7

The ring $R= \mathcal{H}(S^1)$ is both noetherian and factorial. a) Noetherianity follows from Theorème (I,9) page 123 of this Inventiones paper by J.Frisch. He proves that given a complex analytic space $X$ and a compact subset $K\subset X$, the ring $\mathcal{H}(K)$ is noetherian as soon as $K$ is real semi-analytic and has a basis of open Stein ...


7

The first sentence of the problem's statement says that $(\mathbb{Z}/2\mathbb{Z})^7$ is a simple $A$-module; hence, by Schur's Lemma, the ring $A' = \text{End}_A((\mathbb{Z}/2\mathbb{Z})^7)$ is a skew field. By the Artin-Wedderburn Theorem, $A'$ is a field, thus $A' \cong \mathbb{F}_{2^n}$ for some $n$. Since $A'$ acts on $\mathbb{Z}/2^7\mathbb{Z}$, the ...


4

Let $p\in\ker(f)$, that is, $p(t^2-1,t^3-t)=0$. Write $$p(x_1,x_2)=(x_2^2-x_1^2(x_1+1))q(x_1,x_2)+r(x_1,x_2)$$ with $\deg_{x_2}r\le1$. Then $r(x_1,x_2)=a(x_1)+b(x_1)x_2$ and from $p(t^2-1,t^3-t)=0$ we get $a(t^2-1)+b(t^2-1)(t^3-t)=0$. Now conclude that $a=b=0$. (In order to do this look at the degree of polynomials involved in the last equation.)


3

Here's a proof that uses Krull dimension. Observations: $\ker f$ and $\mathfrak a$ are prime. $\ker f$ is prime because $k[x_1,x_2]/\ker f$ is an integral domain, $\mathfrak a$ is prime because is generated by an irreducible element of $k[x_1,x_2]$ which is an UFD, hence it is generated by a prime element. In $\text{Im} f=k[t^2-1,t^3-t]$ the element ...


3

If $x$ is a non-unit, then so is $ux$ for any $u$. This is true in any commutative ring. Otherwise we'd have $$(ux)^{-1}ux=((ux)^{-1}u)x=1$$ This contradicts the assumption that $x$ is not a unit. If the ring is noncommutative we can still obtain the result that $ux$ is not a unit from the fact that the nonunits form an ideal in a local ring.


3

For the second question, the finite rings are precisely those for which every finitely generated module has finitely many submodules. For a ring $R$ and $r\in R$, let $M_r$ be the submodule of $R\oplus R$ generated by $(1,r)$. Then $(1,r)$ is the only element of $M_r$ whose first coordinate is $1$, and so $M_r\neq M_s$ for $r\neq s$, and so if $R$ is ...


3

Take $R=\mathbb{Z}$, $M=\mathbb{Q}\oplus \mathbb{Z}/2\mathbb{Z}$ and $N= \mathbb{Z}/2\mathbb{Z}$, with inclusion $R\to M:n\mapsto (n,\overline{0})$. Then, in $M\otimes_\mathbb{Z} N$, we have that $$ (1,\overline{0})\otimes \overline{1} = (\frac{1}{2},\overline{0})\otimes 2\cdot\overline{1} = (\frac{1}{2},\overline{0})\otimes \overline{0} = 0.$$ Therefore ...


3

Since $p$ is nonzero and prime, $(p)$ is a height one prime, and so $A_{(p)}$ is a one dimensional local UFD, since localizations of UFD's are UFD's. But a one-dimensional UFD is noetherian. Thus, since UFD's are integrally closed, $A_{(p)}$ is an integrally closed noetherian local domain of dimension one, hence is a DVR.


3

Yes, and in fact you do not need $X$ to be an H-space: the coproduct $\psi$ of $H_*(X)$ exists anyway (if the ring of coefficients is a field), and the definition of a group-like element does not involve the product. So take $X$ to be any space, and define $S \subset H_*(X)$ as in your question (the set of classes in $H_*(X)$ satisfying $\psi(a) = a \otimes ...


2

Let me complement the nice and abstract existing answer by a concrete one: Yes, the argument is correct. To see that the ring $A_{(p)}$ is noetherian directly use an argument as you might know it from Euclidean domains. Let $I$ be an non-zero ideal. Let $a \in I$ be non-zero with minimal valuation, say $k$; then show $I = (a) $, as for $b \in I$ you ...


2

A submodule that is generated by two elements. (This isn't exactly a standard or well-known term, but in that context there's nothing else it could sensibly mean.)


2

At the point where you have $\phi(i)\phi(I)$, you deduce only $i\in \phi^{-1}(\phi(I))$, which is equal to $I$ if $\phi$ is injective. To have another example, take the canonical morphism $\mathbf Z/4\mathbf Z\to \mathbf Z/2\mathbf Z$, and the ideal $I=0$ in $\mathbf Z/4\mathbf Z$. Then $\;\phi^{-1}(\phi(I))=2\mathbf Z/4\mathbf Z$.


