I have the following question in a competitive exam , but I failed to answer it.The question is:     Let $\mathcal R = \{f:\{1,2, \dots , 10\} \rightarrow \mathbb Z_2\} $ be the set of all $\mathbb Z_2$-valued functions on the set                $\{1,2,\dots , 10\}$.   Then $\mathcal R$ is a commutative ring with pointwise addition and multiplication of functions.Which of the following statements are correct?     $\mathcal R$ has a unique maximal ideal. Every prime ideal of $\mathcal R$ is also maximal. Number of proper ideals of $\mathcal R$ is 511. Every element of $\mathcal R $ is idempotent.   The only option I was able to answer was option 4 .  Can anyone help me understanding the ring in the question and the options.  Any insight will be happily appreciated. Thank you.
77094
Let $R =\{f:\{1,2,3,4,\cdots ,10\}\to \mathbb{Z}_2 \}$ be the set of  all $\mathbb{Z}_2$ -valued functions on the set $\{1,2,3......10\}$  of the first ten positive integer. Then $R$ is a commutative ring with point-wise addition and multiplication of functions. Which of the following is true?     1) $R$ has a unique maximal ideal.      2) Every prime ideal of $R$ is also maximal.      3 ) The number of proper ideals of $R$ is $511$.      4 ) Every element of $R$ is idempotent.   R is finite commutative ring so  every prime ideal  is maximlal ideal ,so I think option  2 is true   .But for the other options   I am  helpless  .
119369
Let $R$ be a commutative ring and let $I$ be a prime ideal of $R$. Prove that $S = R \setminus I$ is a multiplicative closed subset containing $1$.  Please can anyone help me out here?
60225
Consider $K$ field and consider the ring $R=K[X^2,X^3]\subset K[X]$. It is clear that $R$ is not a Dedekind domain, since with the element $X$ one immediately see that it is not integrally closed. But $R$ is a Noetherian domain and every non trivial prime ideal is maximal. I have a proof for the last two part, but it is a bit heavy and maybe someone can help me finding a better proof! Thanks in advance!
63122
The problem is this:      Suppose $I \subseteq R$ is a nilpotent ideal and there is $r \in R$ with $r \equiv r^2 \pmod I$. Show $r \equiv e \pmod I$ for some $e \in R$ idempotent.   I have spent a few hours rolling around in abstracta with no destination. I believe that if I could write down a concrete example of this, the example could guide me through the abstract definitions and show me where to look for an idempotent in $R$. I have looked at 2x2 matrices and could not find any such examples. Might anybody have one?
36070
The ring of continuous functions on $[0,1]$ to $\mathbb{R}$ has an interesting property:      every maximal ideal of this ring is the subset of all functions vanishing at a common point.    If we follow the argument in the proof of this property, it can be seen that $\mathbb{R}$ can be replaced by $\mathbb{C}$ and still similar property holds for new ring (we have to slightly modify argument).   What if we place $\mathbb{Q}$? Here $\mathbb{Q}$ is disconnected with induced topology from $\mathbb{C}$ and so the only continuous function from $[0,1]$ to $\mathbb{Q}$ are constant (am I correct?). So, the property no longer holds true with replacement of $\mathbb{R}$ by $\mathbb{Q}$.  Question: Are there subfields of $\mathbb{C}$ other than $\mathbb{R}$ which are connected as topological spaces with induced topology from $\mathbb{C}$?
78025
Let $R$ be a commutative ring with identity. I've been trying to prove the following:     If $S \subset R$ is a maximal multiplicative set, then $R \setminus S$ is a prime ideal of $R$.   I have spent some time in it, but nothing is coming to my mind. Any idea/solution will be appreciated.
