Let $R$ be a ring and $I$ the set of non-invertible elements of $R$.   If $(I,+)$ is an additive subgroup of $(R,+)$, then show that $I$ is an ideal of $R$ and so $R$ is local.   $$$$   I have done the following:   Since $(I,+)$ is an additive subgroup of $(R,+)$, we have that $\forall a,b \in I$ : $ab\in I$.   But how can we show that it holds that $ax\in I, \forall a\in I, \forall x\in R$ ?
123834
Problem is this:      Suppose for any element $r$ in a ring $R$ with unity $1$, $r$ or $1-r$ is invertible. Then show that non-invertible elements form an ideal of $R$.   It is crystal clear that $-r$ is non-invertible if $r$ is. So I tried to show the sum of two non-invertible elements is non-invertible, and the product of non-invertible elements with any ring element is non-invertible but I'm stuck.  Any help will be appreciated.
80370
If A and B are ideals of a ring, show that A + B = $\{a+b|a \in A, b \in B\}$ is an ideal  I have the ideal test but no clue as to what to do with it:   $a-b \in A$ whenever $a,b \in A$ $ra$ and $ar$ are in A whenever $a \in A$ and $r \in R$
54222
Let $R$ be a commutative, unital ring and let $I$ be an ideal of $R.$ Prove that if $I$ is maximal then $R/I$ is a field.   This problem is probably trivial for most of you abstract algebra fans out there, but remember that I'm new to the field, so please don't use some next-level math that I can't understand at all. I think this problem could be solved using next-level math very quickly though.   For this problem, I know that fields are commutative, unital, and every nonzero element is a unit/invertible. Also, $I$ is maximal wrt subset, so if there is an ideal that contains $I,$ then it must be $R.$
112879
If $R$ is a ring and $I$ an ideal of this ring, is that true that $(R/I)[x]=R[x]/I$?   Note that can $I$ be identified as a subset of $R[x]$ and we need that in order to $R[x]/I$ makes sense.  I think it's true because every polynomial $(a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n$ of $(R/I)[x]$ can be written as a polynomial of $R[x]/I$ and vice versa since we have this equality: $$(a_0+I)+(a_1+I)x,\ldots,(a_n+I)x^n=(a_0+a_1x+\ldots a_nx^n) +I$$  Is my reasoning right?  Thanks
14998
Q: Let R be a commutative ring with unity. Prove that if A is an ideal of R and A contains a unit, then A=R.  This is my attempt at an answer: It suffices to show that all the elements in R are in A.  Let a be and element of A, and r be an element of R. Since A is an ideal of R then all elements ar and ra must be in A.   We know that A has a unit say 1, then we know 1r and r1 for all r in R, are in A.  Thus all elements in R are in A, therefore A=R.
13698
I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.  I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.  FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?  SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?  Thank you in advance for any insight.
56872
If $U,V$ are ideals of a ring $R$, then which of the following is also an ideal of $R$?   $U+V=\{u+v\mid u\in U,v\in V\}$ $U\cdot V=\{u\cdot v\mid u\in U,v\in V\}$ $U\cap V$   My attempt: I have proved $U+V$ and $U\cap V$ are ideals of $R$.  My problem: I have problem about the second option. I think it is not ideal of $R$. If I am wrong then please prove this otherwise give one counter example to show that $U\cdot V$ is not ideal of $R$.
50854
$$Rad(I)=\{a \in R | \exists n \in \mathbb{N} \text{ such that } a^n \in I\}$$   R is a commutative ring, I is an ideal.   To show that $Rad(I)$ is an ideal of $I$, we have to show that for $a,b \in Rad(I)$ it stands that $a-b \in Rad(I)$ and for $r\in R$ it stands that $a\cdot r\in Rad(I)$, right??   When $a,b \in Rad(I)$ we have that $a,b \in R$ and $a^n, b^m \in I$.   $a-b \in R$.   But how can we show that $(a-b)^x \in I, x \in \mathbb{N}$ ??
61191
Let $I$ and $J$ be ideals of a Noetherian ring $A$. If $JA_P\subseteq IA_P$ for every $P\in \operatorname{Ass}_A(A/I)$, then $J\subseteq I$.   I'm reading Matsumura's Commutative Ring Theory book on my own. This is exercise $6.4$ in that book, and I fail to prove it. Could you help me with this? Thanks in advance.
