Let $A $ be a commutative ring with identity. Let $f,g\in A [x] $. Let $I_1,I_2, J $ be the ideals generated by the coefficients of $f,g,fg $ respectively. Must $J $ be equal to $I_1 I_2$ ?  It is an exercise an Atiyah and Macdonald to show that this true in the case  $I_1=I_2=A $.  Thank you
109459
Define the content of a polynomial (over an arbitrary commutative ring $A$) to be the ideal generated by its coefficients, denoted $c(f)$. I want to show that   $$c(fg) = c(f)c(g).$$      (I was told this is true.)    What I was able to show was that $c(fg) \subseteq c(f)c(g)$ (this is obvious), and that their radicals are the same. My reasoning for the latter was as follows: let $f = a_0 + \dotsb + a_nx^n,$ $g = b_0 + \cdots + a_mx^m,$ and consider the matrix  $$ \begin{pmatrix}  a_0 b_0 &amp; \cdots &amp; a_0b_m\\ \vdots &amp; \ddots &amp; \vdots\\ a_nb_0 &amp; \cdots &amp; a_nb_m \end{pmatrix} .$$  Then $c(f)c(g)$ is generated by the entries of this matrix, while $c(fg)$ is generated by the sums along the diagonals. Let $\mathfrak p$ be a prime ideal of $A$ not containing $c(f)$ or $c(g)$, and let $i,\,j$ be minimal such that $a_i,\, b_j \notin \mathfrak p$.  Then all the terms in the generator of $c(fg)$ corresponding to the coefficient of $x^{i+j}$ are in $\mathfrak p$ except for $a_ib_j$, so that $c(fg) \not\subset\mathfrak p$. It follows that $c(fg) \subseteq \mathfrak p \Longleftrightarrow c(f)c(g) \subseteq\mathfrak p$, and hence $\sqrt{c(fg)} = \sqrt{c(f)c(g)}$.     Can I go all the way and show that in fact $c(fg) = c(f) c(g)$?
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
65669
Let $I$ be any ideal of $\mathbb {Z}[x]$, and $I=\left&lt;{f_1,f_2 , ...}\right&gt;$.  Without loss of generality, $f_1$ has minimal (absolute value) coefficient of constant term (if there exist).  Then, let $g$ be any element of $I$ .  We have for some integer $m$ that $g - mf_1$ is element of $I$ (minimal coefficient and division algorithm).  I guess this process terminate finitely, but I don't know how can I do next step.  ( ex. $g-mf_1, g-mf_1-nf_2, ...$)  Please help me and thanks for in advance.
49646
This is question #66 from        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 (that is, all elements except the additive identity have multiplicative inverses).   III. $R$ is infinite.     Here is my reasoning: Because $R$ is a ring, $R$ is also a additive group with some identity element $0$. We have a theorem that says $0r = 0 $ in any ring, so $\{0\}$ is a right ideal of $R$. Also, $R$ is a right ideal of $R$. Now I have found two different right ideals and there mustn't be any more.     Edit: As mentioned in the comments, the example below is not a ring, so it is not applicable to the problem. I could not fix it by taking additive closure because that introduced more than two ideals.  A possible candidate for $R$ could be the set of $2\times2$ matrices $\{0,I,-I,a,-a\}$ where $a = [[^1_0] ,[^0_0]]$. The only right ideals are $R$ and $\{0\}$. This ring satisfies only property I, but the answer key says that II is the correct answer.
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.
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.
31364
Let A be a commutative ring with 1.   1) Prove that a sum of a nilpotent element and an invertible element is invertible.   2) Prove that if $f=a_0+a_1x+\dots+a_nx^n \in A[x]$  a) $\exists f^{-1}\in A[x] \Leftrightarrow a_0$ is invertible and the other coefficients are nilpotent.  b) f is nilpotent $\Leftrightarrow $ all its coefficients are nilpotent.  p.s. Those are the first two in a series of problems. The rest easily follow from each other. I'm only struggling with the first two.
45456
Let $R$ be a ring and let $I_1,\dots,I_n,J\subseteq R$ be ideals such that $J\subseteq I_1\cup\dots\cup I_n$. Suppose that there exists an infinite field $K$ and a homomorphism $\phi:K\to R$. Prove that $\exists k$ s.t. $J\subseteq I_k$.   Attempt: (Edited)  If $J\subseteq I_1$ then we're done. Otherwise, let $x\in I_1$ and let $y\in J\setminus I_1$. $\forall r\in K, x+ry\notin I_1$ (because otherwise we'll get $y\in I_1$).  $K$ is infinite so we have infinitely many elements of the form $x+ry\in J$. So there's some $k$ for which $I_k$ contains infintely many elements of the former form.  If $n=2$ so $x+y\notin I_1$ implies $$x+y\in I_2\Rightarrow x\in I_1\Rightarrow I_1\subset I_2\Rightarrow J\subseteq I_2$$  Suppose the theorem holds for some $n$. $x+y\notin I_1\Rightarrow \exists k,I_k\ni x+ry$ for infintely many elements as above.
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.
52999
Let $A$ be a commutative ring, $M$ an $A$-module and $N_1, N_2$ two submodules of $M$.  If we have $M/N_1 \cong M/N_2$, does this imply $N_1 \cong N_2$?  This seems so trivial, but I just don't see a proof... Thanks!
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.
27785
So I basically have to prove what is on the title.      Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.   One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.   I am having troubles proving the other inclusion. Any comments?  Thanks
19973
Let $R$ be a commutative ring and $I_1, \dots, I_n$ pairwise comaximal ideals in $R$, i.e., $I_i + I_j = R$ for $i \neq j$. Why are the ideals $I_1^{n_1}, ... , I_r^{n_r}$ (for any $n_1,...,n_r \in\mathbb N$) also comaximal?
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.
98611
Let $R$ be a commutative ring with unity and $M$ an $R$-module.   If $M\oplus R^m\cong R^n$ for some integers $m,n \geq 1$ then must $M$ be finitely generated and free?   Can somebody help me with this? Regards.
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?
63833
Let $R$ be a commutative ring with identity and $I$ and $J$ be ideals in $R$.     Show that $R/I\otimes_R R/J\cong R/(I+J)$ as $R$-modules.   This is what I tried.  Define a map $f:R/I\times R/J\to R/(I+J)$ as $f(\bar r, \bar s)=rs+(I+J)$. This is a well-defined bilinear map.  By the universal property of tensor products, we get an $R$-linear map $F:R/I\otimes_R R/J\to R/(I+J)$ which sends $\bar r\otimes \bar s$ to $rs+(I+J)$.  This map is surjective.  So all we need to show is that the kernel of $F$ is $0$.  But I am unable to do this.