In a commutative ring with an identity, every maximal ideal is a prime ideal. However, if a commutative ring does not have an identity, I'm not sure this is true. I would like to know the counterexamples, if any. The more examples, the better.  EDIT I would like to know the counterexamples other than $2\mathbb{Z}$. The more examples, the better.  EDIT I also would like to know the counterexamples that are not given in  if any, namely an example of a non-prime maximal ideal which does not contain $R^2$.
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A maximal ideal  is always a prime ideal, and the quotient ring  is always a field. In general, not all prime ideals are maximal.      In $2\mathbb{Z}$, $4 \mathbb{Z} $ is a maximal ideal. Nevertheless it is not prime because $2 \cdot 2 \in 4\mathbb{Z}$ but $2 \notin 4\mathbb{Z}$. What is that is misunderstand?
112858
Suppose that $A$ is a local Noetherian ring with principal maximal ideal. Can we prove that every ideal of $A$ is principal?   I tried to exploit the Noetherian property on the set of non-principal ideals obtaining that (if the statement is false) there must exist  a prime non-principal ideal but I can't conclude nothing.
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.
26946
Let $R$ be a commutative Noetherian ring with unit and let $I$ be a fixed ideal. I am sorry if the following turns out to be a very silly question.  1) Suppose $\operatorname{Ass}(R/I)$ are all minimal, then what can we say about $R/I$?  One conclusion is that $I$ has a unique primary decomposition. If $R/I$ is CM, then $\operatorname{Ass}(R/I)$ are all minimal, but I don't think the converse is true. Can someone provide a counterexample?  2) Suppose $\operatorname{Ass}(R/I)$ are all minimal and have the same height. I still dont think we can conclude that $R/I$ is CM. Can someone provide a counterexample? Theorem 2.1.6 of Bruns and Herzog, CM Rings, gives a result in this direction but it needs additional hypothesis. It says that if every ideal $I$ is generated by height $I$ elements and $\operatorname{Ass}(R/I)$ are all minimal, then $R$ is CM.  It would be nice if someone can say something nice about the ring in the special situation described in 1) and 2).  Thanks.
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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 $M_p$ ?
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The question is from the following problem:     Let $R$ be a ring with a multiplicative identity. If $U$ is an additive subgroup of $R$ such that $ur\in U$ for all $u\in U$ and for all $r\in R$, then $U$ is said to be a right ideal of $R$. If $R$ has exactly two right ideals, which of the following  must be true?   I. $R$ is commutative. II. $R$ is a division ring. III. $R$ is infinite.    I know the definition of every concept here. But I have no idea what is supposed to be tested here.    Why is the ring $R$ which has exactly two right ideals special? What theorem does one need to solve the problem above?   Edit. According to the answers, II must be true. For III, $R$ can be a finite field according to mt_. What is the counterexample for I then?
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I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is principal. I already proved that if R and S are ring every ideal in R x S is I x J with ideals in the original ring. But I cant follow from that that $\mathbb Z\times\mathbb Z$ has only principal ideals.      Explicitly, if $I$ is an ideal $(a)\times(b)$ which would be its generator $(c,d)$ in $\mathbb Z\times\mathbb Z$?    Thanks.
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I am trying to prove $(a)$ is a maximal ideal in $\mathbb{Z}$,  if and only if $a$ is prime number.  Now I wrote: assume $a \in\mathbb{Z}$, while it's not prime number we can write as $a=xy$, for some integers $x$ and $y$. then $(a)\subset (x)$ and $(a)\subset (y)$  while if $(a)$ is maximal ideal, it's not exist $(k)$ such that $(a)\subset (k)\subset\mathbb{Z}$. so $(a)$ cannot be maximal ideal. we can say by contradictory if $a$ is prime, $(a)$ is maximal ideal.  is this correct?  and how to prove the other way.
