# Tag Info

4

It is indeed not surjective, for the reason you state. Consider the case of abelian groups: the chain complex $$\cdots \to 0 \to \mathbb{Z} \stackrel{2}{\to} \mathbb{Z} \to 0 \to \cdots$$ is quasi-isomorphic to the chain complex $$\cdots \to 0 \to 0 \to \mathbb{Z} / 2 \mathbb{Z} \to 0 \to \cdots$$ via an obvious chain map, but the only chain map in the ...

3

Here is a general criterion that is useful. Let $\rm\,D\,$ be a domain with fraction field $\,\rm K.$ $$\rm f\,\ is\ prime\ in\ D[x]\iff f\,\ is\ prime (= irreducible)\ in\ K[x]\ and\,\ f\,\ is\ superprimitive$$ $$\rm where\,\ f\,\ is\ {\bf superprimitive}\ in\ D[x]\,\ :=\,\ d\,|\,cf\, \Rightarrow\, d\,|\,c\,\ \ for\ all\,\ c,d\in D^*$$

3

The general strategy here is to take the exact sequence $$0\to \ker \pi\to R^2\overset{\pi}\to I\to 0$$ and find a splitting for $\pi$. Then we'd have $R^2\cong \ker\pi\oplus I$, so $I$ is projective as a direct summand of a free module. You have to implicitly use fractional ideals of $R$, so if you know what that is do some Googling, but if you don't then ...

3

For $1$: Suppose $I$ is a proper finite ideal and $|I|=n$. Take $x\in I$, $x\neq 0$ and think about $\{x,x^2,\ldots,x^n\}\subseteq I$. Suppose none of these powers of $x$ in this set are zero: then $x^k=x^m\neq 0$ for some $m> k$, but then $(x^{m-k}-1)x^k=0$, but the left hand factor must be a unit. See the end? For $2$ look at $(x)$ in $\Bbb R[[x]]$. ...

2

Why not making your life easier? There is no need to calculate with polynomials. We have homeomorphisms $\mathrm{Spec}(k[T]/(T^2)) \cong V(T^2) = V(T) \cong \mathrm{Spec}(k[T]/(T)) \cong \mathrm{Spec}(k) = \{\eta\}$ It maps $\eta$ to the kernel of $k[T]/(T^2) \to k[T]/(T) \cong k$, which is $(T)$. In general, if $A$ is a commutative ring with a maximal ...

2

Hint for hands-on approach: Show that it's enough to take $s = 1$, i.e., given an element of the form $n/1 \in \ker g$, show that $n/1 \in \text{im } f_m$, and then show that this implies that $\ker g \subseteq \text{im } f_m$ (for this last part, notice that $n/s = (n/1) \cdot (1/s)$). High-brow approach: localization is an exact functor. This implies ...

2

You actually didn't use that $M$ is finitely generated. If $x_1,\dots,x_n$ is a system of generators of $M$, then $x_i/1=0/1$ in $S^{-1}M$, so there exists $s_i\in S$ such that $s_ix_i=0$. Set $s=s_1\cdots s_n$ and note that $sx_i=0$ for all $i=1,\dots,n$. Then $sM=0$ since every element of $M$ is a linear combination of the generators.

2

If $R$ is a local normal domain with $\dim R=2$, then every MCM is reflexive. First prove that $M$ is torsion-free. This shows that $M_p$ is free over $R_p$ for any prime $p$ with height $\le 1$. Next, if $p$ is a prime of height $2$ it's obvious that $M_p$ satisfies Serre's condition $(S_2)$. In the end, use Proposition 1.4.1(b) from Bruns and ...

2

The associated ideal sheaf you you define makes sense only for affine schemes. For a projective scheme $X$ with projective coordinate ring $S$, given a homogeneous element $f \in S_+$ the basic open set it defines is $D_+(f)=Spec{ S_{(f)}}$. So the corresponding ideal should be the zero grading of the ideal $I \cdot S_f$. Globally you just need to take the ...

