I am aware that questions related to the title of this post have already been posted but I think that the proof I'm looking for is a bit different. This proof uses the two following results which I was not able to prove either:  1) If $L$ is a finite algebraic extension of $\mathbb{Q}$ and $R \subseteq L$ a finitely generated $\mathbb{Z}$-algebra, there exist positive integers $n_1,...,n_k$ such that $R$ is integral over $\mathbb{Z}[\frac{1}{n_1},...,\frac{1}{n_k}]$.  2) If $\mathfrak{m}$ is a maximal ideal of $\mathbb{Z}[X_{1},...,X_k]$ it must contain a prime number $p\in \mathbb{Z}$.  And then deduce the theorem from the title.
13280
Suppose $k$ is a field that is finitely generated as a ${\mathbb Z}$-algebra.  (That is, $k$ is a quotient of ${\mathbb Z}[X_1,\dots,X_n]$ for some $n$). Does it follow that $k$ is finite?
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$$s=\frac{1-q}{2}\implies q \text { is odd}$. But $q \in 4\mathbb Z.$  Please someone give some hints. how can solve this.  Thank you..
87998
It is well-known that if $S$ is a graded ring, and $f$ is a homogeneous element of positive degree, then there is a bijection between the homogeneous prime ideals of the localization $S_f$ and the prime ideals of $S_{(f)}$, the subring of $S_f$ comprising the homogeneous elements of degree $0$.    This is proposition II.2.5b in Hartshorne, exercise 5.5B in Ravi Vakil's notes $[1]$ (p. $130$ of the February $24$, $2012$ version), and proposition 8.1.21 of Akhil Mathew's notes $[2]$ (p. $136$). Unfortunately I cannot follow any of those proofs to my own satisfaction, perhaps because I'm not well-versed in commutative algebra.  The crux of the proof appears to be to show that, given a homogeneous prime ideal $\mathfrak{p}$ of $S$ not containing $f$, the construction used to obtain a prime ideal $\Psi (\mathfrak{q})$ of $S$ from a prime ideal $\mathfrak{q}$ of $S_{(f)}$ will recover $\mathfrak{p}$ when $\mathfrak{q} = S_f  \mathfrak{p} \cap S_{(f)}$. To be precise, let $\Psi (\mathfrak{q})$ be the homogeneous ideal of $S$ generated by $$\bigcup_{d \in \mathbb{N}} \{ s \in S_d : s / f^d \in \mathfrak{q} \}$$ and let $\Phi (\mathfrak{p}) = S_f \mathfrak{p} \cap S_{(f)}$. It's easy to see that $\Phi \circ \Psi$ acts as the identity on $\operatorname{Spec} A$ (or, for that matter, the set of all ideals of $A$), but I cannot see any obvious reason why $\Psi \circ \Phi$ should act as the identity on the set of prime ideals of $S$ not containing $f$. A detailed proof of this point would be much appreciated.  References  $[1]$ .  $[2]$ Algebraic geometry notes (covers material at the level of the first and second volume of EGA): , and .
123230
I am looking at   and trying understand the proof for the theorem that states:  Given a nonzero ideal I in $P(F)$, there is a monic polynomial $p(t)$ such that   $$I=\left\{q(t)p(t) | q(t) \in P(F) \right\}$$  Proof goes like  Since $I$ contains at least one nonzero element, we can let $p(t)$ be a nonzero element of $I$ of minimum degree. We can show that every element of $I$ is a multiple of $p(t)$. Let $f(t) \in I$. Then there exist $q(t),r(t) \in P(F)$ such that  $$f(t) = q(t)p(t) + r(t)$$ and deg $r(t)$ &lt; deg $p(t)$. Since $p(t) \in I$, so is $q(t)p(t)$, and therefore so is $$r(t) = f(t) -q(t)p(t)$$.  By the choice of $p(t),r(t) = 0$, so $f(t)$ is a multiple of $p(t)$.  What I am having trouble understanding is why $r(t)$ must equal $0$. If someone can explain why it is, I'd appreciate it.
48295
Let $K$ be a field, and $a_1,\dots,a_n \in K$. Prove that the ideal $$(X_1-a_1,\dots,X_n-a_n)$$ is maximal in $K[X_1,\dots,X_n]$.   I tried proving that the only elements outside the ideal are the invertibles of $K$ (I should still prove that this implies maximality, but it shouldn't be too difficult).  Is there a better strategy, or another stategy?
