Let $I$ be an ideal of a ring $R$ such that $I$ is not finitely generated but every ideal properly containing $I$ is finitely generated. Then I is prime.   Based on this result it can be proved that if every prime ideal in a ring $R$ is finitely generated, then $R$ is Noetherian.   Assume by contradiction. Let $I$ be a proper ideal of $R$ which is not finitely generated. Then $I$ is contained in a maximal ideal ideal $M$ of $R$ and hence prime. I am having difficulties in showing that $I$ is finitely generated. I think to show that if $J$ is an ideal that properly contains $I$ is finitely generated.  Would you help me please? Thank you in advance.
117346
Prove that if $R$ is a commutative ring in which all the prime ideals are finitely generated, then $R$ is Noetherian.   Here is what I been told to do: Suppose that $R$ is not Noetherian, and use Zorn’s lemma to obtain a maximal element $I$ in the collection of all ideals of $R$ that are not finitely generated.  Then use the following proposition: Let $I$ be an ideal of a commutative ring $R$, and let $r ∈ R$. If the ideals $I +rR$ and {$s ∈ R : sr ∈ I$} are finitely generated, then $I$ is a finitely generated ideal.  Can anyone help please. thanks a lot.
117131
Suppose $R$ is a commutative ring. Show that every ideal of $M_n(R)$ is of the form $M_n(I)$  where $I$ is an ideal of $R$.   I have spent 30 minutes on this question and I still got nowhere. Can anyone give some hints ?
15341
Let $R$ be a commutative Noetherian ring (with unity), and let $I$ be an ideal of $R$ such that $R/I \cong R$. Then is it true that $I=(0)$ ?  I know that a surjective ring endomorphism of a Noetherian ring is also injective, and since there is a natural surjection from $R$ onto $R/I$ we get a surjection from $R$ onto $R$, but the problem is I can not determine the map explicitly and I am not sure about the statement. Please help. Thanks in advance.
70782
Let D be the ring of integers of a number field K. Let I be a non-zero ideal of D. Why D/I is finite?  Would you help me please? Thank you in advance.
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?
119458
Let  $R$  be  a  ring  such  that  the  only  right  ideals  of  $R$  are  $(0)$  and  $R$.  Prove  that  either  $R$  is  a  division  ring  or  that  $R$  is  a  ring  with  a  prime  number  of  elements  in  which  $ab=  0$  for  all  $a,b\in R$.    I  don't  want  the  proof.  I  am  stuck  at  one  point.  Why  does $R$  have  to  be  finite  here?
75081
Let $R$ be a commutative ring with $1$. Suppose that all ideals $I \neq R$ are prime. Prove that $R$ is a field.   Help me some hints.  Thanks a lot.
19194
Show that an ideal $I$ of $R$ is proper if and only if it does not contain $1$ iff and only if it does not contain any units. ($1$ is the identity element) I'll need to show 3 parts: (1) $\implies$ (2): Let $I$ be a proper ideal and assume $1 \in I$. I want to show that $1 \notin I$. (2) $\implies$ (3): Let $I$ be a proper ideal that doesn't contain $1$ and assume there are units in $R$. By the definition, units of $R$ are elements with multiplicative inverses. So I need to show that there aren't multiplicative inverses. (3) $\implies$ (1): Assume there are no units in $R$ and then show that $I$ is a proper ideal. For this part I was thinking of doing something like this- Let $u \in I$ such that $u$ has a multiplicative inverse. Let $x \in R$. Then \begin{align}u^{-1}x \in R;\\ x=u(u^{-1}x) \in I,\end{align} so $R \subseteq I$ which implies that $I = R$. But this is a contradition so therefore $I$ must be a proper ideal. I just need help completing the proof.
74255
I'm supposed to show in a part of an exercise that if we have a ring $R$ that is a principal ideal domain, then for any ideal $I$ in $R$, $R/I$ will also be a PID.  So $I=(i)$ for some $i \in R$, and $R/I$ is $\{rI |r\in R\}$, and we know:   An ideal $J$ in $R/I$ will be an ideal $J$ in $R$ containing $I$.  So if $I \subset J$ and $I=(i)$, then $J$ should be $(j)$, where $j|i$. but wherefrom do I get the necessity that $J=(j)$, so that $J$ is also principal? and there will be several $j$ that divide $i$, but should it not be only one?  And then I shall use this to show a 1-1 correspondence between ideals of $\mathbb{Z}$ and $\mathbb{N}$.  Can someone support with with the proofs?
123701
Here is the problem I am stuck on: Fix a field $K$ and consider the subring $A \leq K[x,y]$ generated by $K \cup \{x,xy,\dots,\}$. Show that $A$ is not Noetherian.  I figure that taking ideals $I_n = (x,xy,\dots,xy^n)$ should give an infinite strictly ascending chain, which would establish that $A$ is not Noetherian, but I cannot figure out how to show that each $I_n$ is a proper subset of $I_{n+1}$. Any help here?
