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8

Lemma 1: The only unit is the identity. If $e$ is the leftmost idempotent in a factorization of a unit $u$, then $(1-e)u=0$ implies $1-e=0$ and $e=1$. After finitely many steps we have proven $u=1$. Lemma 2: The ring is reduced, that is, it has no nonzero nilpotent elements. The only nilpotent element is $0$. If $x$ is nilpotent, then $1-x$ is a ...

5

I believe this is proven in Chapter 7 of Nestruev's Smooth Manifolds and Observables, but I haven't checked carefully. More precisely, the functor $M \to C^{\infty}(M)$ from smooth manifolds to the opposite of real commutative algebras is fully faithful, meaning that smooth maps $M \to N$ are precisely algebra maps $C^{\infty}(N) \to C^{\infty}(M)$.

4

If $i \neq j$, we deduce that $P_i+P_j$ is not contained in any maximal ideal: Assume $P_i+P_j \subset M$, then $R_M$ has at least two minimal primes - namely $P_i$ and $P_j$ - and is thus not a domain. So $P_i+P_j=R$ for $i \neq j$. This is precisely what we need for the Chinese Remainder Theorem, hence we are done. Note that $\bigcap_{1 \leq i \leq n} ... 4 Recall the ideal correspondence for quotient rings: The ideals of a ring of the form$R/I$are ideals of$R$which contain$I$. This is just like the subgroup correspondence for quotient groups. So you are right to start off by noting the maximal ideals of$k[x,y]$are of the form$(x-a, y-b)$by the Nullstellensatz. Now any maximal ideal$m$of your ring ... 3 These two schemes are certainly not isomorphic to each other, as the first has one dimensional tangent space and the second two dimensional. Clearly any such scheme must be affine, so it is just a case of classifying local rings$A$of vector space dimension three over an algebraically closed field$k$. Let$m$be the maximal ideal. Then$m$indues a ... 3 [I assume all "smooth manifolds" are Hausdorff and paracompact.] Yes, you can recover$M$as a smooth manifold from the ring$C^\infty(M)$. Here's a quick sketch. First, note that we can recover the set of connected components of$M$, since each connected component$N\subseteq M$corresponds to a minimal nonzero idempotent in$C^\infty(M)$, and the ideal ... 3 As Eric Wofsey points out, because$M$as a topological space is the space of homomorphisms$C^\infty(M) \to \Bbb R$, appropriately topologized, we know precisely what the elements of$C^\infty(M)$are as functions on$M$. So we can recover the space$C^\infty(M)_p$of germs at$p$, and hence we can recover the dimension of$M$as the dimension of the space ... 3 Your chain is not a chain of prime ideals, since$1 \in \langle 2,3 \rangle$. Your ring is clearly isomorphic to$\mathbb Z_6[y]$. One easily shows, that this ring is one-dimensional: On the one hand, we have chain of length$1$:$\langle 2 \rangle \subsetneq \langle 2,y\rangle$. Now let$\mathfrak p$be a prime ideal of height$1$,$\mathfrak p \cap ...

3

Yes, it is obvious. Take the long exact sequence of cohomology groups. Anytime, the cohomology of the two outer terms vanishes, the cohomology of the inner term will vanish, too.

2

The long exact sequence of homology (which indeed comes from the snake lemma) reads $$H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})\to H_i(X,\mathbb{Z}/n\mathbb{Z}) \to H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$$ where the maps $H_i(X,\mathbb{Z}) \to H_i(X,\mathbb{Z})$ and $H_{i-1}(X,\mathbb{Z})\to H_{i-1}(X,\mathbb{Z})$ are given by multiplication by $n$ ...

2

Under the assumption that $A$ is noetherian, yes, $\dim A[x_1, \ldots, x_n]_s =\dim A[x_1, \ldots, x_n]$. It is enough to prove that for a noetherian ring $A$ of finite dimension $n$ the ring $B:=A[x]_x$ has dimension $\dim B=n+1$, since by noetherianity $\dim A[x]=n+1$. Indeed let $\mathfrak p_0\mathfrak \subsetneq \cdots \subsetneq \mathfrak p_n$ ...

2

In general this is not true: for instance that $\mathbb{Z}_p \otimes_\mathbb{Z} \mathbb{Z}_q \cong \{ 0\}$ if $gcd(p,q)=1$.

2

Matsumura means the $\mathfrak{m}_i$-adic completion of this ring (when one says, as does Matsumura, "the completion of the local ring," the max-adic completion is intended). But this doesn't in fact matter, since $I^nA_{\mathfrak{m}_i}=\mathfrak{m}_i^nA_{\mathfrak{m}_i}$ for $n\geq 1$, so the $I$-adic completion of $A_{\mathfrak{m}_i}$ coincides with its ...

2

Set $I=I_0$, and $J=\cap_{i=1}^nI_i$. We have $\mathrm{reg}(I)\le1$, $\mathrm{reg}(J)\le n$, and $\mathrm{reg}(I\cap J)\le n+1$. (See Theorem 2.1 from this paper.) Now use the following exact sequences: $$0\to I\to I\oplus J\to J\to0.$$ $$0\to I\cap J\to I\oplus J\to I+J\to0.$$ From the first we get $\mathrm{reg}(I\oplus J)\le n$, while form the second ...

2

I have seen these rings discussed, but never a never with a name attached. The reason is probably because this class of rings is a disjoint union of the following two classes of rings which do not need complicated names and which have rather divergent properties compared to each other: Rings of Krull dimension 0 Domains of Krull dimension 1

1

The book "Introduction to commutative algebra" by "Atiyah-Macdonald" is good terse text. But since you want a Book which is easy to understand, the book "Steps in Commutative Algebra" by "R.Y.Sharp" would be better for you, (it has more explanation). Maybe the book Monomial Ideals by "Herzog-Hibi" is the best for you, if you want to see examples of ...

