Suppose $Q$ is a $P$-primary ideal and $Q'$ is a $P'$-primary ideal such that $P$ and $P'$ are comaximal in the Noetherian ring $R$.  Show that $Q$ and $Q'$ are comaximal.   Proof.  Since $Q$ is a $P$-primary ideal and $Q'$ is a $P'$-primary ideal, then $P=\sqrt{Q}$ and $P'=\sqrt{Q'}$.  Since $P$ and $P'$ are comaximal, then $P+P'=R$.    I also have a proposition that says if $R$ is a Noetherian ring, then for any ideal $I$ some positive power of $(\sqrt{I})$ is contained in $I$.  That is, for positive $p$, $(\sqrt{I})^p \subset I$.  I want to show $Q+Q'=R$.  Here's what I have so far:  $Q+Q'\subset\sqrt{Q}+\sqrt{Q'}=P+P'=R$   As for the other direction I'm having a bit of trouble.  I have:  $R=P+P'=\sqrt{Q}+\sqrt{Q'}$ I want this to be a subset of $(\sqrt{Q} )^q +(\sqrt{Q'})^{q'} \subset Q+Q'$ and then I would be done.    I'm studying the section about Dedekind domains right now in class so I don't know if that helps?
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?
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?
56135
Let $p,q,r$ be prime integers with $q\neq r$. Let $\sqrt[p]q$ denote any root of $x^p-q$ and $\sqrt[p]r$ denote any root of $x^p-r$. Please show that $\mathbb{Q}(\sqrt[p]q)\neq\mathbb{Q}(\sqrt[p]r)$.  By Eisenstein we know the two polynomials are irreducible and so they are minimal polynomials and both have degree $p$.
81640
Let R be a commutative ring, and let P be a prime ideal of R. Suppose that P has no nontrivial zero divisors in it. Show that R is an integral domain.   My proof:  Take $r,s,a \in R$ with $ar = as$, and $p \in P$. Then  \begin{align*} par &amp;= pas\\ a p r &amp;= a p s &amp;&amp; \text{R is commutative} \end{align*} and since $P$ is an ideal of $R$, then $pr, ps \in P$. Since $P$ is an integral domain, then $a p r = a p s \implies p r = p s$, so $r = s$.  I'm not using the fact that $P$ is a prime ideal which leads me to believe that I messed up a step.
45575
I want to prove the  over a field are prime ideals. To be concrete:      For simplicity, let $I=(x_{11}x_{22}-x_{12}x_{21},x_{11}x_{23}-x_{13}x_{21},x_{12}x_{23}-x_{13}x_{22})$ be an ideal of the polynomial ring $k[x_{11},\ldots,x_{23}]$. I have no idea how to prove that $I$ is a radical ideal (i.e. $I=\sqrt{I}$). Could anyone give some hints?   Generally, let $K$ be an algebraically closed field, then $\{A\mid\mathrm{Rank}(A)\leq r\}\subseteq K^{m\times n}$ is an irreducible algebraic set (I first saw this result from ). And I tried to prove this by myself, then I have proved it (when I see the "Segre embedding").      But I have no idea how to show that the "determinantal ideals" are radical ideals (I hope this is true). BTW, is the statement that the determinantal ideals over a field are prime ideals true ?   Thanks.
49019
The question may have been addressed in Math SE before; but my question is a little variation of it, coming because of a very common example for this question.  It is well known that powers of a prime ideal need not be primary and a standard example is to consider the quotient of $k[x,y,z]$ by the ideal $(xy-z^2)$, and in the quotient ring, take the prime ideal $(\bar{x},\bar{z})$; its square is not primary.  Now this example of prime ideal is in quotient of polynomial ring.   Q. Can we construct an example of a prime ideal within a polynomial ring, say $k[x,y]$ or $k[x,y,z]$, whose some power is not primary?
119724
Let $k$ be a field of characteristic $\neq2$ and consider $R=k[x,y]/(y^2-x^3+x)$. Then  (a) Show that $R$ is normal.  (b) Let $P=(x,y)$ be a prime ideal of $R$. Show that $P^2$ is a principal ideal.  (c) Show that $P$ is not a principal ideal.   I came across this in some commutative algebra reading recently. Here are my thoughts so far (really only for part (c)). Assume that $P$ is generated by $\alpha=f(x)+g(x)y$. Then if $\sigma$ is the nontrivial element in the Galois group, $\sigma(\alpha)$ will generate $\sigma(P)=P$, so the norm of $\alpha$ will generate $P^2$. I thought to compare this supposed generator with the one found in part (b) and then arrive at a contradiction.     Any help with this problem would be appreciated!
26946
Let $R$ be a commutative Noetherian ring with unit and let $I$ be a fixed ideal. I am sorry if the following turns out to be a very silly question.  1) Suppose $\operatorname{Ass}(R/I)$ are all minimal, then what can we say about $R/I$?  One conclusion is that $I$ has a unique primary decomposition. If $R/I$ is CM, then $\operatorname{Ass}(R/I)$ are all minimal, but I don't think the converse is true. Can someone provide a counterexample?  2) Suppose $\operatorname{Ass}(R/I)$ are all minimal and have the same height. I still dont think we can conclude that $R/I$ is CM. Can someone provide a counterexample? Theorem 2.1.6 of Bruns and Herzog, CM Rings, gives a result in this direction but it needs additional hypothesis. It says that if every ideal $I$ is generated by height $I$ elements and $\operatorname{Ass}(R/I)$ are all minimal, then $R$ is CM.  It would be nice if someone can say something nice about the ring in the special situation described in 1) and 2).  Thanks.
