If you have an irreducible element say $b$ in a ring, is the ideal $\langle b\rangle$ a maximal ideal?
120742
Possible Duplicate:          I am trying to see whether the ideal generated by irreducible element in a principal ideal domain (PID) is maximal ideal.  Suppose r is irreducible in a PID say D  Let I be an ideal of D containing (r) the ideal generated by r   Since  D is a principal ideal domain, there exist s in D such that I=(s), therefore (r) is a subset of (s).  So,  r=st , for some t in  D   but r is irreducible, this implies that s or t is a unit.  If s is a unit then I= (s)= D .  If t is a unit then (r)= I=(s). But I am not sure this is true, because I do not have reason for saying (r)= I=(s), so that I can conclude and say (r) is maximal.  I need a little help for this. Thanks
10576
A homework question from my algebra class asks:     Show that in a local ring $R$ with maximal ideal $M$, every element outside $M$ is a unit.   My argument is that since $M$ is maximal $R /M $ is a field and so for any $ x \in R \backslash M $, $ x + M $ has a multiplicative inverse, which implies $ x $ is a unit.  I don't see where we need the fact that $R$ is a local ring.
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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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The ideal $I= \langle x,y \rangle\subset k[x,y]$ is not a principal ideal.   I don't know how to consider it. Any suggestions?
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.
65827
In the midst of attempting to prove that for a commutative ring $A$ with identity, and an ideal $I$ of $A$, $I = \operatorname{rad}(I)$, where $\operatorname{rad}(I) = \{x: x^m \in I, m &gt;0\}$, implies that $I$ is an intersection of prime ideals, I've run into a little bit of confusion.   An answer to another question on this site, , indicates a strategy that I am attempting to use. I know that if $x$ is an element of $A\setminus I$, then so too is $x^m$ for any positive integer $m$. I use this fact to consider a subset $X = \{x^m : m &gt; 0\}$, which is closed under multiplication, and use Zorn's lemma to find an ideal of $A$ containing $I$ maximal with respect to the constraint that said ideal, $J$, is disjoint from X.  I want to prove $J$ is prime by contradiction. I suppose that $J$ were not prime; this also means that $J$ is not maximal in $A$. Then there would exist two elements, $f$ and $g$, of $A$ so that $fg$ is in $J$, but neither $f$ nor $g$ is in $J$. Following the prompting of the aforementioned answer, I know that $J + \langle a\rangle$ and $J + \langle b\rangle$ would be ideals strictly containing $J$, and since J is the maximal ideal disjoint from $X$, both $J +  \langle a\rangle$ and $J + \langle b\rangle$ contain at least some elements of $X$ (I am tempted to say all of them, but I'm not entirely sure that's true, in case for example $\langle a\rangle$ started containing powers of $x$ at $x^2$, in which case it isn't necessarily true that $\langle a\rangle$ contains $x$ itself), and so both $\langle a\rangle$ and $\langle b\rangle$ must themselves contain elements of $X$.  And this is where I am stuck. I feel like there is something I should know about multiplicative subsets of rings that would help me arrive at the desired contradiction, but I can't remember what it might be.
110838
Let A be a commutative ring with unity ring, M be a maximal ideal and $b\notin M$ . Show that there exists $q \in M, s\in A$ such that for every $a \in A$ we have $a=sba+qa$. Show that this is true only when $M$ is a maximal ideal.   Suppose if we take $A=2\mathbb Z$ and $M=4\mathbb Z$, $a=6.$ Then by the given statement $\exists$ $s\in A$, $q\in M$ s.t. $6=6(2s+q)$ $\implies$$s=\frac{1-q}{2}\implies q \text { is odd}$. But $q \in 4\mathbb Z.$  Please someone give some hints. how can solve this.  Thank you..
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..
66738
Let $R$ be a ring with unity (not necessarily commutative) and $I$ an ideal of $R$.  Suppose that for every element $a \in I$ there exists an element $c\in I$ such that $ac=a$. Note that $c$ is related to $a$. Now we have the following question:  Can we say that  for every element  $a\in I$ there exists an idempotent element $c\in I$ such that $ac=a$?    Of course we have many examples such that the answer is true for them but in general  we don't know.