1

Why do you think $\operatorname{depth}_SI=3$? I'd say that it is $1$.

1

We have $H^n_{\mathfrak m}(I)=H^n_{\mathfrak m}(S)\ne0$. But $\max\{i:H^n_{\mathfrak m}(S)_i≠0\}=−n$, so for $d≥−n+1$ we have $H^n_{\mathfrak m}(S)_d=0$. This shows that the same conclusion holds for $m≥0$ and $d≥m−n+1$. At this point I guess (but I'm not sure!) they assume $m≥0$ in this lemma (otherwise (c) is automatically satisfied).

1

One implication is easy. For the harder one, suppose $S \not \subset T$. Fix an element $x \in S \setminus T$. Show that the set of all ideals of $R$ that contain $T$ and do not contain $x$ satisfies the condition for Zorn's lemma, so it has a maximal element. Show that such a maximal element $P$ is a prime ideal in $R$, and $SR_P \not \subset TR_P$.

1

Here's a guide to finding the discrete valuations on $\mathbb{C}(x)$. I've put my solutions in spoiler boxes so you can try before you peek. Also, since you didn't mention it, $A$ is a valuation ring of $\mathbb{C}(x)$ iff it is a subring and $z\in A$ or $z^{-1}\in A$ for all non-zero $z\in\mathbb{C}(x)$. In particular, valuations rings are local. The main ...

1

I suppose $R$ and $S$ are local rings with maximal ideals $\mathfrak m$, respectively $\mathfrak n$. If $r_1,\dots,r_n$ is a sop for $R$, then the ideal $(r_1,\dots,r_n)$ is $\mathfrak m$-primary. We wonder if $(r_1,\dots,r_n)S$ is $\mathfrak n$-primary. First note that $\sqrt{(r_1,\dots,r_n)S}=\sqrt{\mathfrak mS}$. But $\mathfrak mS$ is $\mathfrak ... 1 Without loss of generality, assume that$S=\{x_1,\ldots, x_k\}$. Then for any$x_m, m>k$, we have$f_m(x_1,\ldots, x_k, x_m)=0$in$R=\mathbb{k}[x_1,\ldots, x_n]/\mathfrak{a}$, for some non-zero polynomial$f$in the polynomial ring. Let$a_m=a_m(x_1,\ldots, x_k)$be the leading coefficient of$f_m$as a polynomial in$x_m$. Let$s=\prod_{m>k} a_m$. ... 1 I suppose it really depends on what aspects you are interested in, but even for the basic properties, there are many choices. My personal choice is Atiyah-Macdonald's text where the chapters are pretty clearly laid out. Another good resource is Keith Conrad's notes: Found here. 1 If for$\mathbf b=(b_1,\ldots, b_n)\in K^n$we denote with$\phi_{\mathbf b}$the endomorphism of$K[X_1,\ldots,X_n]$given by$\phi_{\mathbf b}(X_i)=X_i+b_i$, then we see that$\phi_{-\mathbf b}\circ\phi_{\mathbf b}=\phi_{\mathbf b}\circ\phi_{-\mathbf b}=\operatorname{id}$because these compositions map$X_i\mapsto X_i+b_i\mapsto X_i$(resp.$X_i\mapsto ...

1

The most important property of polynomial rings is the following: given a ring homomorphism $\varphi\colon K\to R$ and $r_1,\dots,r_n\in R$, there exists a unique ring homomorphism $\hat{h}\colon K[X_1,\dots,X_n]$ such that $\hat{\varphi}(k)=\varphi(k)$, for $k\in K$, and $\hat{\varphi}(X_i)=r_i$, for $i=1,\dots,n$. Here $R$ is any commutative ring. Now ...

1

As an example, take $I = \langle 4\sqrt{-14}\rangle\subset\mathbb Q(\sqrt{-14})$. It's clear that $I = \langle 2\rangle^2\cdot\langle \sqrt{-14}\rangle$. To factorise $I$, it is sufficient to factorise $\langle2\rangle$ and $\langle \sqrt{-14}\rangle$ separately. To factorise $\langle 2\rangle$ we can use the Kummer-Dedekind theorem: $\mathcal O_{\mathbb ... 1 Denote$U_f=\{p\in Spec(R),f$is not in$p\}$. Let$i:R\rightarrow R_f$the canonical map, it induces a continuous map$u:Spec(R_f)\rightarrow Spec(R)$whose image is the open subset$U_f$, Since it is bijective, you have only to show that$u$is open, here$U_f$is endowed with the topology induced by the topology of$Spec(R)$which is equivalent to ... 1 What about$(x_3)/\mathfrak a\subsetneq(x_2,x_3)/\mathfrak a\subsetneq(x_1,x_2,x_3)/\mathfrak a$? 1 The Koszul complex is the right choice, you can show that your polynomials must form a regular sequence. First of all, let$I:=\langle f_1,f_2,f_3\rangle$, then$\sqrt{I}=R_+=(x_0,x_1,x_2)$by assumption and hence,$I$has height$3$. In particular,$I$contains a regular sequence of length$3$. Recall that any inextendible regular sequence is of maximal ... 1 Hint: if$I$is nilpotent, it is contained in the Jacobson radical, you deduce that$N+J(A)N'=N+IN'=M$. You have$I\subset J(A)$, thus$IN'\subset J(A)N'$and$M=N+IN'\subset N+J(A)N'\subset M\$. You can apply the Nakayama lemma. statement 3 https://en.wikipedia.org/wiki/Nakayama_lemma#Statement

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