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In general the map might not be either open nor closed. For example, for $p \in Spec(R)$, $Spec(R_p)$ is the set of all prime ideals that are contained in $p$. So $(2) \in Spec(\mathbb{Z})$ is the set $\{(0), (2)\}$ which is neither open nor closed. if $f : X \rightarrow Y$ is a continuous bijection between topological spaces, then $f$ is a homeomorphism ...

1

Let $V=\mathbb{R}^2$, $U=\mathbb{R}\times 0$, and $W=0\times\mathbb{R}$, and let $L$ be the subgroup of $V$ generated by $(0,1)$ and $(\pi,\pi)$. Then we have $U\cap L=\{(0,0)\}$, so that $(U\cap L)+W=W=0\times\mathbb{R}$. On the other hand, we have $U+W=V$, and $L+W=\mathbb{Z}\pi\times\mathbb{R}$, so that $(U+W)\cap(L+W)=\mathbb{Z}\pi\times\mathbb{R}$.

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In case $I$ is a radical ideal which isn't the irrelevant ideal we can prove this as follows: The condition implies that for each $i$, $$x_i^kf=g_i$$ for some $k\in \mathbb{Z}$. If $k\leq 0$ for some $i$ we are done. If not then for any minimal prime ideal $p$ over $I$, take $i$ s.t. $x_i\notin p$. Then $x_i^kf=g_i\in p$ implies $f\in p$. So $f$ is in ...

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The text says If $L(P_0)$ were generated by global sections, then $P_0$ would be linearly equivalent to some other point $Q \in X$. This goes as follows: Suppose $L(P_0)$ is generated by global sections. Then there is a global section $s \in \Gamma(X, L(P_0))$ which generated $L(P_0)$ at the point $P_0$. The effective Cartier divisor of zeroes $$D = (s)_0 =... 2 I do not believe the statement is true. Take for example f=x_0 and for 0\leq i\leq n let g_i=x_0x_i and define I=(g_0,\ldots ,g_n). Obviously f\notin I since all the generators of I are homogeneous of degree 2 but f is homogeneous of degree 1. On the other hand,$$f(x_0,\ldots,x_{i-1},1,x_{i+1},\ldots,x_n)=g_i(x_0,\ldots,x_{i-1},1,x_{i+1},\...

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If $\mathrm{Hom}(R/I,M)=0$, then there is no nonzero $a \in M$ whose vanishing ideal contains $I$. In other words, the (finite) set of associated primes of $M$ contains no prime ideal containing $I$. By the prime avoidance lemma, the reunion of the associated primes of $M$ does not contain $I$. But the reunion of the associated primes of $M$ is exactly the ...

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Let us suppose that $\operatorname{ht}(a_1,\dots, a_s)\le s-1$. Then there is a prime ideal $\mathfrak{p}$ containing $(a_1,\dots, a_s)$ such that $\operatorname{ht} \mathfrak{p}\leq s-1$. In particular, $\mathfrak p\supseteq(a_1,\dots,a_{s-1})$. If $\mathfrak p$ is not minimal over $(a_1,\dots,a_{s-1})$, then there exists a prime ideal $\mathfrak p'$ such ...

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Well every such ideal $I$ will be contained in a maximal ideal $\mathfrak{m}$ so that $A/\mathfrak{m}\simeq k$. Now take $a_i\in k$ such that $x_i$ corresponds to $a_i$ under this isomorphism. So $\mathfrak{m}=(x_1-a_2,\dots, x_n-a_n).$ Now $I \subset \mathfrak{m}=(x_1-a_1,\dots, x_n-a_n)$ says that for all $f\in I$, $f(a_1,\dots, a_n)=0$.