53351
Prove: Let $R$ a Artinian commutative ring with $1 \neq 0$. If $I$ is prime, then $I$ is maximal.     I got stuck on this. I understand that every ideal is generated by finitely many elements. Here is my approach:  It's enough to prove the following implication: $$ R/I \ \text{is a domain} \quad \Rightarrow \quad R/I \ \text{is a field}$$ Then only finding an inverse is left. So let $x \in R$ in arbitrary. We have to find an inverse element of $x+I \in R/I$, that as an element y so that $x \cdot y \in I+1$. We denote $I = (x_1, x_2, \cdots, x_n)$.  Now I have to find y so that $xy -1 = r_1 x_1+ r_2 x_2 + \cdots + r_n x_n$. This is where I no more knew what to do.   Your advise will be appreciated.
95896
Let $R$ be an integral domain and suppose that every prime ideal in $R$ is principal. Assume that the set of ideals of $R$ that are not principal is nonempty and prove that this set has a maximal element under inclusion. (Use Zorn's Lemma.)  Let $A$ denote the set of all nonprincipal ideals of $R$. I know that $A$ is a collection of sets, so it is partially ordered by set inclusion. I want to say that $A$ has a chain $\{C_\alpha\}_{\alpha\in B}$, where $B$ is an index set, but I want to know why I can say that a chain exists in this case.  I know that every ideal $I\in A$ is nonprincipal, so $I$ is generated by at least two elements. Also, $I$ cannot be prime. Thus, if $ab \in I$, where $a,b \in R$, then $a \notin I$ and $b \notin I$. Would this help me form a chain?  Thank you in advance.
33128
Prove that the set of nilpotent elements of a ring is the intersection of its prime ideals.   I know these two useful facts:  {nilpotent elements}$=\sqrt{0}$  $\sqrt{I}= \bigcap$ of prime ideals containing I   First, I need to show that the nilpotent elements are in fact an ideal and then I need to show it's prime.   Any specific suggestions would be appreciated. I just don't know how to make this proof flow and exactly what leads to what.   I read through this question and comments as well:
108133
Let R be a commutative ring with identity. If every ideal generated by two elements of R is principal, then can we conclude that R is a PID? Also, if every finitely generated ideal of R is principal, can we conclude that R is a PID?
82026
Suppose that $R$ is an integral domain. How to show that if every non-zero prime ideal of $R$ is invertible, then every non-zero ideal of $R$ is invertible?    Actually, I am trying to prove the unique prime factorization of ideals in $R$, so I cannot use the fact that any ideal can be factored as a product of primes.
96329
The statement is:     In a commutative ring with 1, every proper ideal is contained in a maximal ideal.   and we prove it using Zorn's lemma, that is, $I$ is an ideal, $P=\{I\subset  A\mid A\text{ is an ideal}\} $, then by set inclusion, every totally ordered subset has a bound, then $P$ has a maximal element $M$.  My question is why $M$ must contain $I$?
118754
Suppose that R is a commutative ring with unity such that for each $a$ in $R$ there is a positive integer $n &gt; 1$ such that $a^n =a$. Prove that every prime ideal of $R$ is a maximal ideal of R.  Attempt: Suppose $I$ is a prime ideal of $R$. Let $I'$ be an ideal of $R$ such that $I \subseteq I' \subseteq R$.  Given that $\forall~~a \in R, \exists~~ n&gt;1$ such that $a^n=a$.  $\implies $ if $b \notin I$, then, for any positive integer $i,b^i \notin  I$   Now, since $I$ is a prime ideal $\implies$ if $a,b \in R, ab \in I \implies a \in I$ or $b \in I$     Also, this means, for any $b \notin I , ~r \notin I, br \notin I$ and also that $br,(br)^2,(br)^3,..... \notin I$   By the definition of ideal, we know that for any $a \in I, ar \in I,~~~\forall~~r \in R$  I need to prove that either $I = I'$ or $I'=R$  If I want to prove the first one, I must show that $I \subseteq I'$ and $I' \subseteq I$  If I want to prove the second one, then I need to prove that $1 \in I'$. If I somehow know that there exists an element $c \notin I, c \in I'$ whose inverse $d$ exists anywhere in $R$, then, $I'$ being an ideal contains $cd=1$  I know I haven't been able to use the existence of prime ideals nor the existence of prime ideal satisfactorily enough. How do I proceed further. Thank you for your help..  Edit: $R/I$ will be an integral domain but to prove it to be a field, we must prove that every element is invertible with respect to multiplication. $r = r^n \implies r(r^{n-1}-1)=0 \implies ... $ Can I deduce something from here? Thanks..