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6

Hint: Given $a\in R$, consider the chain of ideals $a^n R$.


5

$k[x,y]$ just refers to a $k$-algebra generated by $2$ elements $x, y$, which need not be algebraically independent over $k$ (as $X, Y$ are). Indeed, $x, y$ are the images $X + (F), Y + (F)$ in the quotient ring $k[X,Y]/(F)$.


5

(2) Yes, if $\phi$ is surjective, it is injective. Indeed, the dual morphism $f=\phi^*:\mathbb{A}^n_k\to \mathbb{A}^n_k$ is a closed embedding and since $f(\mathbb{A}^n_k)\subset \mathbb{A}^n_k$ are irreducible of dimension $n$ they are equal and $f$ is an isomorphism (since $\mathbb{A}^n_k$ is reduced ). Thus $\phi$ is an isomorphism too and, in ...


4

Suppose that $F$ was birational to $\mathbb{P}^1$, then because they are both smooth curves over $\mathbb{C}$, we'd have that $F\cong \mathbb{P}^1$. But, note that every affine open of $\mathbb{P}^1$ is a localization of $k[x]$, and so a UFD. That said, $\mathbb{C}[x,y]/(x^3+y^3+1)$ is an affine open of $F$, which is not a UFD. PS: You can show, in fact, ...


4

Two more ways: The differential $X^2dX + Y^2dY = -Z^2dZ$ is holomorphic and nowhere vanishing on your curve. But $\mathbf P^1$ has no such differential. (This one is silly.) Remark that permutations of $X,Y,Z$ or multiplication of either one of $X,Y$ and $Z$ by a cube root of unity preserve the equation. Hence there is a finite group of order $6\times ...


4

If $A$ is noetherian and $B$ is not, then $A$ is a noetherian $A$-module, but $A\otimes_A B \cong B$ is not a noetherian $B$-module.


3

Let me help you see the connection to Artinian rings. A commutative ring with finitely many ideals is Artinian, so if you prove it for Artinian rings, then you've proven it for rings with finitely many ideals. Before you, people have asked this question for finite rings rather than for rings with finitely many ideals. Some of the solutions will work in your ...


3

For your second question: there exist connected acyclic spaces (i.e. with vanishing cohomology in dimension $>0$) which are not contractible (i.e. not homotopy equivalent to a point). For instance, the Poincaré sphere with a point deleted from it is acyclic, but it is not contractible because it has nonvanishing $\pi_1$. (The $\pi_1$ of the non-punctured ...


3

Consider the ring $A = k[x,y]/(x^2, xy) = k[x,y]/\left((x)\cap(x,y)^2\right)$. Geometrically this is the line $x=0$ in the plane, with an embedded point at the origin. It is regular at every maximal ideal other than $\mathfrak m = (x,y)\subseteq A$, hence is CM away from this ideal a fortiori. However, at the origin, the depth is $0$. To see this, let ...


3

These fields are precisely those which can be generated by a single element over their prime field. In fact, $Q(\mathbb{Z}[x]/\mathfrak{p})$ is generated by the class of $x$, and conversely if a field $F$ can be generated by a single element $a$ over its prime field $P$, then let $\mathfrak{p}$ be the kernel of $\mathbb{Z} [x] \to F, x \mapsto a$ and observe ...


2

$f$ being dominant implies that the corresponding ring homomorphism is injective. Thus we have an inclusion of integral domains over $k$, $A\hookrightarrow B$ and the function fields in question are the fraction fields of these domains. Obviously, we have $frac(A)\hookrightarrow frac(B)$. Assuming that $A,B$ are finitely generated as algebras over $k$, we ...


2

This is not true. For example, $\{1\}$ and $\{2,3\}$ are minimal generating sets of the $\mathbb{Z}$-module $\mathbb{Z}$.


2

Let $R$ be the subalgebra of $K^{\mathbb N}$ (a countable direct product of copies of a field $K$; for simplicity one can choose $K=\mathbb Z/2\mathbb Z$) generated by $1$ and $K^{(\mathbb N)}$ (a countable direct sum of copies of the field $K$), that is, $R$ consists of all sequences $(a_n)_{n\ge 0}$ such that $a_n\in K$ and which are constant from some $n$ ...


1

It is actually more general. Whenever $M$ is an $R$ module, and $S\subset R$ a multiplicative subset, $S^{-1}M$ is an $S^{-1}R$ module. One defines the $S^{-1}R$ action in the only possible manner, $$\frac{r}{s}\cdot\frac{m}{t}=\frac{rm}{st}.$$ Now, if $\{m_\alpha\}$ generate $M$ over $R$, then their images in $S^{-1}M$ generate it over $S^{-1}R.$


1

I assume (as rschwieb points out) that the question concerns the category of $\mathbb{Z}$-modules. 1) $\mathbb{Z}_{p^\infty}$ is the injective hull in the case when $n=p^k$ as well. Basically because we have $\mathbb{Z}_p\subseteq \mathbb{Z}_{p^k} \subseteq \mathbb{Z}_{p^\infty} $ and the extension $\mathbb{Z}_{p} \subseteq \mathbb{Z}_{p^\infty}$ is ...


1

The same idea for $\mathbb{Z}$ extends to a polynomial ring over a field: for the ideal $(x) \subseteq k[x]$, $\{x\}$ and $\{x^2, x + x^2\}$ are both minimal generating sets. In general, given a generating set for an ideal $I$, say $I = (a_1, \ldots, a_n)$, one cannot conclude that $I$ can be generated by a proper subset of the $a_i$, even if $I$ is known ...


