algebraic geometry – The field of quotients of the ring of germs of functions.

Let $Ysubseteq mathbb{A}^n$ be an affine variety with affine co-ordinate ring $A(Y)=k(x_1dots, k_n)/I(Y).$

Denote with $mathcal{O}(Y)$ the ring of all regular functions $fcolon Yto k$.

Fix an affine variety $Y$ and a point $Pin Y$. Let $U,Vsubseteq Y$ be open neighbourhoods of $P$.

For $finmathcal{O}(U)$ and $ginmathcal{O}(V)$ define $f sim g$ if there exists a open neighbourhood $Wsubseteq Ucap V$ of $P$ such that $f|_W=g|_W$. Then $sim$ s an equivalence relation between these the set of all such pairs $(U,f)$. The set $mathcal{O}_{P,Y}$ (or $mathcal{O}_P)$ of all equivalence classes is a local ring, called the ring of germs of functions.

For a pair of functions $(U,f)$, $(V,g)$ (as above except we ignore a specific point $P$), define $fsim’g$ if there exists a open set $Wsubseteq Ucap V$ such that $f|_W=g|_W$. Then $sim’$ is an equivalence relation and the set $K(Y)$ of all equivalence classes is a field.

Denote now with $K(R)$ the quotient field of an integral domain $R$.

I have to prove that in general the following is true:


We consider the restriction map $$mathcal{O}_{P,Y} to K(Y)quadtext{defined as}quad (f)mapsto (f)’.$$

I have shown that this map is injective, then results that $$K(mathcal{O}_{P,Y})subseteq K(Y).$$

For the opposite inclusion I was suggested to use the following:

Using the above facts, irreducibility of $Y$ and $$color{red}{K(Y)=bigcup_{Qin Y}K(mathcal{O}_Q)}$$

Question 1. Why is the expression in red true?

show that $$K(mathcal{O}_Q)=K(Y)quadtext{for each}quad Qin Y.$$

Could someone suggest me how to proceed? I have no idea how to deal with this inclusion. Thanks!

ac.commutative algebra – Flatness criterion for $I$-adic ring : $I$-torsion free

Let $R$ be an $I$-adically separtaed and complete valuation ring, with $I$ finitely generated.

It is used a few times in this (book by Bosch) e.g. first lines of pg164 Cor. 5 and Cor. 6 (their condition (V) is what I stated above) that

If an $A$ module has no $I$ torsion then it is flat over $R$.

I don’t see why this is true. Any suggestions / references would be appreciated.

What I thought: We know $A$ is flat over domain $R$ iff it is $R$-torsion free.

If the statement were true: $I$-torsion free $Rightarrow$$R$-torsion free.

This would hold if $I$ is a maximal ideal but otherwise I don’t see why.

ag.algebraic geometry – Cohomology ring of grassmannian and Pieri rule

I am sorry if this question is not for mathoverflow. I asked the same question on stackexchange, but I didn’t get an answer:

Let $X=OG(n,2n+1)$, where $OG(n,2n+1)$ denotes the variety of $n$-dimensional isotropic subspaces of a vector space $mathbb{C}^{2n+1}$ with a nondegenerate symmetric bilinear form.

According to Theorem 2.2 a) (Page 17, Anders Skovsted Buch, Andrew Kresch, Harry Tamvakis, Quantum Pieri rules for isotropic grassmannians,, the cohomology ring of X is given by
$$ H^{*}(X,mathbb{Z})=mathbb{Z}(tau_{1},ldots, tau_{n})/I,$$

where $I$ is the ideal generated by
$$ tau_{r}^{2}-2tau_{r+1}tau_{r-1}+2tau_{r+2}tau_{r-2}+cdots +(-1)^{r}tau_{2r}$$
for $1leq rleq n$.