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Let $I=\mathrm{Ann}(M)$. The question reduces to show that $$\dim R_p/I_p+\dim R/(p+I)\le\dim R/I$$ for $I\subset R$ an arbitrary ideal, and $p\subset R$ a prime ideal. If $I\nsubseteq p$, then we have to show that $\dim R/(p+I)\le\dim R/I$ which is obvious. If $I\subseteq p$, then the inequality becomes $$\dim R_p/I_p+\dim R/p\le\dim R/I.$$ Since $R_p/I_p\... 1 You only need conditions 2) and 3), since 2) implies 1). Let$R$be a commutative ring. The first/second condition tells us that$R$contains at least three ideals$(0) \subset I \subset R$. Now let$J$be any non-zero, proper ideal of$R$. By definition of$I$we have that$J \subset I$. Our goal is to show that$J = I$. Now, certainly$J$contains some non-... 2$(3)$doesn't exclude the possibility that$R\setminus R^*\subseteq\{0\}$. See for instance the zero ring and the fields. However, these are the only obstructions: condition$(3)$in rings that are neither fields nor the zero ring implies that$R\setminus R^*$is the nilradical of$R$and that it isn't zero.$(2)\Rightarrow (1)$tautologically. there are ... 2 Suppose$n$is squarefree and let$p_1 \cdots p_n$be a prime factorization of$n$. The Chinese Remainder Theorem implies$\mathbb{Z}/n\mathbb{Z} \cong \prod_{i=1}^n \mathbb{Z}/p_i\mathbb{Z}$as rings. Each$\mathbb{Z}/p_i\mathbb{Z}$is a field and thus is regular. Products of regular rings are regular, and so$\mathbb{Z}/n\mathbb{Z}$is regular. 2 You already seem to have what you were asking for. Starting from where you left off, let$J’=(b_1,\ldots,b_n)$. Then$(a)=IJ\supseteq IJ’\supseteq (a)$. Thus we have equality all across. By uniqueness of$J$then,$J=J’$is finitely generated. (I haven’t gone further because you indicated this is the subproblem you wanted to settle.) 0 Let$N$be a finitely generated$R$-module and$q\subseteq p$two prime ideals with$q\in Supp(N)$. Since$q_p\in Supp(N_p)$and $$(N_p)_{q_p}\simeq N_q\neq0$$ we must have necessarily$N_p\neq0$, that is,$p\in Supp(N)$. From this and$Supp(M)=Supp(M')\cup Supp(M'')$we conclude that any chain of prime ideals in$Supp(M)$lies completely in$Supp(M')$or$...

