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10

I'm going to make this more geometric and less algebraic. But it all translates to the algebro-geometric setting of divisors if you wish. You should think of a line bundle as a twisted product, and tensor product means that you concatenate or superimpose the twists. For example, thinking of the Möbius strip as a real line bundle $\mathscr L \to S^1$, then ...


8

To my mind, "invertible with respect to the tensor product" is already the correct definition of a line bundle, in full generality. If that isn't the definition you're using I assume you're using something like "locally free of rank $1$," so let me say something about this. Intuitively you should think of a line bundle as literally a bundle of lines; that ...


4

This problem is an instance of the (almost trivial) statement: Let $R$ be a commutative ring and $a \in R$, then $R[Y]/(Y-a) \cong R$. Just set $R = k[X]$ and $a = X^2$.


3

Yes, as a DVR is a local ring there is only one prime number that is not invertible. (Note that a nontrivial ideal cannot include two distinct prime numbers and each non-unit is contained in some maximal ideal.) In more detail: Every non-zero subring of $\mathbb{Q}$ must contain $\mathbb{Z}$ and since a field is not a DVR the ring must not equal ...


3

In fact $a\otimes b\mapsto a\otimes b$ is an isomorphism $A\otimes_DB\leftrightarrow A\otimes_FB$ as spaces or $D$-algebras, for any spaces/algebras $A$ and $B$ over a field $F$ which is the fraction field of a domain $D$. The reason is that $$\begin{array}{ll} \displaystyle \color{Blue}{\frac{1}{y}}a\otimes b & \displaystyle =\frac{1}{y}a\otimes ...


2

If I had to give a tl;dr to my incoherent drivel above, it would probably be as follows. $V \otimes V^*$ has a canonical basis if and only if $V$ is $1$-dimensional. We have a circle. We draw the circle the opposite direction.


2

Like Ben, I do not think you will get a satisfactory answer. More precisely, I will show that computing the degree of the Hilbert polynomial is equivalent to computing the size of the maximal clique of a graph. Since the latter problem is NP-HARD, if you had a description of the Hilbert polynomial which was explicit and short enough to easily read off the ...


2

I doubt you will get a satisfying answer. The reason is that it (e.g., the Hilbert-Poincaré series) depends deeply on the divisions between the (factors of the) $m_i$. The following works for explicit and not too long sequences $m_1,\dots m_l$, but it's not good enough for a general answer. $\newcommand{\HP}{\mathrm{HP}}$ We can and should make use of the ...


2

It is false: take any DVR $V$, $\pi$ a uniformising element and $K$ its field of fractions. Then $K/\pi K=0$, but $K$ can't be a finitely generated $V$-module, since this would mean $K$ is integral over $V$ and hence $K=V$, which is impossible since by definition, a DVR is not a field.


2

In short, no, I do not think the proof is complete. It looks like you are trying to show $A(\mathbb{A}^1)\cong A(V(y-x^2))$ and then note $A(\mathbb{A}^1)\cong k[t]$ and $A(V(y-x^2))\cong k[x,y]/(y-x^2)$. Although the claim that the algebraic set $\{(x,y)\in \mathbb{A}^2: x=t, y=t^2\}$ is the zero set of $y-x^2$ is true, how do you know it is exactly the ...


2

Let $x\in S$. Then there are two chances: $x\in A$, or $x\notin A$. If the latter happens then $x^{-1}\in A$ and then $x^{-1}\in\mathfrak m$, so $x^{-1}\in\mathfrak m_S$. But $x\in S$, so $1=xx^{-1}\in\mathfrak m_S$, a contradiction.


2

In $\mathbb{C}[x,y]$ we have $(x)\cap(x+y,x-y) = (x)\cap(x,y) = (x)$, but $(x^2+xy)+(x^2-xy) = (x^2,xy)$.


2

Hint $\ $ Distributivity easily yields that a finitely generated ideal $\,=1\,$ if it contains a cancellable element $\rm\:z\:$ that is $\rm\:lcm$-coprime to the generators, e.g. for a $2$-generated ideal $\rm\:(x,y)$ $$\rm\ \begin{array}{} (x)\cap(z)\ =\ (xz)\\ \rm (y)\cap(z)\ =\ (yz)\end{array}\ \ \ and\ \ \ (x,y)\supseteq (z)\ \ \Rightarrow\ \ (x,y) = ...


1

In $\mathbb Z[x]$ you can try the following example: $\mathfrak a=(x)$, $\mathfrak b=(x-2)$ and $\mathfrak c=(x+2)$.


1

Consider $R = \mathbb{Z}$ and $A=C= \mathbb{Z}/(p)$ and $B=\mathbb{Z}/(p^2)$ and the short exact sequence $$0 \to \mathbb{Z}/(p) \to \mathbb{Z}/(p^2) \to \mathbb{Z}/(p) \to 0$$ where the first morphism is given by $1\mapsto p$, the second is the projection mod$(p)$. Clearly for $p$ prime, this does not split, since $\mathbb{Z}/(p^2)$ is not isomorphic to ...


1

Yes, one can avoid the use of Nullstellensatz. Let $I=(Y-X^2)$, $\alpha:k[X,Y]\to k[X,Y]/I$ be a natural homomorphism. Then we have two obviuous facts: $\alpha(k[X])=k[X,Y]/I$; $k[X]\cap I=\{0\}$. It follows, that restriction $\alpha'=\alpha|_{k[X]}$ epimorphic and injective, hence $\alpha':k[X]\to k[X,Y]/I$ is an isomorphism. Verification of statement ...


1

Here is a collection of hints from my comments to OP: For (1), think about the analogous question: "is $\mathbf{Q}$ finited generated as a $\mathbf{Z}$-algebra (!)?" A more explicit answer to (1): Recall that $F[t]$ is a principal ideal domain (much like $\mathbf{Z}$). We will show that $F[t]$ has infinitely many primes. If $K$ is a finite ...


1

First of all, if $I$ is a regular sequence, then the Hilbert function of $R/I$ is very easily computed, let me know if you want to know more on this. For the more general case, let me introduce the basics of what is known as Apolarity Theory, an extremely powerful tool to compute the Hilbert function of $R/I$ in the case where $R=k[x_0,...,x_n]$ and $I$ is ...


1

Here is a small computation that will help: We are given: \begin{align*} &\qquad \left(\bigoplus_{n=0}^k N_nt^n \right)\left(R + \sum_{j=1}^\infty I^jt^j\right) \\ &= \left(\bigoplus_{n=0}^k N_nt^n \right)\left(I^0t^0 + \sum_{j=1}^\infty I^jt^j\right) \\ &= \left(\bigoplus_{n=0}^k N_nt^n \right)\left(\sum_{j=0}^\infty I^jt^j\right) ...


1

Let $A$ be a noetherian ring which is regular in codimension $1$. That means $A_P$ is a discrete valuation ring for all $P$ with $\mathrm{height}\, P = 1$. The ring $A'=k[X,Y]$ fulfills this condition. In fact every prime of height one is principal $P=(f)$ with $f$ irreducible and the valuation of $P$ is given by the exponent of $f$ in the prime factor ...


1

One way to describe the algebraic closure is that it is in some sense a "maximal" algebraic extension: it's an algebraic extension into which every other extension embeds. So it seems to me like the following question is a more basic one that should be answered first: What's an algebraic extension of commutative rings? There are various ways to answer ...


1

$aL=0$ and $bN=0$ implies $(ab)x=0$ for all $x\in M$:



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