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4

Usually you would say that a one-dimensional noetherian UFD is a Dedekind domain and for Dedekind domains UFD and PID is the same thing. Let us recap the proof on an elementary level: First of all we show that every prime ideal is principal: Let $0 \neq \mathfrak p$ be a prime ideal and $0 \neq f \in \mathfrak p$. Since we have an UFD, we can factorize $$f ...


3

In representation theory people often use the word representation as a synonym for module, and for good reason. A representation of a Lie algebra is the same thing as a module for its universal enveloping algebra, a representation of a quiver is the same as a module over its path algebra, etc. One important example is that a representation of a finite ...


3

For a given group $G$ and field $F$, the category of group representations of $G$ (in the sense of group homomorphisms from $G$ into $GL(n,F)$) is equivalent to the category of finite dimensional modules over the group ring $F[G]$. Similarly, the category of $F$-algebra representations of an algebra $A$ (in the sense of ring homomorphisms from $A$ into ...


3

Hint $\ $ The overring is uniquely determined by the set of elements inverted, which is uniquely determined by the set of primes inverted. This yields a bijection between such overrings and subsets of primes of $A$. In further detail, the set $\,S\subset A\,$ of elements that become units (invertibles) in the overring is a saturated monoid $\,S,\,$ i.e. ...


2

A counterexample is $\mathbb{Q}$ considered as a $\mathbb{Z}$-module. Definitely torsion-free but not free.


2

You're starting from the wrong side. Let $f\colon M\to N$ be a homomorphism of $A$-modules. You can compose it with the canonical mapping $j\colon N\to \hat{N}$ and show it's continuous when $M$ is endowed with the $\mathfrak{a}$-adic topology. Indeed, if $U$ is an open neighborhood of $0$ in $\hat{N}$, we have that, for some $n$, ...


2

Start with $$|R/\mathfrak{m}^k|=|R/\mathfrak{m}|\cdot|\mathfrak{m}/\mathfrak{m}^2|\cdots|\mathfrak{m}^{k-1}/\mathfrak{m}^k|$$ and use that $\mathfrak m^{i-1}/\mathfrak m^i$ is an $R/\mathfrak m$-vector space, so isomorphic to a direct sum of copies of $R/\mathfrak m$.


2

The statement is incorrect as attested by the case $n=3, d=2$ and the linearly independent polynomials $$f_1=x_1^2,\quad f_2=x_1x_2,\quad f_3=x_2^2 $$ The jacobian determinant is identically zero. Indeed its third column is zero, because the $f_i$'s do not not depend on $x_3$.


2

It depends on the ring. For instance, if $R=\mathbb{Z}_{(p)}$ is the localization of $\mathbb{Z}$ at the prime ideal $p$, then the maximal ideal is principal, so isomorphic as a module to the ring itself, hence projective. In case the ring is $\mathbb{Z}/4\mathbb{Z}$, the maximal ideal is not projective. Added from comment. If your aim is to discuss ...


2

Over a local commutative ring, projective modules coincide with free modules, so the question is whether $\mathfrak m$ is free. If it is free it must be of rank one, because two elements of $a,b\in A$ are necessarily $A$-linearly dependent: $a\cdot b-b\cdot a=0$ (duh!) Freeness of dimension one means that for some $m\in \mathfrak m$ the $A$-linear map ...


2

You must keep in mind Proposition 10.12: If $$0 \to M' \to M \to M'' \to 0$$ is an exact sequence of $A$ modules and $\hat{M}', \hat{M}, \hat{M}''$ are their ${\mathfrak a}$-adic completions then $$0 \to \hat{M}' \to \hat{M} \to \hat{M}'' \to 0$$ is exact too. This gives you first $\widehat{A^n} = \hat{A}^n$ by considering the sequence $$0 \to A^{n-1} ...


1

Let $R=K[x_1, \ldots, x_{n-1}]$, and $\mathfrak p$ a maximal ideal of $R$. Then $(R/\mathfrak p)[x_n]\simeq R[x_n]/\mathfrak p[x_n]$. In $R[x_n]/\mathfrak p[x_n]$ the ideal $\mathfrak m/\mathfrak p[x_n]$ is principal. That's all.


1

If u plug $im(d)$ in place of $M$ you will get splitting of $d$ (by universal propery of $im(d)$). More precisely $\Omega_{A/R} = Im(d)\oplus \Omega'$. Its easy to see that for nonzero $\Omega'$ this contradicts uniqness property.


1

By the construction of Kahler differential, we know that $\Omega_{A/R}$ can be seen as $F/E$, where $F$ is generated by symbols $da$,here $a\in A$, $E$ is generated by symbols $dr$($r\in R$) and $d(a_1+a_2)-da_1-da_2$ as well as $d(a_1a_2)-da_1 \cdot a_2-a_1\cdot da_2$ ($a_1,a_2\in A$). It's obvious that the map is surjective. Then ...


1

Let $S=K[X_1,\dots,X_{n-1}]$. Now consider the polynomial extension $S\subset S[X]$. The question becomes Why a maximal ideal of $S[X]$ lies over a maximal ideal of $S$? This holds in the more general frame of finitely generated algebras over a field, and a proof can be found here. (However, one can not extend the property too much since even for $S$ ...


