Why are ideals in valuation rings totally ordered? I was thinking of defining a valuation $v:I \rightarrow \text{min}\{v(x): x \in I\}$, would this work?
122390
How do I prove the fact that for any valuation ring $V$ the ideals are totally ordered under inclusion?
2234
Let $A$ be a Noetherian ring and let $x \in A$ be an element which is neither a unit nor a zerodivisor. Why the ideals $xA$ and $x^{n}A$ (where $n \in \mathbb{N}$) have the same prime divisors? i.e. $$\operatorname{Ass}(A/xA) = \operatorname{Ass}(A/x^{n}A).$$  (Matsumura, Commutative Ring Theory, Exercise 6.3)
122471
please first read these then answer my question  Let $D$ be a division ring and r, n be two positive integer numbers such that $0 ≤ r ≤ n$. Then $H_r(D)$ denotes the left ideal of $M_n(D)$ containing all matrices whose $j$th column is zero, for every $r &lt; j ≤ n$.     Theorem A: Let $D$ be a division ring and $n$ be a positive integer number.   Then for every left ideal $I$ of $M_n(D)$, there exist an invertible matrix $P$ and integer $r$,   $0 ≤ r ≤ n$, such that $I = PH_r(D)P^{−1}$.   Now from theorem A how I can conclude that each left ideal of $M_2(D)$ is minimal? I'm totally confused with this
79815
I would like to compile all questions I have encountered with Ideals in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions and ask if there are any gaps.  Question is to see if :     the ideal $\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=0\}$ is maximal ideal in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions.   What I have done so far is :  consider $\eta :\mathcal{C}[0,1]\rightarrow \mathbb{R}$ where $f\rightarrow f(0)$   $Ker (\eta) = \{ f\in \mathcal{C}[0,1] : f(0)=0\}$  I could see that for each $r\in \mathbb{R}$, I would set $f$ such that $f(x)=x+r$  So, $\eta$ is surjective.  So, I would get $\mathcal{C}[0,1]/\mathcal{I}\cong \mathbb{R}$  As $\mathbb{R}$ is a field so ia the quotient $\mathcal{C}[0,1]/\mathcal{I}$ which means that $\mathcal{I}$ is an ideal in $\mathcal{C}[0,1]$.    Now Another Question on same idea :  Question is to see if :     the ideal $\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=f(1)=0\}$ is maximal ideal in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions.   I thought of using same ideas as above but i am unable to choose perfect $\eta$  I believe that this is not even a prime ideal..  I set  $f(x)=x$ and $g(x)=x-1$ and consider $f(x)g(x)=(x)(x-1)$ then,   $(fg)(0)=f(0)g(0)=0.(-1)=0$   $(fg)(1)=f(1)g(1)=1.0=0$  So, $fg\in \mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=f(1)=0\}$ but the neither $f$ nor $g$ is in $\mathcal{I}$.    I have two questions:   I see that $0$ does not play any role in the question $\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(0)=0\}$ I mean I would have same case if i replace $0$ by any real number.  So, I would like to say that $$\mathcal{I}=\{f\in \mathcal{C}[0,1] : f(r)=0\}$$  is a maximal ideal in the ring $\mathcal{C}[0,1]$ of all continuous real valued functions.  And one more thing I would claim is that      If the collection $\mathcal{I}$ is such that all elements of $\mathcal{I}$ has more than $1$ common zero then $\mathcal{I}$ is not a maximal ideal and not even a prime ideal..   I would like to learn more about this kind of ideals in $\mathcal{C}[0,1]$.  I would be thankful if some one can assure that what  have done is sufficient to conlcude what i have concluded and I would be thankful if some one can suggest some material to read regarding this.  Thank you
111123
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 $R$, $$AB \subset P \Rightarrow A \subset P \mbox{ or } B \subset P. $$
2337
Why can positive rationals be not well ordered?  If we define the relation to be greater than(&gt;), then every subset will have a least element.  Or why are positive or even integers not well ordered?  By the same logic we can always find a least element in any subset.  I know I am wrong at some very fundamental point, but please explain it to me.
