Let $A$ be a Noetherian local ring such that your Krull dimension is greater than 1, show that $Spec(A)$ has an infinity many points. I have been trying with the following idea; I take a chain of ideals in it, there must be a maximal $Max_A$ and a minimal $Min_A$, and ideals between them, I suppose there is a finite number of these, but I still can't arrive the conclusion. I would like to get some recommendation or other way for the proof. Thanks for all.
123849
If $R$ is a Noetherian commutative ring with unity having finitely many height one prime ideals, one could derive from the "Principal Ideal Theorem", due to Krull, that $R$ has finitely many prime ideals (all of height less than or equal to $1$). It may comes to mind that, in general,      if $R$ is a commutative Noetherian ring with unity of finite Krull dimension $d$ has finitely many prime ideals of height $n$ with $0&lt;n≤d$, then $n=d$.    My question: is this generalization valid?   I shall appreciate any help.
47818
Let $k$ be an algebraically closed field. Nullstellensatz states that the maximal ideals of the polynomial ring $R=k[X_1,\dots,X_n]$ are precisely those of the form $\langle X_1-a_1,\dots,X_n-a_n\rangle$, with $(a_1,\dots,a_n)\in k^n$.     What if we are working with infinitely many variables, say $\{X_i\}_{i\in I}$, being $I$ an infinite set? Again, the ideals of the form $\bigl\langle\{X_i-a_i: i\in I\}\bigr\rangle$, with $(a_i)_{i\in I}\subseteq k$ are maximal (even if $k$ is not algebraically closed), but what about the converse?
82517
I would like to know, why $ \mathfrak{p} A_{\mathfrak{p}} $ is the maximal ideal of the local ring $ A_{\mathfrak{p}} $, where $ \mathfrak{p} $ is a prime ideal of $ A $ and $ A_{\mathfrak{p}} $ is the localization of the ring $ A $ with respect to the multiplicative set $ S = A -\mathfrak{p} $ ? Thanks a lot.  N.B. : I have to tell you that I'm not very good at Algebra, so please, be more kind and generous in your explanation, and give me a lot of details about this subject please. Thank you.
13664
Let $S$ be the ring of all continuous functions from $[0,1]$ to $\mathbb R$. How to prove that all maximal ideals of $S$ have the form $M_{x_0}=\{f\in S \mid f(x_0)=0\}$?   Thanks in advance.
77286
If for a ring $A$ every localization $A_\mathfrak{p}$ by a prime $\mathfrak{p}\subseteq A$ is noetherian, is it true that $A$ is noetherian?   I believe not but I can't find a good counterexample.
98625
Let $K$ be an algebraic extension field (not necessarily finite) of $\mathbb{Q}$. Let $\mathscr{O}_K$ be the integral closure of $\mathbb{Z}$ in $K$.  Then, is $\mathscr{O}_K$ Noetherian?  If $K$ is a finite extension of $\mathbb{Q}$, then $\mathscr{O}_K$ can be shown to be Noetherian. However, if we do not assume that finiteness condition, is it still true?  (It can be shown that $\mathscr{O}_K$ is an integrally closed domain of Krull-dimension $1$. Hence, if it is Noetherian, it is a Dedekind domain.)
110838
Let A be a commutative ring with unity ring, M be a maximal ideal and $b\notin M$ . Show that there exists $q \in M, s\in A$ such that for every $a \in A$ we have $a=sba+qa$. Show that this is true only when $M$ is a maximal ideal.   Suppose if we take $A=2\mathbb Z$ and $M=4\mathbb Z$, $a=6.$ Then by the given statement $\exists$ $s\in A$, $q\in M$ s.t. $6=6(2s+q)$ $\implies$$s=\frac{1-q}{2}\implies q \text { is odd}$. But $q \in 4\mathbb Z.$  Please someone give some hints. how can solve this.  Thank you..
41218
I know $\mathcal C([0, 1])$ has all maximal ideals of the form $M_p=\{f\in \mathcal C([0, 1]) : f(p) =0,\  p\in [0, 1] \}$. My question is little bit different.  If I replace $[0, 1]$ by $(0, 1)$  then how can I show that this ring has infinitely many prime ideals not of the form $M_p$ ?
27682
Let $R$ be a commutative domain.   Prove that the Jacobson radical of $R[X]$, i.e. the intersection of all maximal ideals, is the zero ideal.  Thank you.