1

Hint: Here's one with minimal set $3$: $\langle x^2,xy,y^2\rangle$.


1

There is a theorem of Issac which says that if every prime ideal is principal then so is every ideal.


1

Suppose $\mathrm{Ann}(r_2rm)\not\subseteq Q$. Then there is $a\in \mathrm{Ann}(r_2rm)$, $a\notin Q$. Now $\mathrm{Ann}(arm)\subseteq Q$: $b\in\mathrm{Ann}(arm)\Rightarrow b(arm)=0\Rightarrow (ba)rm=0\Rightarrow ba\in P\Rightarrow ba\in Q\Rightarrow b\in Q$. Since $P\subseteq\mathrm{Ann}(arm)$ and $P$ is maximal with some properties that $\mathrm{Ann}(arm)$ ...


1

$\newcommand{\Ass}{{\mathrm{Ass}}}$ I assume $R$ noetherian. We write $I=Q_1 \cap \cdots \cap Q_r$ as the representation of $I$ as an intersection of primary ideals. Then $(I:x) = (Q_1:x) \cap \cdots \cap (Q_r:x)$ as this is a general property of intersection of ideals. Now $(Q_i:x)$ is equal to $R$ if $x \in Q_i$ and $(Q_i:x) \subseteq \sqrt{Q_i} = P_i$ ...


1

Edit: I found an even easier way to explain it, so I have revised the answer. The definition of $\mathfrak{m}$-adic completion is always completion (i.e. convergent sequences, since this is a metric space) relative to the $\mathfrak{m}$-adic topology or, more simply, the $\mathfrak{m}$-adic metric on $V$. Recall that this topology is defined by having ...


1

Since CM rings which are regular in codimension one are normal, you should expect singularities in codimension one. So, just take the $R = k[x, y]/y^3 - x^2$ [cuspidal singularity]. This is a complete intersection, so CM. To see that the node is not normal, the fraction field is $k(t)$ where $x = t^3$ and $y = t^2$, and note that $t$ is in the integral ...


1

Let $S_n$ denote Serre's condition, that is, a finitely generated $R$-module $M$ is $S_n$ if $$\operatorname{depth} M_p \geq \min(n,\dim M_p)$$ for every $p\in \operatorname{Spec}R$ and let $R_n$ mean that $R_p$ is regular for every $p\in \operatorname{Spec}R$ of height at least $n$. Then $R$ is normal iff it is $S_2$ (as a module over itself) and $R_1$ and ...


1

$f_i$ must be functionally independent.


1

It might be easier to approach this problem the following way: We have the following free resolution of $I$: $$0 \to R \to R \oplus R \to I \to 0,$$ where the first map is given by $1 \mapsto (-y,x)$ and the second map is given by $(1,0) \mapsto x, (0,1) \mapsto y$. After tensoring with $R/I$ we get the following exact sequence: $$0 \to Tor_1(I,R/I) \to ...


1

I'll assume the isomorphism is as $R$-modules (or the result is false, see Can $R \times R$ be isomorphic to $R$ as rings?). Suppose $R/A\oplus R/B\cong R/AB$, in particular is cyclic, and that $(x+A,y+B)$ is a generator. Then, for some $r\in R$, $(1+A,0+B)=r(x+A,y+B)=(rx+A,ry+B)$, so $$ 1=rx+a,\qquad ry\in B $$ for some $a\in A$. Then $y=y1=ay+rxy\in A+B$. ...


1

Let $R=K[X,XY,XY^2,\dots, XY^n,\dots]=K+XK[X,Y]$, and $I=(X)$. Let $P$ be a prime ideal containing $X$. Then we have $(XY^m)^2=X(XY^{2m})\in P$, thus $(X,XY,XY^2,\dots, XY^n,\dots)\subseteq P$. But $(X,XY,XY^2,\dots, XY^n,\dots)$ is maximal, so $(X,XY,XY^2,\dots, XY^n,\dots)=P$. We have $(0)\subsetneq P'\subsetneq P$, where $P'=(XY,XY^2,\dots, ...


1

I will only try amd answer question 1 above. For the sake of convenience, I will assume that $A,B$ are integral domains, finite type over an algebraically closed field $k$. Furthermore, we are given an inclusion of rings $f : A \hookrightarrow B$ such that $B$ is module finite over $A$. Definition 1 (Standard definition): Let $\mathfrak{m}$ be a ...


1

Let $m$ be the maximal ideal of $R$. We have $A\otimes_Rk\simeq A/mA$. You want to show that every maximal ideal of $A$ contains $mA$, or equivalently that every maximal ideal of $A$ contracts to $m$. Let $M$ be a maximal ideal of $A$. Then $M\cap R$ is a prime ideal of $R$, and the extension $R/M\cap R\subset A/M$ is finite. Since $A/M$ is a field we get ...


1

You have a map $\phi:S^3\to S$ given by $\phi(e_1)=x^2$ and so on. (In this case $e_1=(1, 0, 0)$ and so on.) The matrix of this map is (as in the vector spaces case) the following: $(\phi(e_1)\ \phi(e_2)\ \phi(e_3))$, so your guess is correct.



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