97596
Is it true that for a ring $R$ and a noninvertible element $x\neq 0$ in it, can always find a ring $R'$,  such that $R\leqslant R'$  and $x$ is invertible in $R'$ ?  If $R$ is an integral domain, the answer is of course that $R'$ is the field of fractions. But what happens, if $R$ is "less" than an integral domain ?
106146
I have been thinking a bit about Morita equivalence   and I would like to know whether it also applies to subrings such as right or left ideals.  And, if so, how specifically? Could you illustrate if with the example of 2x2 matrices with zero entries in the bottom row? Since Morita equivalence concerns non-commutative rings, I am assuming it should apply, but I am just guessing.  Thanks in advance.
80299
I want to prove the following result:      Let $R$ be a ring and $M$ a maximal ideal in $R$. If $P$ is a prime ideal in $R[x]$ that (strictly) contains $M[x]$, then $P$ is a maximal ideal in $R[x]$.    I have an idea how to prove that, but I'm not quite sure if the argument is totally valid; maybe someone has a cleaner proof. My argument is like that:   $R[x]/M[x]$ is isomorphic to $(R/M)[x]$, that is a PID (since $R/M$ is a field). So, any prime ideal in this ring is maximal; If $P$ contains $M[x]$, then $P$ corresponds to a (prime??) ideal in $R[x]/M[x]$ (this needs more clarification); The prime ideal (that corresponds to $P$) in $R[x]/M[x]$ is then maximal, and this guarantees that $P$ is maximal in $R[x]$.   I'll appreciate any comment.
90416
For valuation rings I know examples which are Noetherian.      I know there are good standard non Noetherian Valuation Rings. Can anybody please give some examples of rings of this kind?    I am very eager to know. Thanks.
26230
Suppose that $k$ is an algebraically closed field. Then what do the prime ideals in the polynomial ring $k[x,y]$ look like?   As far as I know, the maximal ideals of $k[x,y]$ are of the form $(x-a,y-b)$ where $a,b\in k$. What can we say about the prime ideals? Are there similar results? And what about $k[x,y,z], k[x,y,z,w]$ and so on. Would someone be kind enough to give me some hints or referrence on this topic? Thank you very much!
115283
Let $R$ be an associative ring with $1$. Prove that $R$ is a division ring if and only if $R$ has no proper left ideals.    Clearly, if $R$ is a division ring and $I\neq\{0\}$ is a left ideal, then every nonzero $z\in I$ satisfies $1=z^{-1}z\in I$, hence $I=R$.  Now if $z\in R$ is nonzero and $R$ has no proper left ideals, then $I:=\{rz:r\in R\}$ is equal to $R$ and there exists $r\in R$ such that $rz=1$. However, we still need to prove $zr=1$. How can we do this?  Thank you.
20966
How  can I show the following?     Show that, if $a$ and $b$ are elements of a ring $R$ and $I$ is an ideal of $R$, then $$a+I=0+I \iff a \in I$$    I am so interested to know the proof.Thanks and good night all!
45873
$A$ and $B$ are ideals of a ring ${R}$ such that $A\cap B=\{0\}$. Prove that $st=0$ for every $s\in A, t\in B$.  I have no idea how to solve it. please help.
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.
66721
Let $aR$ be a nonzero ideal in a PID $R$. Show that $R/aR$ is a ring with only finitely many ideals.  Honestly, I do not know how to start. Appreciate any tips.
72037
I am doing some exercises on commutative algebra and came along the following expressions, which were not elaborated on. Is someone familiar with them?  The first is for $p$ a prime number $\mathbb{Z}_{(p)}$, where the exercise is to determine all ideals. Do you think this is just $\mathbb{Z}/p\mathbb{Z}$?  The second one is for $P$ a prime ideal the expression $P_P$, where $P_P$ lies in $R_P$ for $R$ a ring.  I never came across this notation before so I have to ask here.