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Let $R$ be a (not necessarily commutative) ring and suppose $R$ has a maximal left ideal $I$ consisting of nilpotent elements, that is for $x\in R$ there exists some $n\in\mathbb{Z}_{\geq 0}$ such that $x^{m}=1$. I want to show that in this case, $I$ is the unique maximal left ideal.  To attempt to prove this, assume $J$ is some other maximal left ideal of $R$. Since $x^{m}=0$, we have $x^{m}\in J$. This is where I get stuck.  In the commutative case we can note that $I$ is necessarily a prime ideal, so either $x\in J$ or $x^{m-1}\in J$. Iterating this process shows that $x\in J$, then $I=J$ by maximality.  In the non-commutative case I can't seem to proceed. This is probably really easy but I'd appreciate a hint.
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Let $R$ be a commutative ring with $1$ and $I$, $J$ ideals in $R$.   For a prime ideal $P$, let $I_P=(R-P)^{-1}I$ be the localization of $I$ at $P$.   Question:     If $I_P=J_P$ for all prime ideals $P$ then is $I=J$?   My first attempt at a solution: take $x\in I$ then for each prime $P$ write $x/1=y/s$ for some $y\in J$ and $s\in R-P$. Then for some $t\in R-P$, $xst=yt\in J$ where $y,s,t$ all depend on $P$. From this can I deduce that $x\in J$?   Is there a better way to show this?  Thank you!
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.
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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.
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Let $R$ commutative ring with unity, $S\subseteq R$ subring, $p$ minimal prime ideal of $S$. Show there exists a minimal prime ideal $q$ in $R$ with the property that the contraction $q^c=q\cap S=p$.  First of all, I am not sure whether the minimality of $q$ refers to the prime ideals in $R$ or to the prime ideals with such contraction property, if the latter is the case, then we only need to show the existence of such prime ideal, minimal one would be given by Zorn's lemma. Second, if we drop the minimality condition, then the proposition clearly doesn't hold ($\mathbb{Z}\subseteq\mathbb{Q}$ for instance), so this condition must be crucial here.
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I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
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In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal  Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which every ideal $A$ is of the form $\langle a \rangle = \{ar~~ | ~~r \in R\}$  Let $ \langle a \rangle $ be a prime ideal $\implies R/A $ is an integral domain.  A finite integral domain is a field . Hence, if we prove that $R/A$ is finite, then $R/A$ is a field $\implies A$ is a maximal ideal.  Now, $\langle a \rangle = \{ar~~ | ~~r \in R\}$  Since, R is an integral domain, there are no zero divisors and cancellation is allowed $\implies ar_1 = ar_2 \implies r_1=r_2 \implies ar_i$ maps to a different member of $R$ for each different $r_i \implies \langle a \rangle$ represents the elements of $R$ in some random order.  $\implies \langle a \rangle = R$ and hence, $R/A \approx {0}$ is finite and hence $A$ is maximal.  Is my attempt correct?
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Algebra, by T. W. Hungerford, page 127. Theorem 2.15. ''If $P$ is an ideal in a ring $R$ such that $P \neq R$ and for all $a, b \in R$ \begin{equation} \label{1}  ab \in P \Rightarrow a \in P \mbox{ or } b \in P, \hspace{2cm} (1) \end{equation} then, $P$ is prime. Conversely if $P$ is prime and $R$ is commutative, then $P$ satisfies condition $(1)$''  Remark: Commutativity is necessary for the converse.   If I understood it correctly, there should exist a ring $R$ which is not commutative, with a prime ideal $P$ with the following property: $ab \in P$ and $a \notin P$ and $b \notin P$.   This is the example I had asked for in the previous version of my question.   PS: Definition of prime ideal, according to Hungerford:  ''An ideal $P$ in a ring $R$ is said to be prime if $P \neq R$ and, for any ideals $A, B$ in $R$, $$AB \subset P \Rightarrow A \subset P \mbox{ or } B \subset P. $$
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Let $A$ be a commutative ring with $1$. Let $I$ and $J_k$, $k=1,\dots,n$ be ideals of $A$ with $I\subseteq \cup _{k=1}^n J_k$. Then I have obtained the following:  (1) If $J_k$, $k=1,\dots,n$, are prime ideals, then there exists some $j$ such that $I\subseteq J_j$.   (2) If $A$ is a principal ideal ring, then there exists some $j$ such that $I\subseteq J_j$.   I want some counterexamples that for all $k=1,\dots,n$, $I$ is not contained in $J_k$. How to get the counterexamples?