This questions comes under the section about Zorn's Lemma. But I'm really lost to how maximality here implies prime.  Edit: R is commutative.
34488
I've been doing some old exam problems and I've come across a problem that I've answered, but my gut is telling me that there's something I'm glossing over.     Let $R$ be a commutative ring with identity and let $U$ be an ideal that is maximal among non-finitely generated ideals of $R$. I wish to show that $U$ is a prime ideal.   Assume that $U$ is not prime. Let $x, y\not\in U$ be such that $xy\in U$. $U$ is contained in a maximal ideal $M$ and $xy\in M$, so either $x$ or $y$ is in $M$; assume $x\in M$. The condition $U\subset M$ then implies that there is a ring homomorphism $$\varphi: R/M\to R/U$$  Since $R/M$ is a field, $\varphi$ is injective. Hence, $\varphi(x)\in U$. This is a contradiction, so $U$ must be prime.  The thing that worries me is that I never explicitly used the hypothesis that $U$ was not finitely generated or the result that $M$ must be finitely generated.
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?
99873
Let $A$ be a ring (commutative unital), and $\mathcal I$ be a nonempty family of proper ideals of $A$.  I will say that $\mathcal I$ has property $\dagger$ if for any $\mathfrak a\in\mathcal I$ and any $xy\in \mathfrak a$, one of $\mathfrak a+(x),\mathfrak a+(y)$ is in $\mathcal I$.  In particular, any maximal (w.r.t. inclusion) element of $\mathcal I$ is prime.  Does $\dagger$ (or maybe $\dagger$ + hypothesis of Zorn's lemma) have a name (a prime family, perhaps, as I suggest in the title)? As a side question, are there some interesting criteria for checking that a given family of ideals has property $\dagger$?  It seems to me that posets with property $\dagger$ are rather abundant in commutative algebra (e.g. in proof of Cohen's characterization of Noetherian rings), but I've yet to see $\dagger$ discussed on its own, or any name for it.
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..
120944
I have been trying to prove the statement in the title, however I seem to get stuck at a certain point. Let $M$ be the maximal left-ideal of $R$. Then consider $M.r$ for $r \in R$. If $M.r \neq R$, we have that $M.r \subseteq M$. Now if this assumption would be true, we would have that $M$ is a right ideal as well. However when trying to prove that $M.r \neq R$ for all $r \in R$ I don't get very far. I know that if we assume equality, one would get $m.r=1$ for one $m \in M$. However I don't know how to go from here. Thank you in advance!
23759
I started studying about localization of rings and i am stuck at a question about $Spec(R_{S})$ where $R$ is a commutative ring and $S$ a multiplicatively closed set, so $R_{S}$ is also known as localization.   So, have to specify how the prime ideals of a certain $R_{S}$ look like. Given is the ring $R = \mathbb{Z}$, some $k\in \mathbb{Z}$, the localization is $\mathbb{Z}_{k}$. Our multiplicatively closed set is $S=\left \{ 1, k, k^{2}, \dots \right \}$ and $\mathbb{Z}_{k} = \left \{ \frac{r}{k^{n}} : r\in \mathbb{Z}, n\in \mathbb{N}  \right \}$.  How to find the prime ideals of $\mathbb{Z}_{k}$? Can anybody help me, please? Thank you very much!
15896
Here is the theorem and proof in my textbook.     Thm.  Let $P$ be an ideal of a commutative ring $R$ with identity $1$.  Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.   Proof.   Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$.  Then $P$≠$R$ implies $1+P≠0+P$. Hence $R/P$ is a commutative ring $R$ with identity.  Assume that $(a+P)(b+P)=0+P$.  Then $ab+P=0+P$ and $ab∈P$.  By the definition of a prime ideal P we get $a∈P$ or $b∈P$.  That is, $a+P=0+P$ or $b+P=0+P$.  Thus $R/P$ is an integral domain.   Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors.  Hence $P≠R$.  Assume $ab∈P$.  Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$.  Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$.  So $a∈P$ or $b∈P$.  Thus $P$ is a prime ideal.    I am having problems understanding proof with "Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$.".  I know that $R/P$ has an identity. (cause it is ID)  But, why doesn't "$P$" have an identity?  Isn't there a case $P=R$?  Please help me understanding. I think $P≠R$ should be with condition of Thm..  Thank you in advance.
98754
Consider the subring $R\subset\Bbb Q$ , $R=\{\frac ab\ | \ a,b\in\Bbb Z, b\text{ odd}\}$  I am struggling with the following questions:  (1) Prove that the ideals of $R$ are $\{0\}$ and $2^nR$ for $n\ge 0$  (2) Prove that $R$ has 2 prime ideals and 1 maximal ideal  For (1) I can see that $2^nR$ is an ideal for $n\geq0$ but I'm not sure how to show all ideals other than $\{0\}$ are of this form  With question (2),  $2^nR\subsetneq 2^{n-1}R\subsetneq...\subsetneq 2R\subsetneq R$  so $2R$ is clearly the only maximal ideal and is therefore prime, but I am not sure where to go from here.  Any suggestions?