In particular, if $n=4$, then the ideal $I$ is generated by the following four elements

tau_{1}^{2}-tau_{2},quad tau_{2}^{2}-2tau_{3}tau_{1}+tau_{4}, quad tau_{3}^{2}-2tau_{4}tau_{2},quad tau_{4}^{2}.tag{*}label{*}$$

But if I apply Pieri rule for X (Theorem 2.1, Page 16, Anders Skovsted Buch, Andrew Kresch, Harry Tamvakis, Quantum Pieri rules for isotropic grassmannians) to $tau_{2}cdot tau_{2}$, I get the following relation

$$tau_{2}^{2}-2tau_{3}tau_{1}-tau_{4} tag{**}label{**}$$

Therefore, combining (ref{*}) and (ref{**}), I get $2tau_{4}=0$ in $H^{*}(X,mathbb{Z})$. I seems that some computation is wrong, but I don’t know where I made a mistake.

Thank you.

abstract algebra – Boolean ring prime ideals

I’m asked 2 question:

1- Prove or disprove: In every Boolean ring prime ideals are maximal.

2- Prove or disprove: If $R$ is Boolean and $p$ is a prime ideal of $R$, then the ring homomorphism $Rto R_p$ is always finite.

try for 1:
$bar R=R/p$ is a boolian domain. boolian domain is a field with 2 elements. (if $x in bar R$, then x(1-x)=0 so x=0 or x=1 ). therefore R/p is field and $p$ is maximal. is this proof true?

—- for 2, i dont know i should prove or disprove. i need help and clue please.

abstract algebra – For any ring $R$, $exp : R[[X]] to R[[X]]$ is easily well-defined using Taylor series?

Let $S = R((x))$ be the ring of formal power series over a ring $R$.

Take any series $s in S$, we have that $dfrac{s^n}{n!} in S$ as well. This is since if $a + b + c + d = e$ is a way of summing the exponents of terms of $s$ to an exponent $e$ of $s^4$, then there are $4!$ associated ways which are the permutations of $(a,b,c,d)$. Thus we can always divide out this multiplicity.

Can we conclude that $e^s in S$ or only are its finite partial sums?

I think it can be concluded because, if $a_n x^n$ is a term of $e^s$ then $a_n = sum_{k = 1}^n sum_{i_1 + i_2 + dots + i_k = n} prod_{j=1}^k s_{i_j}$ (or something)… Is this reasoning correct?

co.combinatorics – Integral face ring of the triangulation of a sphere

The integral face ring of a (finite) simplicial complex $K$ on $m$ vertices is the quotient ring


where $mathcal{I}_K$ is the ideal generated by the square free monomials $v_{i_1} v_{i_2}dots v_{i_n}$ where $I={i_1,…,i_n}$ is a minimal missing face of $K$. That is, $I notin K$, but for all $J subsetneq I$, $J in K$.

It was proven by Stanley that a simplicial complex is a generalised homology sphere (that is, a homology manifold with the homology of a sphere) if and only if the face ring is Gorenstein.

Is there a similar characterisation of when $K$ is a genuine triangulation of a sphere, in terms of the algebra of $K$?

ag.algebraic geometry – How to create a toric variety whose Cox ring has a specific grading?

If one wanted to obtain a fan for a toric variety of dimension $ n>1 $ whose Cox ring is $ mathbb{Z}^{2} $ graded with weights $ {(a_{i},b_{i})}_{i=1}^{n+2} $, then one could let $ B $ be the $ (n+2)times 2 $ matrix whose $ i $-th row has entries $ (a_{i},b_{i}) $. After computing the nullspace, one ends up with an $ n times (n+2) $ matrix $ A $ with entries in $ mathbb{Z} $. The $ n+2 $ columns of $ A $ are the rays for a fan in $ mathbb{R}^{n} $. If the rays are $ {u_{rho_{1}},dots,u_{rho_{n+2}} } $, then maximal dimensional cones $ sigma $ are of the form $ operatorname{Cone}(u_{rho_{i_{1}}},dots,u_{rho_{i_{n}}}) $. The fan $ Sigma $ is then obtained from the maximal cones and their faces. From $ Sigma $ one obtains an ideal $ B(Sigma) = langle x^{widehat{sigma}} rangle_{sigma in Sigma} $ where $ x^{widehat{sigma}} $ is $ prod_{i mid rho_{i} notin sigma} x_{i} $. From here the quotient of $ mathbb{A}^{n+2}_{mathbb{C}} setminus Z(B(Sigma)) $ by the $ mathbb{G}_{m}^{2} $ action which sends $ x_{i} $ to $ z_{1}^{a_{i}}z_{2}^{b_{i}}x_{i} $ is isomorphic to the variety $ X_{Sigma} $ obtained from the fan $ Sigma $. As a result, the Cox ring of $ X_{Sigma} $ has the desired grading.

