I am just wondering if Product of two Ideals is again an Ideal.  Let $I$ and $J$ be ideal of Ring, then is $IJ=\{ij: i\in I, j\in J\}$ still an ideal?
121059
My textbook says that the product of two ideals $I$ and $J$ is the set of all finite sums of elements of the form $ab$ with $a \in I$ and $b \in J$. What does this mean exactly? Can you give examples?
105012
Let $R$ be a non-commutative ring, and $S\subset R$  some subset. Let $I_S$ be the smallest two-sided ideal of $R$ such that $I_S\supseteq S$. Is it true that (if $R$ is unital ring) $I_S$ consist only of elements of the form: $$\sum_{s\in S}\left(\sum_{k=1}^{N_s}x_{k}sy_{k}\right),$$ or (if $R$ isn't unit ring) $I_S$ consist only of elements of the form: $$\sum_{s\in S}\left(m_{s}s+xs+sy+\sum_{k=1}^{N_s}x_{k}sy_{k}\right),$$ where $N_s\in\mathbb{N}$, $m_{s}\in\mathbb{Z}$ and $x, y, x_{k}, y_{k}\in R$.  Thanks a lot!
111960
Let $R$ be a ring and let $J$ be an ideal of $R$. Assume $J$ contains   a unit of $R$. Prove $J=R$, rigorously.   Again I feel here as if the statement is intuitive so I am not able to make it rigorous. If you could show all steps in the proof that would be nice.
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.
39872
What are the proper ideals, prime ideals, maximal ideals of $\mathbb{Z}[i]$, the ring of Gaussian integers.    Check whether $(1+i)$ is prime or maximal ideal.     Can someone help me please. I have no idea how to crack this problem. Thanks for your time.
78666
I've been trying to prove that given a ring $R$ and a collection of ideals $\{I_{\beta}\}_{\beta \in B}$ in $R$, the set $$\sum_{\beta \in B} I_{\beta}=\{f_1+\cdots+f_r:f_j \in I_{\beta_j}, \mbox{ for some } I_{\beta_j}\}$$ is an ideal and is the smallest ideal containing $\cup_{\beta \in B} I_{\beta}$.  I've already proved when I only have two ideals, and so, using induction, I've seen it for finite sums. But I don't see how to generalise that to an arbitrary family of indexes (maybe infinite) $\beta$.   Could someone please give me a hand? Thank you.
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.
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.
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.
123212
Is it true that an ideal $I$ in a commutative ring is primary iff $Rad(I)$ is prime?  If not, what are some nice counterexamples?
103617
If $I$ is ideal, then $I$ is a subring. But, if $R$ is a subring, $R$ is not necessarily an ideal. In other words, the converse of the first statement does not always hold.  Examples of ideal in $\mathbb{Z}_6 = \{0,1,2,3,4,5\}$, where $I$ denoted an ideal:   $I = \{ \overline{0} \}$ $I = \{ \overline{0} , \overline{2} , \overline{4} \}$ $I = \{\overline{0} , \overline{3} \}$   I have two questions here:   How to find an example of a subring that is not necessarily an ideal? Are there any examples in $\mathbb{Z}$? How do we prove that $I = \{ \overline{0}, \overline{3} \}$ is an ideal of $\mathbb{Z}_6$, above?
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..
81375
In $\mathbb{F}[x]$, where $\mathbb{F}$ is a ﬁeld, let $J$ be the set of elements of polynomials that have coefficients that add to zero (so $a_0 + a_1 + ... + a_n = 0$). Show that $J$ is an ideal of $\mathbb{F}[x]$. I know that the proof of this statement is meant to be very short, but I don't know how to go about it.
95421
Is the ideal generated by $2$ and $x$ in $\mathbb{Z}[x]$ all polynomials except for the odd constants?  And is this ideal a non-maximal, prime ideal?
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.
120231
Say I have some function $f:R \rightarrow S$ such that $f$ is a ring homomorphism and $J$ is an ideal of $f$. $I = f^{-1}(J)$ is an ideal of $R$, but I don't really understand why. $J$ being an ideal means that $I$ is non-empty, that much makes sense.  But how can I show that addition/absorption hold if $f$ isn't surjective, nor do I know if $f^{-1}$ is even a ring homomorphism? I've seen a very similar question has been asked, but the only response appeared to just prove that these things held in $J$.  For instance, if I want to prove addition holds, I could take $i_1, i_2 \in I$ so that for $j_1, j_2 \in J$, $i_1 = f^{-1}(j_1)$, $i_2 = f^{-1}(j_2)$. But then I can't really add these elements together because I don't know if addition is preserved in $f^{-1}$. What gives?
74255
I'm supposed to show in a part of an exercise that if we have a ring $R$ that is a principal ideal domain, then for any ideal $I$ in $R$, $R/I$ will also be a PID.  So $I=(i)$ for some $i \in R$, and $R/I$ is $\{rI |r\in R\}$, and we know:   An ideal $J$ in $R/I$ will be an ideal $J$ in $R$ containing $I$.  So if $I \subset J$ and $I=(i)$, then $J$ should be $(j)$, where $j|i$. but wherefrom do I get the necessity that $J=(j)$, so that $J$ is also principal? and there will be several $j$ that divide $i$, but should it not be only one?  And then I shall use this to show a 1-1 correspondence between ideals of $\mathbb{Z}$ and $\mathbb{N}$.  Can someone support with with the proofs?