0

Recall $\mathrm{Supp}(M)=\{\mathfrak{p} \text{ prime ideal in }A\mid M_{\mathfrak{p}}\neq 0\}$. Now observe that $$\mathrm{Supp}(M)=\mathrm{Supp}(M')\cup \mathrm{Supp}(M'').$$ Let $\mathfrak{p}_0\subset\cdots\subset \mathfrak{p}_n$ be a maximal chain of prime ideals in $\mathrm{Supp}(M)$. Assume $M'_{\mathfrak{p}_s}\neq 0$ for all $s=0,\dots, n$. Then $\... 0 Replace$x^2=y^3$in$4y^2+x^4+5x^2y = y^2(y^2+1)(y^2+5)$, if$a\not=0\Leftrightarrow b\not=0$, it's clear that$b$is not a repeated$0$of$y$, and$x^2 - b^3=0$has no repeated zero either. Therefore,$A_{\mathfrak m}\simeq \mathbb C[x,y]/\langle x-a, y-b\rangle\simeq\mathbb C$has dimension$1$. When$a=b=0$, then after the localization, the original ... 4 Use the Chinese remainder theorem,$A = A/(\prod \mathfrak m_i)\simeq A/(\mathfrak m_1^{n_1})\times\cdots\times A/(\mathfrak m_i^{n_i})$. It's enough to show$A/\mathfrak m_j^{n_j}$is Artinian. (if you assume$\mathfrak m_i$are distinct, then$A$is just a finite product of fields, hence Artinian, without assuming it's already Noetherian.) This follows ... 7 One characterization you could give is that a Dedekind domain$R$is the ring of integers in some number field iff it is finitely generated and torsion-free as an abelian group. These conditions imply that$R$has characteristic$0$and its field of fractions is finite dimensional over$\mathbb{Q}$, i.e. a number field. Since$R$is finitely generated as a ... 3 I like the functor of points perspective. You want to think of affine space as$k^n$, or more generally$R^n$for a ring$R$. The "universal affine space"$\operatorname{Spec} k[x_1, ... , x_n]$, as a scheme over$k$, has the property that for any$k$-algebra$R$, there is a natural bijection of sets $$\operatorname{Hom}_{\textrm{k-schemes}}(\... 1 Addressing your questions in order: For knowing that the norm lives in \mathbb{Z}\cap P, note that it lives in P since one of its factors lives in P, and P is an ideal, and lives in Z since its an algebraic integer, that lives in the image of the norm map, which is \mathbb{Q} (one way to see this is to view the norm via matrices, we are taking ... 0 I think the statement has a counterexample. Suppose A = \mathbb{F}_2 and B=A^3. Then C = B, which cannot be expressed as a quotient of A[X] as A[X] only has two A-valued points. If B is local, then the statement is true. In this case, k'(\subseteq B\otimes_A k) is a field because B\otimes_A k is local. k' is finite separable over k so ... 3 The association is (a_1, \dots, a_n) \to (x_1 - a_1, \dots, x_n - a_n). The subspace topology of maximal ideals then coincides with the Zariski topology on the points. For your second question, let$$A = k[x,y]/(x^3 - y^2 +xy) \; .$$By setting t = y/x we get t(t-1) = \frac{y^2 - yx}{x^2} = x, and t^2(t-1) = \frac{y^3 - y^2 x}{x^3} = y \frac{x^3}{x^3} ... 1 Lets first reduce to a simpler case, since B/A is finite, we can break it up into iterated extensions, each dominated by extensions of the type A'\subset A'[x]/\langle f(x)\rangle, for f(x) monic of finite degree with coefficients in A':$$A\subset A[x_1] \subset A[x_1,x_2] \subset ... A[x_1,..x_n]=B$$So it suffices to prove the claim in this case, ... 4 No, it is not true. It is possible to construct several examples using numerical semigroups. A numerical semigroup S is simply a submonoid of \mathbb{N} such that \mathbb{N}\setminus S is finite. It is easy to see that every numerical semigroup has a unique minimal system of generators, which is finite. In this case I write S=\langle s_1, \dots, s_\mu\... 0 \operatorname{ht}(a_1,\cdots, a_s)=\min \operatorname{ht} \mathfrak{p} where \mathfrak{p} is a prime ideal containing (a_1,\cdots, a_s). If \operatorname{ht}(a_1,\cdots, a_s)<s, then there is a prime ideal \mathfrak{p} containing (a_1,\cdots, a_s) such that \operatorname{ht} \mathfrak{p}\leq s-1<s. Hence \mathfrak{p} is one of the \... 2 For any prime ideal \mathfrak{p} of R, the residue field at \mathfrak{p} is defined to be R_{\mathfrak{p}}/\mathfrak{p}R_{\mathfrak{p}}, which one can show is equivalently the quotient field Q(R/\mathfrak{p}) of the integral domain R/\mathfrak{p}. In the case where \mathfrak{p} is maximal, then, the residue field of R at \mathfrak{p} is ... 1 Let F be your free A-module, and let A^n = Ab_1 \oplus \ldots \oplus Ab_n (we could use F here, but this is confusing). Define a map f: A^n \to F by f(b_i) = e_i. By assumption, the e_i generate F so this map f is surjective. Since F is free, hence projective, there exists a map g: F \to A^n such that fg is the identity and A^n = F ... 3 A finitely generated as a \Bbb Z-algebra exactly means the structure morphism \operatorname{Spec} A\to\operatorname{Spec}\Bbb Z is of finite type by definition. Since \Bbb Z is regular, proposition 14.103 applies. To check this, simply verify that every localization of \Bbb Z at a prime ideal is a regular local ring: you just have to check \Bbb ... 0 Let M be a flat A-module, S\subset A a multiplicative set and N\to N' an injective homomorphism of A-modules. We would like to show that the induced homomorphism$$N\otimes_A S^{-1}M \to N' \otimes_A S^{-1}M$$is injective. Now, recall that$$S^{-1}M\cong S^{-1}A\otimes_A M.$$Hence$$N\otimes_A S^{-1}M\cong N\otimes_A M \otimes_A S^{-1}A\to N'\... 1 What we want to show is equivalent to showing that$A/I$only has finitely many prime ideals. Because$(0) \subsetneq I$and$(0)$is a prime ideal (as$A$is an integral domain), we have that$\mathrm{dim}(A/I) \leq 0$. Also, being a quotient of a Noetherian ring,$A/I$is Noetherian. Thus$A/I$is (at most) zero-dimensional and Noetherian, thus$A/I$is ... 1 Let$G = \{ x,y \}$and$G' = \{ x,y+1 \}$in$\mathbb{Q}[x,y]$with the lexicographic monomial ordering. The sets of generators$G$and$G'$are Groebner bases of$I=\langle G \rangle$and$I' = \langle G' \rangle$respectively. The product set$G \cdot G' = \{ x^2, xy+x, xy, y^2+y \}$is not a Groebner basis of the product ideal$I \cdot I'$. Indeed,$x = (...