45456
Let $R$ be a ring and let $I_1,\dots,I_n,J\subseteq R$ be ideals such that $J\subseteq I_1\cup\dots\cup I_n$. Suppose that there exists an infinite field $K$ and a homomorphism $\phi:K\to R$. Prove that $\exists k$ s.t. $J\subseteq I_k$.   Attempt: (Edited)  If $J\subseteq I_1$ then we're done. Otherwise, let $x\in I_1$ and let $y\in J\setminus I_1$. $\forall r\in K, x+ry\notin I_1$ (because otherwise we'll get $y\in I_1$).  $K$ is infinite so we have infinitely many elements of the form $x+ry\in J$. So there's some $k$ for which $I_k$ contains infintely many elements of the former form.  If $n=2$ so $x+y\notin I_1$ implies $$x+y\in I_2\Rightarrow x\in I_1\Rightarrow I_1\subset I_2\Rightarrow J\subseteq I_2$$  Suppose the theorem holds for some $n$. $x+y\notin I_1\Rightarrow \exists k,I_k\ni x+ry$ for infintely many elements as above.
51045
Can someone give a hint or a solution for showing that a scheme has Krull dimension $d$ if and only if there is an affine open cover of the scheme such that the Krull dimension of each affine scheme is at most $d$, with equality for at least one affine scheme in the cover?
123479
Proposition. Let $R$ be commutative ring with $1_R$. We assume that $R$ is an Artinian ring and $M_1,\dots,M_n$ its maximal ideals. Then         $R/\mathrm{Jac}(R)\cong (R/M_1)\times \dotsb \times (R/M_n)$.   The ring $R$ is isomorphic to the direct product of a finite number of Artinian local rings.      Proof. 1. It's obvious that $M_i+M_j=R,\ \forall 1\leq i \neq j \leq n$. So, from Chinese Remainder Theorem, we have  $$R/\bigcap_{i=1}^{n}M_i = R/\mathrm{Jac}(R) \cong (R/M_1)\times \dotsb \times (R/M_n),$$ as we wanted.    Since $R$ is an Artinian ring, we have $\mathrm{Jac}(R)^m=\{0_R\}$, for some $m\in \Bbb Z^+$. But,  $$\{0_R\}\subseteq M_1^m\dotsb M_n^m=(M_1\dotsb M_n)^m\subseteq (\bigcap_{i=1}^{n}M_i=\mathrm{Jac}(R))^m=\{0_R\}.$$ So, if we apply CRT we will take  $$R\cong R/\{0_R\}\cong R/M_1^m\dotsb M_n^m\cong (R/M_1^m)\times \dots \times (R/M_n^m).$$   Questions.   1) Are these thoughts complete and correct?  2) Why $R/M_i^m$ are artinian local rings?  3) Could you please elaborate some examples as an application?
68593
If $I$ is a maximal ideal of $R$, why is $R/I$ a field?  I'm trying to use the fact that $I$ is maximal to show that $R/I$ only have ideals $\{0\}$ and $R/I$. Can anyone help me with this method. Many thanks.
36865
How to prove a commutative, with unit, Noetherian ring $A$ only has finitely many minimal prime ideals via the following step?   I have proved:     Step. All radical ideals of Noetherian ring $A$ can be expressed as an intersection of finitely many prime ideals.   How to continue?
65174
Let $g$ be meromorphic on a region $R$ of $\mathbb{C}$, where $R$ is all but finitely many points of $\mathbb{C}$---that is, $R$ = $\mathbb{C}\setminus S$ for some finite set $S$. Suppose $\lim_{z \rightarrow s} |g(z)| = \infty$ for all $s \in S$ and $g$ takes infinity to infinity. Then, somehow $g$ is rational.  Any help would be appreciated.
121070
I want to understand why the following is true:     Let $S \subseteq R$ be commutative rings with $1$ and assume that $R$ is finitely generated as an $S$-module by at most $k$ elements. For every maximal ideal $M$ of $S$ there are at most $k$ maximal ideals of $R$ lying over $M$.   So let $T$ be a maximal ideal of $R$ lying over $M$ then $T \cap S = M$. Now it can be shown that $R/T$ is a finitely generated $S/M$-module. Hence $R/T$ is Artinian and thus it has finitely many maximal ideals. From here I have two questions:  1) Why from this follows that there can be only finitely many $T$ lying over $M$? I suspect they are using some correspondence which I don't see.  2) Why exactly at most $k$ ?  Can you please explain?
101021
As the title suggests, I am asked to prove that, given a Noetherian scheme $(X,\ \mathcal{O}_{X})$ and any open subset $U\subseteq X$, $\Gamma(U,\ \mathcal{O}_{X}):=\mathcal{O}_{X}(U)$ is a Noetherian ring.  Up to now, I have been able to show that the result is true if $U$ is an affine open subset of $X$, i.e. $U\simeq\mathrm{Spec}(A)$ for some ring $A$ (and this is actually true when $X$ is just locally Noetherian). I have also shown that, given $U$ as above, $(U, \mathcal{O}_{X\vert U})$ is a Noetherian scheme as well, which should then allow me to reduce the problem to the case $U=X$. So, when all is said and done, I should try to prove that the ring $\mathcal{O}_{X}(X)$ is Noetherian. However, I can not go any further and I am stuck here.  Any help or suggestion would be grately appreciated.  Thank you.