40799
One very convenient property of $\mathbb{R}$ as a ring is that there is an upper bound for the degree of irreducible polynomials in $\mathbb{R}[x]$, as      If $f\in\mathbb{R}[x]$ has degree larger than $2$, then $f$ is reducible.   However, the proof as I know depends highly on the fact that $\mathbb{C}$ is algebraically closed, and the very nice property: $z\in\mathbb{C}$ is in $\mathbb{R}$ if and only if $z=\bar{z}$.  This makes a generalization to other integral domains rather difficult. So the problem is      For what kind of integral domain $R$, we have a finite upper bound on the degree of irreducible elements in $R[x]$?   Some most familiar examples are ruled out: in $\mathbb{Z}[x]$, $\mathbb{Q}[x]$ and $\mathbb{F}_{p}[x]$, there is not such a bound. Unfortunately these exhaust all integral domains about which I have a working knowledge.  Another result might help is , which says if we can find a prime ideal $\mathfrak{p}$ in $R$ such that $\mathfrak{p}^2\neq\mathfrak{p}$, then by picking $a\in\mathfrak{p}^2\backslash\mathfrak{p}$ we have an irreducible $a+x^d$, where $d$ can be arbitrary, and hence the upper bound is not possible.  So we only need to focus on domains where $\mathfrak{p}^2=\mathfrak{p}$ for all prime ideals. This seems to be a quite strong restriction but I am not sure what to make of it.  Can someone give a hint?  Thanks!
36939
Let $K$ be a field and $A=K[x_1,x_2,x_3,...]$. Prove that the ideal $I:=\langle x_i: i \in \mathbb N\rangle$ is not finitely generated as $A$-module.   I have no idea what can I do here, I mean, suppose $I$ is finitely generated, then there is a subset $S \subset I$ with $S=\{x_{i_1},...,x_{i_n}\}$ such that $\langle\{x_{i_1},...,x_{i_n}\}\rangle=I$. How can I arrive to a contradiction? Any suggestions would be appreciated.
53896
I need help proving  this stunning result.  Let me state it this way.  Let  $\mathbb L$ be a field and $\mathbb K$ be a field with $\mathbb K \subset \mathbb L$  Let $A$ be the set of algebraic numbers ie $A=\{a \in \mathbb L, \exists P \in \mathbb K [X], P(a)=0 \}$  I managed to prove that $A$ is a ring with the following reasoning:  If $a,b \in A$ then $ \mathbb K[a]$ is a ring.  And $ \mathbb K \subset \mathbb K[a] \subset \mathbb K[a][b] $  Some finite-dimension argument(ask details if needed) proves that $\mathbb K[a][b]$ has finite dimension as a vector space over the field $\mathbb K$  Reminding that $\mathbb K[a+b] \subset \mathbb K[a][b]$ and $\mathbb K[ab] \subset \mathbb K[a][b]$ proves that they both have finite dimension, hence $a$ and $b$ algebraic.  How should I proceed with inverses ?
75004
Let $I$ be a homogeneous ideal of a graded ring $S$, $I\ne S$. I want to show that there exists a homogeneous prime ideal which contains $I$.    I proved the following:  Let $T$ be the set of all homogeneous ideals which contain $I$. Then, by Zorn's lemma, there exists a maximal element of $T$, say $P$. I will claim that $P$ is prime. Suppose that for homogeneous elements $a,b \in S$, $ab \in P$ but, $a\notin P$. Then $\langle a \rangle + P$ is a homogeneous ideal which contains $I$. It contradicts by maximality of $P$ so, $a\in P$.  Is it right???
101356
Let $\mathbb{K}$ a field with a discrete non archimedean absolute value and $A$ its ring of integers. Then $A$ is Noetherian, every ideal is generated by a power of $\pi\in A$, a uniformizer, and its unique maximal ideal is $\mathfrak{M}=\pi A$.  For instance, $\mathbb{Q}_p$ is such a field. Its residue field is $A/\mathfrak{M}\cong\mathbb{Z}/p\mathbb{Z}=\mathbb{F}_p$, the finite field of $p$ elements.   Is there any field $\mathbb{K}$ whose residue field is not finite?
15177
Given an algebraic number field $K/\mathbb Q$, let $R = \mathbb Z_K$ be the ring the algebraic integers of that fields. Is it possible to say how many prime ideals there are in $R$? I suspect we always have infinitely many, as this is certainly the case for $K=\mathbb Q$ (because then $R = \mathbb Z$).   I was already able to show that an Ideal $I\subseteq \mathbb R$ ($I\neq0$) is prime if and only if it is maximal. One idea would have been showing that you can generate infinitely many fields via $R/I$ but I was never able to show what $R/I$ would be isomorphic to, as I do not really know what the prime ideals $I$ look like.   So have no idea how to go from there, and I'd be happy for every hint!
77776
An exercise in my algebra course book asks to prove that if $R$ is a , then $R[[x]]$ is a , where $R[[x]]$ is the ring of formal power series over $R$. After some failed attempts at  proving the  I visited Wikipedia, which :     If $R$ is Noetherian, then so is $R[[x]]$; this is a version of the Hilbert basis theorem.   This is very useful, as $R$ is a PID and hence Noetherian. Unfortunately we only saw (without proof) Hilbert's basis theorem in the form     If $R$ is Noetherian, then so is $R[x]$.   I'm not sure how to conclude the Noetherianity of $R[[x]]$ from this. I know that $R[x]$ is a UFD because $R$ is (we saw this without proof), and have been trying to conclude the ACC in $R[[x]]$ from the ACC in $R[x]$, without success. I can prove the ACC if $R$ is a field, because then every $a_kx^k+a_{k+1}x^{k+1}+\cdots$ with $a_k\neq0$ is associate with $x^k$, hence every ideal is of the form $(x^k)$. (In fact this readily proves that $R[[x]]$ is a UFD.) If $R$ is not a field, say it has some non-invertible element $r$, then there are many more ideals such as $(r)$ or $(r+x)$.   I'm also facing difficulties at identifying the irreducible elements of $R[[x]]$. (I wish to prove that they are all primes in order to conclude the uniqueness of factorisation.) I have figured that they either take the form $ux+Px^2$ for some unit $u\in R^\times$ and some $P\in R[[x]]$ or have a non-zero constant term which is not invertible in $R$. The elements of the form $ux+Px^2$ are indeed primes, the difficulty lies in those with non-zero constant term.  Any hints at proving $R[[x]]$ is a UFD?