2

$1-ux$ can not be in the maximal ideal as $x$ is and hence $ux$ is. And since we have a local ring, any element not in the maximal ideal must be a unit.


2

The matrix $A$ corresponds to an endomorphism $\varphi_A$ of the $K$-vector space $K^n$. Then, by Fitting's Lemma, there is a decomposition $K^n=V_1\oplus V_2$ such that $\varphi_A$ restricted to $V_1$ is nilpotent, and $\varphi_A$ restricted to $V_2$ is invertible, that is, there is a $K$-basis of $K^n$ such that the matrix associated to $\varphi_A$ in this ...


2

The symbol $\bar r$ denotes the equivalence class of $r$ in $\text{Ann}_R(M)$. The point here is that you want to define the action of the quotient on the module. This is well-defined because if $\bar{r_1}=\bar{r_2}$, this means that $r_1-r_2\in\text{Ann}_R(M)$, so for any $m$ we have $(r_1-r_2)m=0$, i.e. $r_1m=r_2m$.


2

It is IMHO never a good idea to try and prove that some structure is not some other structure using the axioms only. After all, in many cases the said structure also IS that other structure, so a proof "generally speaking" is doomed to fail. Here you cannot rule out the possibility that, in addition to what you describe as a substitute to the associativity, ...


1

Note that an $R$-module is the same as an abelian group $M$ together with a ring homomorphism $R\longrightarrow\operatorname{End}_\mathbb{Z}(M)$. The kernel of this homomorphism is precisely $\operatorname{Ann}_R(M)$ (actually this is how one should define the annihilator, so one does not have to check it is an ideal). Now the homomorphism theorem yields a ...


1

When a projection map $R\to S$ is in play, the notation $\bar{r}$ often denotes the image of $r$ in $S$ under the projection map. E.g. the elements of $\Bbb Z/n\Bbb Z$ are the residues $\overline{0},\overline{1},\overline{2},\cdots,\overline{n-1}$. A priori one would have to wonder if $\overline{r}m:=rm$ is well-defined, since $\overline{r}=\overline{s}$ is ...


1

It seems like your best bet for both questions will be to consider finite rings and their finitely generated modules. These at least will be closed under products.


1

For any polynomial $f\in \Bbb Z[X]$ define $\phi(f)=f(0)$. Here $\phi$ is a surjective morphism from $\Bbb Z[X]$ to $\Bbb Z$. Consider the ideal $I=\langle X+2\rangle\subset \Bbb Z[X]$. Clearly $2\in \phi(I)$ and $\phi(3X+2)=2$, but $3X+2\notin I$.


1

[Note: After the comments to this answer and the question, there have been major changes. Thanks to Najib Idrissi, the situation is much clearer now.] There are only slight differences in the definitions and strictly speaking, Hatcher's is more general. In addition to (equivalents of) Hatcher's axioms, May wants a Hopf algebra to be flat (a minor technical ...


1

The tool here is the fundamental theorem of algebra applied to $\Bbb Z[i]$. Indeed, the factorization of $\gamma^n$ is a product of primes whose exponents are multiples of $n$. Each of this primes is a factor of either $\alpha$ or $\beta$ but not both. Then, each prime in the factorization of $\alpha$ is raised to an exponent which is a multiple of $n$, and ...


1

Your claim will work fine. But you need to change the thing you're tensoring over back into a subalgebra! $F[\pi_0 X]$, the sub-$F$-algebra generated by $\pi_0(X)$, is the right choice. It might be easiest to justify by just showing that for $S$ a multiplicative subset of a commutative $F$-algebra $A$ and $G$ the free group on the monoid $S$, ...


1

Yes. More generally: Let $A = \bigoplus_{i\in\mathbb{Z}} A_i$ be a graded domain und $f\in A\setminus \{0\}$ homogeneous. If $f$ factors in $A$ as $f=gh$ then $g,h$ are homogeneous. Proof: Since we are in a domain and $f\neq 0$, the factors $g,h$ are non-zero as well. Let $g$ have non-zero component of lowest degree $d_{min}$ and of highest degree ...


1

I always use your method to solve these equations, but your result is just the standard formula for the Chinese remainder theorem. If $\gcd(a-b,AB) = 1$:   $\gcd(qB-pA,AB) = \gcd((x-pA)-(x-qB),AB) = 1$.   Thus $\gcd(A,B) = 1$ and hence $A_B^{-1},B_A^{-1}$ exist [which you already assumed]. [Note that the converse is clearly not true! $A,B$ ...



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