74118
Show that $I=(X_{k_1}^{a_1},...,X_{k_s}^{a_s})$ is $(X_{k_1},...,X_{k_s})$-primary.   I noticed  that $$\sqrt{({X_{i_1}}^{a_1},...,{X_{i_k}}^{a_k})}=\sqrt{({X_{i_1}}^{a_1})+\cdots+({X_{i_k}}^{a_k})}=\sqrt{\sqrt{(X_{i_1})^{a_1}}+\cdots+\sqrt{(X_{i_k})^{a_k}}}=\sqrt{(X_{i_1})+\cdots+(X_{i_k})}=\sqrt{(X_{i_1},...,X_{i_k})}=(X_{i_1},...,X_{i_k}).$$ I have to show that $I$ is $P$-primary (where $P=(X_{k_1},...,X_{k_s})$), i.e., if $u v\in I$ then $u\in I$ or $v\in P$. I tried to prove that if $u\notin I$ and $v\notin P$ then $uv\notin I$. (There is a similar question, but the answer is too hard for me.)
98341
Suppose $I$ is the ideal $(xy, yz, zx)$ in $R =\Bbb R[x, y, z]$. I want to compute the primary decomposition of $I$.    I have viewed many post on this topic, as I suspect, the primary decomposition of $I$ would be:  $$ (xy, yz, zx)= (x,y)\cap(x,z)\cap(y,z),$$  but I cannot show how.
98754
Consider the subring $R\subset\Bbb Q$ , $R=\{\frac ab\ | \ a,b\in\Bbb Z, b\text{ odd}\}$  I am struggling with the following questions:  (1) Prove that the ideals of $R$ are $\{0\}$ and $2^nR$ for $n\ge 0$  (2) Prove that $R$ has 2 prime ideals and 1 maximal ideal  For (1) I can see that $2^nR$ is an ideal for $n\geq0$ but I'm not sure how to show all ideals other than $\{0\}$ are of this form  With question (2),  $2^nR\subsetneq 2^{n-1}R\subsetneq...\subsetneq 2R\subsetneq R$  so $2R$ is clearly the only maximal ideal and is therefore prime, but I am not sure where to go from here.  Any suggestions?
111549
If $R$ is a   UFD, then every prime ideal of height 1 in $R$ is principal.   I find this is a direct consequence of Kaplansky's theorem. But the proof of Kaplansky's theorem is difficult for me.  So I wonder how to prove the statement directly.
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?
33128
Prove that the set of nilpotent elements of a ring is the intersection of its prime ideals.   I know these two useful facts:  {nilpotent elements}$=\sqrt{0}$  $\sqrt{I}= \bigcap$ of prime ideals containing I   First, I need to show that the nilpotent elements are in fact an ideal and then I need to show it's prime.   Any specific suggestions would be appreciated. I just don't know how to make this proof flow and exactly what leads to what.   I read through this question and comments as well:
121085
Let $p,q$ be primes, $p≠q$, then I have to show that $\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})$  So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient.  This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields.  Could someone maybe give me a hint about how to proceed? Thanks a lot!!!
101318
Let $S$ be a subring of $\mathbb{Q}$ containing $\mathbb{Z}$. Prove that it is a principal ideal domain.  So here is what I tried. Take any ideal $I\subset S$. Take any two elements, say $a=p/q, b=r/s$ in $I$. Rationalize denominators (we can do that since $\mathbb{Z}\subset S$. Now we have $aqs=p=K(r)=K(bqs)+R$, that is, just apply the Euclidean Algorithm. Now, $a-Kb=R/qs\in I$, so take the set of all such remainders $R/qs$.   There are two possibilities. Either this set has a smallest positive element, or it does not.   Suppose it has a smallest such element, call it $x=m/n$. Let $a=p/q\in I$. We show $a=lx$. For suppose not. Rationalizing denominators with $n, q&gt;1$, $p=Lm+Z$, so then $Z&lt;m$ so $Z/qn&lt;m/n=x$, giving a contradiction.   Now though, I am stuck on the case where there is no such smallest element :(  Any thoughts?  EDIT: For those marking it as a duplicate: The first answer in the linked question, as mentioned in the comments, feels somewhat unnatural and requires all this extra machinery. On the other hand, the second answer given is incomplete: The poster says "Now show that the ideal generated by $t=\frac{t}{1}$ in R is the ideal you started out with." But if $\frac{r}{s}\in I$, then there exists $k$ such that $r=kt$, so $\frac{r}{s}=\frac{k}{s}\cdot \frac{t}{1}$, but $\frac{k}{s}$ is not necessarily in $S$.
63122
The problem is this:      Suppose $I \subseteq R$ is a nilpotent ideal and there is $r \in R$ with $r \equiv r^2 \pmod I$. Show $r \equiv e \pmod I$ for some $e \in R$ idempotent.   I have spent a few hours rolling around in abstracta with no destination. I believe that if I could write down a concrete example of this, the example could guide me through the abstract definitions and show me where to look for an idempotent in $R$. I have looked at 2x2 matrices and could not find any such examples. Might anybody have one?