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2

You already showed that $IJ\subset I\cap J$. So one side is obvious. That is, if $I\cap J\subset P$ then $I J\subset P$. For the other side, assume $I J\subset P$ and assume there is $x\in I\cap J$ such that $x\not \in P$. Then since $P$ is prime $x^2\not\in P$, but this is a contradiction since $x^2\in IJ\subset P$.


0

You have to use Hilbert Nullstenllensatz thats is a polynomial say $p$ in $C[x,y]$ have a zero that a point $(a,b)\subset \mathbb{C}^2$,where it vanish in this case it imply the ideal generated by the polynomial $(p)\subset (x-a,x-b)$. Now you can prove.......


2

To add to Eric's answer above, it doesn't even require the existence of a multiplicative identity to shoot this down. See below: $2[a\odot(b \cdot c)] \\ = a\odot(b \cdot c) + a\odot(b \cdot c) \\ = (a+a) \odot (b \cdot c) \\ = [(a+a) \odot b] \cdot [(a+a) \odot c] \\ = [(a \odot b) + (a \odot b)] \cdot [(a \odot c) + (a \odot c)] \\ = (a \odot b)(a \odot ...


5

One such operation is $a\odot b=0$ for all $a,b$. I claim this is the only such operation. Indeed, we have $$a\odot c=a\odot(1\cdot c)=(a\odot 1)\cdot (a\odot c),$$ $a\odot 1$ must be either $0$ or $1$ for each $a$. But if $a\odot 1=1$, then $(a+a)\odot 1=2$, which is impossible. So in fact $a\odot 1=0$ for all $a$, and now the equation above tells us ...


5

It follows that this is false from the same result about algebraic numbers. Every algebraic number can be written as $\frac{\alpha}n$ where $n$ in an integer and $\alpha$ is an algebraic integer. So if this was true for algebraic integers, it would be also true for algebraic numbers. The fact that there are algebraic numbers that cannot be expressed in ...


3

As suggested in the comments, I will try to write an answer (although I'm a little against this). First of all, I may accidentally forget to write the word "compact" in "compact Riemann surface". So I'm assuming that all Riemann surfaces are compact. Actually everything fails if it's not compact. For instance, the correspondence between curves and Riemann ...


3

To understand the explicit structure of group algebras like this, a useful idea is that of a skew group ring, which is a ring theoretic analogue of a semidirect product. Let $G$ be a group acting by ring automorphisms on a ring $A$. Then the elements of the skew group ring $A\ast G$ are formal finite sums $\sum_{g\in G}a_gg$ with $a_g\in A$, with ...


0

Though it is in general true that no element of a maximal ideal has an inverse: if an ideal contains a unit $u$, then it must be the whole ring (because $u(u^{-1}x)$ must be in the ideal for every $x$), I don't know that this is a valuable interpretation.


11

Here's an example to consider. Take a field $F$ and form $R=F\times F$. The ideal $I=F\times \{0\}$ satisfies simultaneously that $F\cong I\cong R/I$ (ring isomorphisms.) Should $I$ be an anti-field? It is not really clear that is a fruitful viewpoint... It is certainly an interesting perspective on things. In my experience, though, I don't see that it ...


3

I don't know much about this topic, but I did a quick calculation in Magma on $A = {\mathbb F}_3G$. It appears to be a direct sum of a simple algebra of dimension $18$ and a uniserial algebra of dimension $3$ with $3$ trivial composition factors. If you do this over the field of order $9$, then the $18$-dimensional algebra splits into a direct sum of two ...


0

Given a ring $R$, right $R$-modules are denoted with $R$ as a right index as in the following: $$M_R$$ this notation recalls you that you can multiply elements of $M_R$ with elements of $R$ on the right i.e. multiplication is a map $$M_R \times R \longrightarrow M_R\ \ \ (m,r) \mapsto mr$$ Now, the right $R$-module $R_R$ consists of the additive group ...


2

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 ...


0

More generally, $f^{-1}(f(I))=I+ \ker f$ Indeed, $f^{-1}(f(I))$ certainly contains $I$ and $\ker f=f^{-1}(0)$, and so $f^{-1}(f(I)) \supseteq I+ \ker f$. On the other hand, if $x \in f^{-1}(f(I))$, then $f(x) \in f(I)$ and so $f(x)=f(i)$ for some $i \in I$. But then $k=x-i \in \ker f$. Thus, $x=i+k \in I+ \ker f$ and $ f^{-1}(f(I)) \subseteq I+ \ker ...