What if instead of wanting to find an explicit fan of a toric variety of dimension $ n>1 $ whose Cox ring is $ mathbb{Z}^{2} $ graded, one wants to find an explicit fan of a toric variety of dimension $ n>1 $ whose Cox ring is $ operatorname{Hom}(mathbb{Z}/langle M rangle mathbb{Z}, mathbb{C}^{ast}) times operatorname{Hom}(mathbb{Z}/langle N rangle mathbb{Z}, mathbb{C}^{ast}) $ graded with weights $ (overline{a_{i}}, overline{b_{i}})_{i=1}^{n} $? Is there a similar algorithm for obtaining the fan for such a variety?

match the ring structure with the definition of congruence of 2 numbers modulo an integer.

match the ring structure with the definition of congruence of 2 numbers modulo an integer.

hello I must make 3 sum tables, multiplication such as the integers module 6 an example like this
$ begin {array} {l}
text {In} \
begin {matrix} {| c | c | c | c | c |}
hline * & overline {0} & overline {1} & overline {2} & overline {3} \
hline overline {0} & & & & \
hline overline {1} & & & & \
hline overline {2} & & & 2 & \
hline overline {3} & & & 0 & \
end {array} \
end {array}$

I do not know if you can give me some guide on how to solve for example how that 2 and that 0 come out

How can I create a simple dropdown to allow users the select an attribute (like ring size) from a Commerce 2 product page?

This seems like it should be simple but despite reading docs and watching videos I cannot figure this out.

I have a Commerce 2 product variation type that I need to add a dropdown on to allow users to select a ring size. This will not affect the price.

I’ve been trying to use attributes but it doesn’t seem to work correctly.

I’ve created a new attribute type (Ring Size) and add the sizes to it. But then I have to go into the variation edit form and it only allows me to select one of the ring sizes and then this one ring size shows up in a dropdown on the product page.

I’d like to have a list of ring sizes and for them to all be available on the product page for the user to select from.

Any direction on how to achieve this would be greatly appreciated. See the enter image description herescreenshot for an example of what I’m trying to accomplish.

ring theory – Generalizing valuations

Let $F$ be a field, and let $Gamma$ be a totally ordered abelian group. A valuation is a group homomorphism $v : F to Gamma$ such that $v(x + y) geq min(v(x), v(y))$. On the other hand, a valuation ring $O subseteq F$ is a ring such that for every $x in F^times$, we have $x in O$ or $x^{-1} in O$. It is classical that the valuations on a field (up to equivalence) are in bijective correspondence with the valuation rings of that field.

Valuation theory also has important applications in number theory (since valuations correspond to prime ideals in a number field), and to algebraic geometry (because they correspond to points in affine space, among other reasons).

My question is: what work has been done to generalize valuations? The definition of a valuation makes sense without any modification if $Gamma$ is an abelian lattice-ordered group instead of an abelian ordered group, and there are plenty of integral domains people care about that are not valuation rings. Have people studied “valuation-like” structures where the definition of a valuation is tweaked or generalized in some way? Have people studied “valuation-like rings” where that definition is tweaked or generalized in some way? Are there established correspondences between generalized valuations and generalized valuation rings?