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Equality of two completions

I have the following question. Suppose $R$ is Noetherian ring, $I$ is ideal in $R$ and $S$ is multiplicatively closed set. Let $(I^n\colon\langle S\rangle) = \varphi^{-1}(I^nS^{-1}R),$ where $\varphi\...
abcd1234's user avatar
1 vote
1 answer
26 views

Localization: $(x) R_{\mathfrak{m}}=R_{\mathfrak{m}}=\mathfrak{a} R_{\mathfrak{m}}$ for $x\in \mathfrak{a}$ but $x\notin \mathfrak{m}$

I have a commutative ring $R.$ Let $\mathfrak{a}$ be a non-zero ideal in this ring and $\mathfrak{m}$ be a maximal ideal. Also, let $x$ be a non-zero element of $\mathfrak{a}$ such that $x\notin \...
Haldot's user avatar
  • 830
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167 views

Is testing whether the nonunits of $K[x,y]_{(x,y)}[1/y]$ form an ideal a messy undertaking or am I going about it wrong?

I'm studying the ideals of $R = K[x,y]_{(x,y)}[1/y]$: the polynomials in $1/y$ with coefficients in the localization of $K[x,y]$ at the ideal $(x,y)$. I've managed to identify that its units are given ...
shintuku's user avatar
0 votes
1 answer
118 views

Intuitively, why $\operatorname{spec}(S^{-1}A) \cong \lbrace \mathfrak{p} \in \operatorname{spec} (A)| \mathfrak{p} \cap S = \emptyset \rbrace$?

If $A$ is a commutative ring, $S$ is a multiplicative subset of $A$, I’d like to understand intuitively why is there a bijection between $\operatorname{spec}(S^{-1}A)$ and $\lbrace \mathfrak{p} \in \...
dahemar's user avatar
  • 1,788
0 votes
1 answer
106 views

Extension and contraction of prime ideals by ring homomorphism

Let $A$ and $B$ be commutative rings with $1 \neq 0$. Let $\varphi$ be a ring homomorphism with $\varphi(1) = 1$. We consider the extension and contraction of $\varphi$. Let $P \subset A$ be a prime ...
Kazune Takahashi's user avatar
1 vote
1 answer
110 views

Questions of (9.3) from Atiyah's Introduction to Commutative Algebra

I don't understand ii)$\Longleftrightarrow$iii), but I know (9.2),(4.8),and(3.11). Could someone give me the complete proof? DEDEKIND DOMAINS Theorem 9.3. Let A be a Noetherian domain of dimension one....
Elizabeth Amanda's user avatar
1 vote
1 answer
95 views

If $I$, $J$ are ideals, $S^{-1}(I \cap J)=S^{-1}I \cap S^{-1}J$

Let $I$, $J$ are ideals of a commutative ring $R$, $S$ a multiplicative subset of $R$. I want to show that $S^{-1}(I \cap J)=S^{-1}I \cap S^{-1}J$. This is my attempt: ⊆ : A typical element of $S^{-1}(...
tjdominic's user avatar
4 votes
1 answer
97 views

Why is for an algebraically closed field $k$, $\mathbb{A}^1_k \setminus \{0\} \simeq \operatorname{Spec} k[x, x^{-1}]$

Let $k$ be an algebraically closed field and $\mathbb{A}^1_k:=\operatorname{Spec}(k[x])$ be the affine 1-space. Why is then $\mathbb{A}^1_k \setminus \{0\} \simeq \operatorname{Spec}(k[x,x^{-1}])$ ? ...
Paul Joh's user avatar
  • 559
1 vote
0 answers
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Correspondence of heights in localization

It is a well know fact that given a ring $A$ setting $S := A\setminus P$, where $P$ is a prime ideal in $A$, there is a corrispondence between the prime ideals of $A_P$ (this is the notation I use for ...
jacopoburelli's user avatar
0 votes
1 answer
18 views

Prove $U^{-1}R \cong R[\{X_s\}_{s \in U}]/(sX_s - 1).$

Let $I = \langle \{sX_s - 1 : s \in U\}\rangle.$ I'm looking for an alternate proof of $U^{-1}R \cong R[\{X_s\}_{s \in U}]/I.$ The proof I found is sending $r/u \to rX_u,$ proving this is a well-...
Display name's user avatar
  • 5,230
0 votes
1 answer
55 views

Localization an ideal by maximal ideal [duplicate]

I try to show this, suppose $R$ be a commutative ring with $1$ and $A$ be an ideal in $R$. Show if $A_{M}=0$ for all maximal ideal $M$, then $A=0$. My idea this, If $a\in A$ then for every maximal ...
Fati's user avatar
  • 71
1 vote
1 answer
136 views

Localization of $\mathbb{Z}$ at the powers of $15$

I am trying to determine the maximal ideals of $\mathbb{Z}[1/15] =\big\{\frac{a}{15^n}\;|\; a \in \mathbb{Z},\; n\geq 0 \big\}$. This is the integers localized at the powers of $15$. My attempt was to ...
slowspider's user avatar
  • 1,065
0 votes
1 answer
26 views

Why this follows for ideals in localisation?

Let $R$ be a PID. Let $I$ be an ideal in $S^{-1}R$(the localisation). Then there exists an ideal $J$ in $R$ so that $JS^{-1}R=I$ I am sure I might be missing something but I really can't seem to wrap ...
Annalisa's user avatar
  • 911
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0 answers
155 views

showing a ring is isomorphic to a localization of $R$ at $S$

Let $\mathbb{K}$ be a field. The domain of $f = \frac{g}h\in \mathbb{K}(x), $ where $g,h\in \mathbb{K}[x]$ is defined to be the set $Dom(f) := \{c\in \mathbb{K} : h(c) \neq 0\}.$ Let $S(c) := \{f\in \...
Alfred's user avatar
  • 869
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0 answers
153 views

Ideals in localisation of graded ring

Let $k$ be a field and $I$ a homogeneous ideal of $k[x_1,...,x_n]$. Consider the graded ring $R=k[x_1,...,x_n]/I$ with unique homogeneous maximal ideal $\mathfrak m=(x_1,...,x_n)R.$ Let $J$ be a ...
uno's user avatar
  • 1,560

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