2

Let $(R,\mathfrak m)$ be a local noetherian ring with $\operatorname{edim} R\le \operatorname{depth} R + 1$. Then $R$ is Gorenstein. (Kaplansky, Commutative Rings, Exercise 1, page 163.) Induction on $\operatorname{depth} R$. If $\operatorname{depth} R=0$, then $\operatorname{edim} R\le 1$ and therefore $\dim R\le 1$. In the case $\dim R=1$ we get ...

1

There are many possible answers. Since you posted this in algebraic geometry, I will give you a geometric answer. Suppose that you wanted to understand a regular affine curve $C=\text{Spec}(A)$. A natural thing that you might want to do is study line bundles over $C$ since this tells you quite a bit of geometric information about $C$. The standard way to ...

1

Let $k = \sup(k_1, \dots, k_n)$. We can write $1 = \sum a_i x_i$ by hypothesis. Therefore $$1 = 1^{nk} = (\sum a_i x_i)^{nk}$$ By the binomial theorem this sum can be written as a sum of elements of the form $b_{a_1, \dots, a_n} x_1^{a_1} \dots x_n^{a_n}$ (where the $b_?$ are some coefficients we don't care about) with $a_1 + \dots + a_n = nk$. So ...

1

(This should just be a comment, but too many of these questions are left without posted answers...) The notation $\text{im}(\varphi) + PR$ refers to the sum of the two submodules $\text{im}(\varphi)$ and $PR$ (these are subsets of $P$, not $P/PR$). Explicitly, \text{im}(\varphi) + PR = \{\varphi(f) + \sum_{\text{finite}} pr \mid f \in F, p \in P, r \in R ...

1

Let $α$ be the generator of $m$, i.e. $m = (α)$. Hint: First try to prove for any nonzero nonunit $x ∈ R$ that $(x) = (α^n)$ for some $n ∈ ℕ$, then use the fact that any ideal $I ≠ 0$ is finitely generated to conclude what you want to show. I’ve already done that, but now realized you maybe wanted to do this yourself. But I will leave below what I already ...

1

There is not a whole lot to say here. You are given that $\{0\}$ is semiprime, and you are asking when it is prime. You could say that $R$ is a domain iff $\{0\}$ is primary. fpqc's example demonstrates this nicely since $(xy)$ is a semiprime but not prime ideal of $k[x,y]$ and also not in the localization at $(x)$ or $(y)$. Locality never really comes into ...

1

Another (general) approach: localize at the maximal ideal $m = (X_1, X_2, X_3, X_4)$. The number of generators can only drop, so it's enough to show that $I_m$ requires more than $2$ generators, where $I = (X_1, X_2) \cap (X_3, X_4)$ is your ideal. But a local ring has a well-defined notion of minimal number of generators, which is also a vector space ...

1

First question: I can't see why the author (and the OP) starts with a Dedekind domain since then $A[y]$ is not necessary factorial (if $A=\mathbb Z[\sqrt{-5}]$ it isn't even a GCD domain) in order to ensure that $A[y]/(f)$ is a domain, that is, $f$ is a prime element of $A[y]$. However, if $A$ is a UFD, then all is okay and we can move on. Since $f$ is monic ...

1

It's not easy to explain a diagram chase in writing, but I'll try: Let $x \in Ш^1(k_S, S \setminus T, A')$, map it to $y \in H^1(k_S | k, A')$ then to $z \in \prod_S H^1(k_\mathfrak{p}, A')$. The image of $z$ in $\prod_{S \setminus T} H^1(k_\mathfrak{p}, A')$ is zero by exactness of the first row, so actually $z \in \prod_T H^1(k_\mathfrak{p}, A')$. $z$ is ...

1

I don't know if this is the simplest example, but it's the one that I thought to try first. Let $K=\mathbb{Q}(\sqrt{-15})$. Then, $L=\mathbb{Q}(\sqrt{-3},\sqrt{5})$ is the Hilbert class field for $K$ and so, in particular, $\mathcal{O}_L/\mathcal{O}_K$ is unramified. But, $\mathcal{O}_K$ is a Dedekind domain, and thus $\mathcal{O}_L$ is flat over ...

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