4708
I am trying to prove the following: Let $R=\mathbb{Z}[i]$, $a, b \in \mathbb{Z}, ab \neq 0$. Show that $a+bi$ is an irreducible element in $R$ if and only if $a^2 + b^2 ∈ \mathbb{Z}$ is prime. Since $R$ is a Euclidean domain, irreducible elements are prime. I think this is relevant, but am stuck on where to go from here.
47042
In commutative rings we have the following  Theorem. $R$ is Noetherian if and only if each prime ideal of $R$ is finitely generated.  From this Theorem I am looking for commutative rings $R$ in which every maximal ideal is finitely generated but $R$ is non Noetherian.  Question: Is there a straightforward example of a commutative ring $R$ so that each maximal ideal is finitely generated, but $R$ is non Noetherian?  Thank You
104072
Question: Suppose that $R$ is a commutative ring without zero-divisors.   Show that the characteristic of $R$ is either $0$ or prime.   I have established that every element in a commutative ring $R$ without zero divisors have the same additive order $n$.   Now, if no such additive order n exists, then the characteristic of $R$ is $0$.  Obviously, if a finite additive order exists, Char of $R$ is finite.  How do I show that Char of $R$ is prime? It probably involves lagrange's theorem and the order of the element in $R$.  Hint is appreciated.  Thanks in advance.
14965
Let $R$ be commutative ring with identity that contains a field $K$ as a subring. If$ $R is a finite dimensional vector space over the field $K$, prove that every prime ideal in $R$ will be maximal.   My idea was to prove the integral domain $R/p$ (if $p$ is a prime ideal) was a field as for any ideal $P$, $R/P$ will be a field iff $P$ is maximal in $R$. But how can I use the fact that $R$ is finite dimensional over $K$? I don't understand.
73289
Let $\Bbb R^+$ be the set of positive real numbers. Use Zorn's Lemma to show that $\Bbb R^+$ is the union of two disjoint, non-empty subsets, each closed under addition.
121305
How can I prove Zorn's lemma is equivalent to Axiom of choice?
42097
In a set of old lecture notes, I came across this corollary:  Corollary: Any maximal ideal is a prime ideal.  While I do not have the proof for this, what immediately comes to mind is whether this corollary is predicated on the fact that...  Theorem: For any commutative ring $R$ and ideal $A$, $R/A$ is an integral domain iff $A$ is a prime ideal Theorem: For any commutative ring $R$ and ideal $A$, $R/A$ is an field iff $A$ is a maximal ideal  Recall: Any finite integral domain is a field. So, from the two theorem above, we can reconcile the fact that a maximal ideal is a prime ideal (and vice versa) on the supposition that the integral domain in question is finite. Or is there another proof for the claims in my notes?
110995
The proof of the following proposition is not completely clear to me. I get everything up until the bold part and I have a feeling some crucial steps are omitted, can anybody help clear this up?     Let $R$ be an integral domain. If every two elements of $R$ have a greatest common divisor, then every irreducible element in $R$ is prime.   Proof:  Consider an irreducible element $p \in R$ and $x,y \in R$ such that $p\vert xy$. Suppose now that $py$ and $xy$ have a greatest common divisor $z$ in R. We want to conclude from this that $p \vert x$ or $p \vert y$. This is obvious if $xy = 0$, so we may assume that $xy \neq 0$. Then $z \neq 0$. As both $p$ and $y$ divide each of $py$ and $xy$, we have that $z = pu = yv$ for certain $u,v \in R$. Using the cancellation law with $\boldsymbol z \boldsymbol \neq \boldsymbol 0$, we obtain that $\boldsymbol v \boldsymbol \vert \boldsymbol p$. As $p$ is irreducible, either $v \in R^\times$ (i.e. the set of invertible elements of $R$) or $v \sim p$ (i.e. $Rv = Rp$). If $v \sim p$, then $p \vert x$. If $v \in R^\times$, then $p$ divides $v^{-1} pu = v^{-1} z=y$.
12382
Let $A$ be a commutative ring. Suppose $P \subset A$ is a minimal prime ideal. Then it is a theorem that $P$ consists of zero-divisors.  This can be proved using localization, when $A$ is noetherian: $A_P$ is local artinian, so every element of $PA_P$ is nilpotent.  Hence every element of $P$ is a zero-divisor. (As Matt E has observed, when $A$ is nonnoetherian, one can still use a similar argument: $PA_P$ is the only prime in $A_P$, hence is the radical of $A_P$ by elementary commutative algebra.)  Can this be proved without using localization?