114500
Is there any way to show that for $a,b \in F$ , the ideal $\langle x-a , y-b\rangle$ is maximal in  $ F[x,y]$ , by showing that the quotient $F[x,y]/\langle x-a , y-b\rangle$ is a field ? Is the quotient  $F[x,y]/\langle x-a , y-b\rangle$ isomorphic with $F$ ? ( Here $F$ is a field )   I can show that the map $g : F[x,y] \to F$ given by $g( p(x,y))=p(a,b)$ is a surjective ring homomorphism with $\langle x-a , y-b\rangle \subseteq \ker g$ but I cannot show that the kernel is exactly that ; PLease help, Thanks in advance
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Let $K=F[x]/(p(x))$, where $F$ is a field, $p(x)\in F[x]$ is an irreducible polynomial.  What are all the maximal ideals in $K[y]$?  Attempt: I have an attempt, but not entirely sure it is correct.  Since $(p(x))$ is a maximal ideal in $F[x]$, this means $K$ is actually a field.  Thus $K[y]$ is a PID.  So maximal ideals in $K[y]$ are $(g(y))$ for irreducible polynomials $g(y)\in K[y]$?
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I am looking for a ring element which is irreducible but not prime.    So necessarily the ring can't be a PID. My idea was to consider $R=K[x,y]$ and $x+y\in R$.  This is irreducible because in any product $x+y=fg$ only one factor, say f, can have a $x$ in it (otherwise we get $x^2$ in the product). And actually then there can be no $y$ in $g$ either because $x+y$ has no mixed terms. Thus $g$ is just an element from $K$, i.e. a unit.  I got stuck at proving that $x+y$ is not prime. First off, is this even true? If so, how can I see it?
52104
A subring $R$ of a field $K$ is said to be a valuation ring of $K$ if for each $x$ $\in$ $K^{*}$ we have either $x$ $\in$ R or $x^{-1}$ $\in$ $R$. How can I show that the valuation ring has a unique maximal ideal?
117024
An ideal $M$ of a commutative ring $A$ (with unity) is maximal iff $A/M$ is a field.  This is easy with the correspondence of ideals of $A/I$ with ideals of $A$ containing $I$, but how can you prove it directly?  Take $x + M \in A/M$.  How can you construct $y + M \in A/M$ such that $xy - 1 \in M$?  All I can deduce from the maximality of $M$ is that $(M,x) = A$.
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If $I$ and $J$ are ideals in a ring $R$ with 1 which are  co-maximal, i.e $I+J = R$, show that $I^m$ and $J^n$ are co-maximal for all $m,n$ in $\mathbb{N}$  Work done:     Should I proceed using    By taking $$K=I^m+J^m \neq R$$  Let $$S=\{I,J \in R \mid I^m+J^n \neq R \}$$ be the required partially order set ordered by inclusion then  $$I+J \subseteq I^n+J \subseteq I^n+J^m \subseteq \ldots $$  Then, I think this chain doesn't have an upper bound...as $I^m+J^n \neq R$ for all $ m,n \in \mathbb N$. Am I right??  Thanks for any help!!
116224
Show that if $P$ is a prime ideal of a  commutative ring $R$ with unity then $P[x]+\langle x\rangle $ is a prime ideal of $R[x]$. Here $P[x]$ consists of all polynomials whose coefficients are in $P$.  I tried to show it using the fact that if $f(x)g(x)\in P[x]+\langle x\rangle $ then either $f(x)\in P[x]+\langle x\rangle $ or $g(x)\in P[x]+\langle x\rangle $. But I don't know how to show it. Please help me out.
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Let $R$ be a PID, and $r\in R- \{0\}$. Prove that   $\langle r\rangle$  maximal $\iff r$ irreducible.   "$\Leftarrow$"Easy.  "$\Rightarrow$"If $J=\langle r \rangle$ then we will prove that $r$ is irreducible. If $r=ab$, we want to prove that $a\in U(R)$ or $b \in U(R)$.  If we take the ideal which is generated by $\langle a\rangle$ then (because $J$ is maximal)$$\langle a\rangle \subseteq\langle r \rangle \iff r\mid a \iff a=kr, k\in R \Longrightarrow r=krb\iff r(1-kb)=0_R \iff kb=1_R$$ so $b\in U(R)$. Same way if we work with $\langle b \rangle$.  Is this proof right?