1

The answer is no. For example, let $R=\mathbb{Z}[x]\big/\langle 2x,x^2\rangle$; for convenience denote the image of $x$ in $R$ as $c$. The maximal ideals of $R$ are the ideals of form $\langle p,c\rangle$ for some prime $p\in\mathbb{Z}$, so the Jacobson radical of $R$ is $\langle c\rangle$ and hence $R\big/J\cong\mathbb{Z}$. In particular, if $A$ is the $1\... 3 Yes, you can deduce the more general form from the Jacobson radical form. Here's a proof; let$S=1+I=\{1+a:a\in I\}$.$S$is a multiplicatively closed set, so consider the localizations$S^{-1}A$and$S^{-1}M$. First I claim that the ideal$S^{-1}I$is contained in the Jacobson radical of$S^{-1}A$; to see this, fix any maximal ideal$P$of$S^{-1}A$. If we ... 2 In your notation,$g_1Y^2\in(X,Z,X-YZ)=(X,Z)$, so$g_1\in(X,Z)$. Then$a\in(XY^2,Y^2Z,X-YZ)\subseteq(XY,X-YZ)$. 4 There is only one:$0 + T\mathbb F_4[[T]]$. Obviously this is a maximal ideal, as the quotient is$\mathbb F_2$, a field. Now it's only necessary to show that$1+T\mathbb F_4[[T]]$are all invertible, therefore they cannot be contained in any maximal ideal. This is essentially to compute the power series expansion$\dfrac{1}{1 + a_1T + a_2T^2 + a_3T^3+\cdots}...

2

Since $I$ is generated by finitely many elements $a_{1},\dots,a_{n}$, we have $(0:_{M}I)=\bigcap\limits_{i=1}^{n}(0:_{M}a_{i})$ and $(0:_{R}I)=\bigcap\limits_{i=1}^{n}(0:_{R}a_{i})$, thus we can reduce the problem to the case that $I=(a)$ is a principal ideal. We have a short exact sequence $$0\longrightarrow (0:_{R}a)\longrightarrow R\overset{x\mapsto xa}{\... 3 If a\in (X)\cap (X^2, Y), then a=fX and a=g_1X^2+g_2Y. Now equate$$fX=g_1X^2+g_2Y.$$So X(f-g_1X)=g_2Y. Since k[X, Y] is a UFD, X|g_2, say g_2=g_3X. So a=g_1X^2+g_2Y=g_1X^2+g_3XY\in (X^2, XY). 0 If by \mathrm{codim}~I you mean \operatorname{ht}I, then yes. More general, if R is a commutative noetherian ring and f_1,\dots,f_r is a regular sequence, then \operatorname{ht}(f_1,\dots,f_r)=r. One ca prove this by induction on r. The main step (which I leave you as an exercise) is to show that \operatorname{ht}\mathfrak p/(f)=\operatorname{ht}... 3 The two rings are not isomorphic. An easy way to see this is to note that both rings have a unique maximal ideal generated by (the residue classes of) X and Y. Then consider for both rings the ideal of elements that are annihilated by all elements of the maximal ideal. For the first ring it is a 2-dimensional vector space generated by X^2 and Y ... 2 It is not necessarily true, simply because we do not always have I\subset J^n. What can be said is that$$\bigl(J/I\bigr)^n=\bigl(J^n+I\bigr)/I\simeq J^n/\bigl(J^n\cap I\bigr).$$1 If so then any A-module is flat if and only if it is projective if and only if it is free. See (3.G) in Commutative Algebra by Hideyuki Matsumura. 2 Here is a counter-example. Consider the blow-up$$X = \operatorname{Bl}_0 \mathbb A^2 \to \mathbb A^2, and let $E \subset X$ be the exeptional curve. According to [1, Exc. II.4.12 (b.2)], the local ring $R_v = \mathcal O_{X, \varepsilon}$ at the generic point $\varepsilon \in E$ is a valuation ring which dominates the local ring $R = \mathcal O_{\mathbb A^... 1 This follows from the fact that the nilradical of a commutative ring coincides with the intersection of all its prime ideals. Thus if$x$is in every prime ideal, then$x$is nilpotent. A proof is available on this wiki page. The basic idea is to use Zorn's lemma, more or less in the same way of showing the existence of maximal ideals. As an alternative ... 0 First let me mention a helpful characterization of codimension 0 (= minimal) primes of a ring: A prime$P$is minimal iff for each$r \in P$there exists$s \notin P$such that$sr$is nilpotent. This is a good exercise which I leave to you. First we check that a reduced Noetherian ring has$(1)$and$ (2)$.$(1)\$ is straightforward because for a reduced ring ...

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