0

Let $x\in f^{-1}(f(I))$. It follows that $f(x)\in f(I)\Rightarrow \exists y\in I$ such that $f(x)=f(y)\Rightarrow f(x-y)=0\Rightarrow x-y\in\ker(f) \Rightarrow x-y\in I$ and since $y\in I$, it follows that $x\in I$ which proves the second inclusion.


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

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 ...


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 $\;\phi^{-1}(\phi(I))=2\mathbf Z/4\mathbf Z$.


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 $3X+2\notin I$.


8

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 ...


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! $A,B$ ...


0

Throughout, I assume you are looking at left $R$-modules. Since $R$ is a finite-dimensional $\mathbb{Z}_p$-module, then using the duality $$ D={\rm Hom}_{\mathbb{Z}_p}(-, \mathbb{Z}_p): {\rm mod}\ R^{op} \to {\rm mod}\ R, $$ where $R^{op}$ is the opposite algebra, is a good way to answer your question. More precisely, projective $R^{op}$-modules will be ...


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 ...


5

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.)


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 $S$, ...


0

These are all standard results: $(1)$ Given an $R$-module $N$ and a submodule $N'$, there is a canonical bijection between the collection of submodules of $N''=N/N'$ and the collection of submodules $M$ of $N$ such that $M\supseteq N'$. $(2)$ A module $S$ is simple if it has precisely two distinct submodules. $(3)$ Finally, an ideal $\mathfrak m$ is ...


0

This is wrong because you can't multiply an element of $R/M$ by an element of $J$. They're separate rings. The proof of this I'm familiar with supposes $M$ is maximal and constructs the ideal $$J=\{m+rx:m\in M,r\in R\}$$ where $x$ is an element such that $x+M\neq M$ (or $x\notin M$). Then this ideal contains $M$ by letting $r=0$, and it contains $x$ by ...


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

$1-ux$ 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.


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.


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.)


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 ...


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 ...


0

I asked exactly the same question two years ago: see here. Also see my comment to the answer, which explains $(2)$. Hope this helps.


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 ...


4

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 ...


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.


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. $r_1m=r_2m$.


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

[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 ...


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 ...


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.


0

If you happen to be following the question, the news is that a solution was given on mathOverflow by Keith Kearnes. He showed that one can actually prove a lemma that in such a ring, each prime contains its annihilator, which was a pleasant surprise. Armed with this lemma, it is easy to prove that the nilradical contains its own annihilator.


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 ...


0

To elaborate on Orat's suggestion: If $R=2\Bbb Z$, $\begin{bmatrix}R&0\\R&0\end{bmatrix}$ is a left ideal of $M_2(R)$, and $\begin{bmatrix}R&R\\0&0\end{bmatrix}$ is a right ideal of $M_2(R)$. You could let $R$ be any ring, really. It doesn't have anything to do with identity. To give a different construction entirely, take a field ...


0

We chose polynomials $f(x)$ and $g(x)$ such that $$f(x)(x^2-4) + g(x)(x^3-3x+3)=1$$ and for this it is necessary the equality of degrees. So $d(f)+2=d(g)+3\iff d(f)=d(g)+1$. Chose the simplest possibilities $f(x)=ax^2+bx+c$ and $g(x)=dx+e$ and calculate the coefficients.It follows $(a+d)x^4+(b+e)x^3+(-4a+c-3d)x^2+(-4b+3d-3e)x+(-4c+3e)=1$ Solving the ...


1

The first two lines are fine. Then you are finished, though you may want to write $x+3=\frac{x+3}{5}\cdot 5+0$. Work backwards from the second line. We get $5=x^2-4 -(x-3)(x+3)$. But from the first line we have $x+3=x^3-3x+3-x(x^2-4)$. Substituting we get $$5=x^2-4-(x-3)(x^3-3x+3-x(x^2-4)).$$ This simplifies to $$5=(3-x)(x^3-3x+3)+(x^2-3x+1)(x^2-4).$$ If ...


1

Just use the bog-standard extended Euclidean algorithm, as given by Blankinship (a terse description here). If you look closely, it assumes only that the elements manipulated have a division algorithm, i.e., form an Euclidean domain. Polynomials over any field form an Euclidean domain, so the exact field is irrelevant.


1

1. "Yes" to the second question. More precisely, he is regarding $B$ as a $\left(B,A\right)$-bimodule, where the left $B$-module structure is obvious (i.e., given by multiplication) and where the right $A$-module structure is the one you define. "No" to the first question. He is tensoring a $\left(B,A\right)$-bimodule with a left $A$-module. This yields a ...


4

Nothing is wrong with it, and indeed this is an accepted definition. For any $A$-algebra $B$ it makes sense to talk about the set of elements in $B$ integral over $A$. See http://stacks.math.columbia.edu/tag/00GI.



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