118281
[Update: Based on the hints provided by @zcn and @whacka, I believe I have found a solution. See my answer below.]  Below, $R$ is a commutative ring with $1$.  In John J. Watkins' Topics in Commutative Ring Theory, the author observes that if $a$ and $b$ are associates in $R$, then they generate the same ideal: $(a) = (b)$. This is clearly true, because $a = ub$ and $b = va$ for some units $u$ and $v$, and so $a \in (b)$ and $b \in (a)$, whence $(a) \subset (b)$ and $(b) \subset (a)$. Indeed, we didn't even require $u$ and $v$ to be units to reach this conclusion.  The author then claims that if $R$ is not a domain, then the converse may not be true: we may have $(a) = (b)$ even if $a$ and $b$ are not associates. As a "counterexample" he gives $R = \mathbb{Z}/(6)$, with $a = 2 + (6)$ and $b = 4 + (6)$. Clearly these elements generate the same ideal: $(a) = (b) = \{0, 2 + (6), 4 + (6)\}$. The author asserts that $a$ and $b$ are not associates, but this is false: $5 + (6)$ is a unit, and $(2 + (6))(5 + (6)) = 4 + (6)$, so $a$ and $b$ actually are associates.  I figured that the author botched the example but that surely the basic fact must be true: if $R = \mathbb{Z}/(n)$ then there may be nonassociates $a,b \in R$ with $(a) = (b)$. However, I was not able to find a counterexample and eventually resorted to writing a computer program and found that there are no counterexamples in $\mathbb{Z}/(n)$ for $n \leq 100$.  So now I'm starting to think that there is no counterexample in $\mathbb{Z}/(n)$ at all, but I have not yet been able to prove this.  My questions:   If I am correct that $(a) = (b)$ in $\mathbb{Z}/(n)$ implies that $a$ and $b$ are associates, I would appreciate a nudge in the right direction toward a proof. (But if the proof is elementary, please give only a hint, not a solution :-) I'll post my solution once I have one. In either case, could you please point me to a genuine counterexample? If there is one in $\mathbb{Z}/(n)$ for some $n$, that would be ideal (no pun intended), otherwise a counterexample in any commutative ring with $1$ would be great.
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?
108915
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?
6308
Here's what I've come up with. Let $\mathcal B$ be a generating set of $\mathcal F(R)$. Let us assume that $\mathcal B$ is a finite set, thus: $\lt  \mathcal B \gt = \lt \{b_1, ..., b_n \}\gt, n \in N$. Let  $\mathcal f \in F(R).$ Thus: $\mathcal f = \alpha_1 b_1 + ... + \alpha_nb_n,$ where $ \alpha_1,..., \alpha_n \in R$ are not all zeroes.  If  $\mathcal F(R)$ is a vector space, so there can be a function  $\mathcal g \in F(R)$ such as  $\mathcal g \neq \lambda f, \lambda \in R$. Thus  $\mathcal f+g \in F(R)$  but  $\mathcal f+g \neq \alpha_1 b_1 + ... + \alpha_nb_n$. If is $\mathcal g$ is to belong to $ \lt \mathcal B \gt$ we must add a new vector to $\mathcal B$.  Repeating the above mentioned procedure, we can find a function $\mathcal g\in F(R)$ such as $\mathcal h \neq \mathcal\lambda g$. Thus, we can find infinite functions and thus need infinite base vectors.  Is this proof right?
62747
Let $\mathbb Z$ be the ring of integers. The question asks to show  that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.
82956
I'm trying to work out the following past paper question and I've got stuck.  $R$ is an integral domain and $S = R[t]$, the polynomial ring in one variable over $R$. We have that $Q$ is a prime ideal strictly contained in another prime ideal $P$ with  $$ P \cap R = Q \cap R  = P_0. $$ We also have $Y = R\setminus P_0$.  I have shown that $P_0S$ is a prime ideal, and that after localisation we have $$ \frac{SY^{-1}}{P_0SY^{-1}} \simeq F[t]    $$ where  $$ F = \frac{RY^{-1}}{P_0Y^{-1}}. $$  This is a field since $P_0 = R\setminus Y$, so $P_0Y^{-1}$ is a maximal ideal of $RY^{-1}$.   What I now have to show is that $P_0S  =  Q$, and the question says that i should do this using the above isomorphism of polynomial rings, but I can't figure it out.  A hint or solution would be great, thanks!
28057
Let $R$ be a ring of the real functions with "+" and "x" and let $I_1:=\{f(x)\in R\mid f(1)=0\}$, $I_1$ is a maximal ideal?   I think yes, but don't know how to prove it.
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.
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.
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?
98946
Let $R_1 = \mathbb{Z}$, and $R_2 = \mathbb{F}_4$. Find all ideals in $R_1\times R_2$. Which of them are principal ideals? Which are prime? Which are maximal?   I know the ideals in $\mathbb{Z}$ and $\mathbb{F}_4$, but I am having trouble finding the ideals in the cartesian product of both of these rings.  If someone could help me out it would be appreciated.