102960
Let $p$ be a prime number. Let $R$ be the subset of all rational numbers $m / n$ such that $n \neq 0$ and $n$ is not divisible by $p$. Show that $R$ is a ring. My Proof-trying. $R1$. Associativity for under addition is clear. $R2$. $\dfrac {m} {n}+0=0+\dfrac {m} {n}$, so there exist the identitiy element for under addition. $R3$. $\dfrac {m} {n} + \dfrac {-m} {n}=\dfrac {-m} {n} +  \dfrac {m} {n}$, so there exist inverse element. $R4$. $\dfrac {m} {n} + \dfrac {m'} {n'}=\dfrac {m'} {n'}+\dfrac {m} {n}$, so commutativity. So $R$ is an additive abelian group. $R5$. $\dfrac {m} {n} ( \dfrac {r} {s} + \dfrac {t} {v})=\dfrac {mr} {ns} + \dfrac {mt} {nv}$ and $( \dfrac {r} {s} + \dfrac {t} {v})\dfrac {m} {n}=\dfrac {rm} {sn}+\dfrac {r} {s}+\dfrac {tm} {vn}$ $R6.$ $R$ is associativity for under multiplication, clear. $R7.$ $1 \dfrac {m} {n}=\dfrac {m} {n} 1 = \dfrac {m} {n}$, so there exist identity element for under multiplication. Can you check my answer? If false, can you write true answer?
98023
Suppose $R$ is commutative ring with unity. For ideals $I$, $J \subseteq R$, the ideal quotient $(J:I)$ is $$(J:I) := \{x\in R \, : \, xI \subseteq J\}$$ Let $S\subset R$ be a multiplicative set. When does localization at $S$ commute with taking quotients, i.e. when does the equality $$(S^{-1} J : S^{-1} I) = S^{-1} (J:I)$$ hold? More generally, when is it true that $$S^{-1} \text{Ann}_{R} (M) = \text{Ann}_{S^{-1} R} (S^{-1} M)$$ for an $R$-module $M$?
116224
Show that if $P$ is a prime ideal of a  commutative ring $R$ with unity then $P[x]+\langle x\rangle $ is a prime ideal of $R[x]$. Here $P[x]$ consists of all polynomials whose coefficients are in $P$.  I tried to show it using the fact that if $f(x)g(x)\in P[x]+\langle x\rangle $ then either $f(x)\in P[x]+\langle x\rangle $ or $g(x)\in P[x]+\langle x\rangle $. But I don't know how to show it. Please help me out.
108915
The question is from the following problem:     Let $R$ be a ring with a multiplicative identity. If $U$ is an additive subgroup of $R$ such that $ur\in U$ for all $u\in U$ and for all $r\in R$, then $U$ is said to be a right ideal of $R$. If $R$ has exactly two right ideals, which of the following  must be true?   I. $R$ is commutative. II. $R$ is a division ring. III. $R$ is infinite.    I know the definition of every concept here. But I have no idea what is supposed to be tested here.    Why is the ring $R$ which has exactly two right ideals special? What theorem does one need to solve the problem above?   Edit. According to the answers, II must be true. For III, $R$ can be a finite field according to mt_. What is the counterexample for I then?
17478
So, I was trying to find a counter-example that shows not every local ring's lattice of ideals is a chain. I think $F[[x_1,\cdots,x_n]]$ is a good counter-example but I'm not able to show that $F[[x_1,\cdots,x_n]]$ is a local ring. I read somewhere that $F[[x]]$ is indeed a local ring.  So here comes the question:  If $R$ is a local ring, what can we say about $R[[x]]$? Is it a local ring too?  I'm looking for a simple proof that doesn't use commutative algebra and localizations to show that. An elementary undergraduate level proof is appreciated.
40727
Let $R$ be a ring and $I,J$ ideals of $R$. Let $$IJ = \left\{ \sum_{i=1}^n a_ib_i \big\vert\ n \geq 0, a_1,\cdots, a_n \in I, b_1,\cdots,b_n \in J \right\}$$ i) Show that IJ is an ideal of R and $IJ \subset I \cap J$. ii) Let $K$ be another ideal of $R$. Show that $I(J + K) = IJ + IK$.  i) In order to show that $IJ$ is an ideal, I need to show that $\forall r \in R\ \forall s \in IJ: r s r^{-1} \in IJ$, right? Let $r \in R$ be arbitrary, $s\in IJ = (a_1b_1)+\cdots+(a_kb_k)$. $rsr^{-1} = r( (a_1b_1)+\cdots+(a_kb_k))r^{-1} = (\underbrace{ra_1}_{\in I}b_1)+\cdots+(\underbrace{ra_k}_{\in I}b_k) \in IJ$, as I is an ideal of R, and therefore any $r\cdot a_i$ is, right? So what about $IJ \subset I \cap J$? If I consider a single summand $a_ib_k$, that the result need to be in I as well as in J, because both are ideals of R. But why does this apply for the whole sum? ii) I have no idea, do you have a hint for me?
112681
Let $R$ be a ring. Prove that each element of $R$ is either a unit or a nilpotent element iff the ring $R$ has a unique prime ideal.  Help me some hints.
95896
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.