122757
Let $R\subset T$ be two commutative rings, and $T$ is integral over $R$. Let $\mathfrak m\in \operatorname{Max} R,\mathfrak n\in\operatorname{Max}T$ such that $\mathfrak m=\mathfrak n\cap R$. Show that $R_\mathfrak m\subset T_\mathfrak n$, i.e. the canonical map $\{\frac ru\mid r\in R,u\in R\setminus\mathfrak m\}\to\{\frac tv\mid t\in T,v\in T\setminus\mathfrak n\}$ must be an injection.   I feel confused because the exercise 4 from Chapter 5 in the book of Atiyah and Macdonald just assume that the localization is a ring extension. And I don't know if the condition "integral over" is necessary.
76124
Let $R = \mathbb{Z}_{(2)}$ be the localization of $\mathbb{Z}$ at the prime ideal generated by $2$ in $\mathbb{Z}$. Then prove that the ideal generated by $(2x-1)$ is maximal in $R[x]$.
55777
Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$. Let $M$ be a finitely generated module. Then there exists an element $x\in M$ such that $\frac{x}{1}\not\in\mathfrak m_iM_{{\mathfrak m}_i}$ for every $i=1,\dots,n$.    I cannot prove that such an element there exists. I was trying to prove it by induction on $n$. If $n=1$ it is true by Nakayama, so suppose it is true for $n-1$. Then for every $i$ I can find an $x_i\in M$ such that $\frac{x_i}{1}\not\in \mathfrak m_jM_{\mathfrak m_j}$ for every $j\neq i$. If $\frac{x_i}{1}\not\in \mathfrak m_iM_{\mathfrak m_i}$ for some $i$ we are done, so suppose $\frac{x_i}{1}\in\ m_iM_{m_i}$ for every $i$. I don't know how to go on, could you help me?
43049
prove $(n)$  prime ideal of $\mathbb{Z}$ iff $n$ is prime or zero    Defintions  Def of prime Ideal (n) $$ ab\in (n) \implies a\in(n) \vee b\in(n) $$ Def 1]  integer n is prime if $n \neq 0,\pm 1 $ and only divisors are $\pm n,\pm 1$   Def 2 of n is prime] If $n\neq0,\pm1$  only divisors of n are $\pm1,\pm n$ $$ n|ab \implies n|a \vee n|b $$    $\Rightarrow $] (Prime Ideal $(n)$ of $\mathbb{Z} $$\Rightarrow$ $n$ prime or zero)  Now consider the case where $(n)\neq (0)$. that is $n\neq 0$   Using the def of prime Ideals $$\begin{aligned} ab \in (n) \implies a \in(n) \vee b \in (n) \end{aligned}  $$  Well, If an element $x\in(n) \iff x=q*n \iff n|x$   $$n|ab \implies n|a \vee n|b$$  So, $n$ is prime ,nonzero.   In the case that $(n)=(0)$ clearly $n=0$ since there are no zero divisors in $\mathbb{Z}$  $\Leftarrow$] (n is prime or zero  $\Rightarrow $ $(n)$ is a prime ideal of $\mathbb{Z}$)  Consider the case where $p is prime$ so $n\neq 0, \pm1$  $$ \begin{aligned}  n|ab &amp;\implies n|a \vee n|b \\ ab \in (n) &amp;\implies a \in (n) \vee b\in(n) \end{aligned}$$  in the case $n=0$, since $\mathbb{Z}$ has no zero divisors  $$ab=0 \implies a=0 \vee b=0 $$ So (0) is a prime ideal.     Concern if this prove holds, also would be surprised if this question is not out there in this site. Did a search and clicked on similar questions and could not find it.And of course any other ways to prove it.
55691
Prove that if $I$ is a prime ideal of $R$ then $I[x]$ is a prime ideal of $R[x]$.   This is homework. I have been trying to assume that there is an $fg$ in $I[x]$ such that neither $f$ nor $g$ is in $I[x]$. Hence $f$ and $g$ have at least one coefficient not in $I$. I was trying to show that $fg$ would then have a coefficient not in $I$, obtaining a contradiction. But I don't see how to control the terms. If $f$ and $g$ had only one coefficient not in $I$, then I think I could use the properties of the ideal to complete the proof. The problem is not being able to know for certain which if any of the coefficients of $f$ and $g$ are in $I$.  Perhaps my approach is wrong to begin. Please help. Even a hint in the right direction will much appreciated.