1

Using your notation: we may assume from the outset that $m = n$ (if $n > m$, $\dfrac{a}{f^m} = \dfrac{af^{n-m}}{f^n}$). The condition $\dfrac{ag^n}{(fg)^n} = \dfrac{bf^n}{(fg)^n}$ in $A_{fg}$ means that some power of $fg$ annihilates $ag^n - bf^n$ in $A$, say $(fg)^N(ag^n - bf^n) = 0$, i.e. $g^{N+n}af^N = f^{N+n}bg^N$. Choose $c, d$ such that $1 = ...


1

$\Bbb Z$ is a free, hence projective, hence flat $\Bbb Z$ module. $\Bbb Z$ modules are flat iff they are torsion-free, and $\Bbb Q_\Bbb Z$ is torsion-free. (And this is another reason $\Bbb Z_\Bbb Z$ is flat.) $\Bbb Z$ modules are injective iff they are divisible, so to produce a nonflat, noninjective module, it suffices to think of a nontorsion-free module ...


1

If $P$ is a prime ideal then the residue ring $k[x_1,...,x_n]$ is an integral domain and naturally a $k$-algebra (quotients of $k$-algebras by ideals are also $k$-algebraS). So if we assume the residue ring is finitely generated, then we have the hypotheses of the second so the residue ring is a field. It is then a classical result of algebra that in a ring ...


1

The proposed "closed" subsets $D(\mathfrak a)$ consisting of the prime ideals not containing an ideal $\mathfrak a$ do not in general form a topology, because an arbitrary intersection of such subsets is no longer of the required form. Example: Take $R=\mathbb Z$ and consider the "closed" subsets $D_p=D(p\mathbb Z)\subset \text {Spec}(\mathbb Z)$ for ...


1

Take $S=k[x,y],\quad R=k[x,y,z]/(yz-x)$. Take $I=(x),J=(y)$. Note that $z\in(I^e:J^e)\setminus(I:J)^e$. Or an alternative formulation of the same counterexample: Take $k[x,y]\subset k[y,\frac{x}{y}]$, with the same ideals - $I=(x), J=(y)$, and observe $x/y$.


1

Let $f:R\to\operatorname{Hom}_K(R,K)$ be an $R$-module monomorphism. This is also a $K$-vector space monomorphism. Since $R$ is a finitely dimensional $K$-vector space, $\dim_KR=\dim_KR^*$, and thus every monomorphism of $K$-vector spaces $R\to R^*$ must be an isomorphism.


1

Yes. If $S=R/K$, then $I=I'/K$ and $J=J'/K$. If $I', J'$ are not coprime, then there is a prime ideal $P$ such that $I'+J'\subseteq P$, and thus $I+J\subseteq P/K$. (Note that $P\supseteq K$ since $I',J'\supseteq K$.)


1

Write $1=i+j$ with $i \in I, j \in J$. Find preimages $u,v$ in $f^{-1}(I)$ resp. $f^{-1}(J)$. Then $1 \equiv u+v \bmod \ker(f)$. Since $\ker(f) \subseteq f^{-1}(I)$, it follows that $1 \in f^{-1}(I) + f^{-1}(J)$. (Notice that the proof by user26857 uses the existence of enough prime ideals, which is independent from ZF.)


1

Recall that, for $S$ a commutative ring with $1$, an element in $S[[x]]$ is a unit if and only if its constant term is a unit (see the wikipedia page for instance). Since $$R:=k[[T_1,\dots,T_n]]=k[[T_1,\dots,T_{n-1}]][[T_n]],$$ by induction an element in $R$ is a unit if and only if it has nonzero constant term, and $R$ is a local ring with maximal ideal ...


1

The definite source is Maximal Orders by I. Reiner. My hopefully mostly accurate recollections below. If $G$ is cyclic of order $n$, then $\Bbb{Q}[G]\cong \Bbb{Q}[x]/\langle x^n-1\rangle$. The polynomial $x^n-1=\prod_{d\mid n}\Phi_n(x)$ is a product of pairwise distinct irreducible cyclotomic polynomials $\Phi_n(x)$. By the Chinese remainder theorem we thus ...


1

Given any ascending sequence of sets $X_1\subset X_2\subset X_3\cdots\subset X$ with $\bigcup_{n\ge1}X_n=X$ and any collection of functions $g_i:X_i\to Y$ such that $g_i|_{X_{i-1}}=g_{i-1}$, we can form the map $g:X\to Y$ defined by the relation $g(x)=g_i(x)$ whenever $x\in X_i$. Each function $g_i$ is a bigger and bigger "glimpse" of $g$. One can check ...


1

Let $N_n=\sum_{i=1}^{k}c_iR$; then $D$ is the union of the increasing family of submodules $N_n$ and as such it is its direct limit with inclusions as transition maps. If you consider the restriction $h_n$ of $g_n$ to $N_n$, the given condition translates into the fact that $h_n\colon N_n\to B$ is a family of morphisms compatible with the inclusion maps, so ...


1

If $(R,m)$ is Noetherian local, then $m$-adic completion is exact. Moreover $(f)\widehat{R} \cong \widehat{(f)}$ for any $f \in m$. Thus the exact sequence $$0 \to (f) \to R \to R/(f) \to 0$$ induces an exact sequence $$0 \to f\widehat{R} \to \widehat{R} \to \widehat{R/(f)} \to 0$$


1

Any (!) collection of subsets $\mathcal{B}$ of a set $X$ provides a subbase for a topology on $X$. The open subsets are the unions of the finite intersections of sets in $\mathcal{B}$. The axioms for a topology are easy to check. In the exercise, we have $X=|A|$ (the underlying set of a commutative ring $A$) and $\mathcal{B} = \{\mathfrak{p} \subseteq X : ...



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