I want to find an example of ideal $Q$ such that $\sqrt{Q}$ is prime, but $Q$ is not primary.    It is clear that our domain would not be PID because $\sqrt{Q}$ should not be maximal ideal.
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?
82026
Suppose that $R$ is an integral domain. How to show that if every non-zero prime ideal of $R$ is invertible, then every non-zero ideal of $R$ is invertible?    Actually, I am trying to prove the unique prime factorization of ideals in $R$, so I cannot use the fact that any ideal can be factored as a product of primes.
46459
Every principal ideal domain $D$ satisfies  (ascending chain condition on principal ideals) Proof. Let $(a_1) ⊆ (a_2) ⊆ (a_3) ⊆ · · ·$ be a chain of principal ideals in $D$. It can be easily verified that $I = \displaystyle{∪_{i∈N} (a_i)}$ is an ideal of $D$. Since $D$ is a PID, there exists an element $a ∈ D$ such that $ I = (a)$. Hence, $a ∈ (a_n)$ for some positive integer $n$. Then $I ⊆ (a_n) ⊆ I$. Therefore, $I = a_n$. For $t ≥ n$, $(a_t) ⊆ I = (a_n) ⊆ (a_t)$. Thus, $(a_n) = (a_t)$ for all $t ≥ n$.  I have prove $I$ is an ideal in the following way:- Let $ x,y\in I$. Then there exist $i,j \in \mathbb{N}$ s.t. $x \in (a_i)$ &amp; $y \in (a_j)$. Let $k \in \mathbb{N}$ s.t $k&gt;i,j$. Then  $x \in (a_k)$ &amp; $y \in (a_k)$. as $(a_k)$ is an ideal $x-y \in (a_k)\subset I$ and $rx,xr \in (a_k)\subset I$. So $I$ is an ideal. Is it correct?
12703
Let $\alpha:A\to B$ be a ring homomorphism, $Q\subset B$ a prime ideal, $P=\alpha^{-1}(Q)\subset A$ a prime ideal. Consider the natural map $\alpha_Q:A_P\to B_Q$ defined by $\alpha_Q(a/b)=\alpha(a)/\alpha(b)$. Suppose that $\alpha$ is injective. Then is $\alpha_Q$ always injective?    I think so, but I'm clearly being too dense to prove it! My argument goes as follows.  Let $\alpha(a)/\alpha(b)=0$. Then $\exists c \in B\setminus Q$ s.t. $c\alpha(a)=0$. If $B$ is a domain we are done. If not we must exhibit some $d\in A\setminus P$ s.t. $da=0$. Obviously this is true if $c =\alpha(d)$. But I don't see how I have any information to prove this!  Am I wrong and this is actually false? If so could someone show me the trivial counterexample I must be missing?   Many thanks!
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!!!
105949
Let $R \subset S$ be an integral extension of domains and $\mathfrak p \subset R$ a prime ideal. Can it be the case that there are infinitely many distinct primes ${\cal P} \subset S$ such that ${\cal P} \cap R=\mathfrak p$?    Certainly this is impossible if $S$ is a Dedekind domain, because the primes lying over $\mathfrak p$ are the primes of $S$ occurring in the factorization of $\mathfrak p$ over $S$.  I don't have much of an intuition for integral ring extensions that aren't number fields, so past this I'm not particularly sure.
18583
In a commutative ring, can I say   An ideal $\mathfrak q$ in a ring $A$ is primary if $\mathfrak q \neq A $ and if $ xy \in \mathfrak q \Rightarrow $ either $ x \in \mathfrak q$ or $y^n \in \mathfrak q $ for some $n &gt; 0$ or either $ y \in \mathfrak q$ or $x^m \in \mathfrak q $ for some $m &gt; 0$.  Why I am asking this question is if I use the above definition, and use the ideal $P^2$ in the question , then $\bar{x}\bar{y} = \bar{z}^2 \in P^2 \text{ and } \bar x^2 \in P^2 $ gives a contradiction to the fact that $P^2$ is not primary.  Where my understanding went wrong?
15038
A maximal ideal  is always a prime ideal, and the quotient ring  is always a field. In general, not all prime ideals are maximal.      In $2\mathbb{Z}$, $4 \mathbb{Z} $ is a maximal ideal. Nevertheless it is not prime because $2 \cdot 2 \in 4\mathbb{Z}$ but $2 \notin 4\mathbb{Z}$. What is that is misunderstand?
67788
Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime. If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with a ring homomorphism $R\to F$. Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f\colon R\to F$, where $F$ is a field, such that $f(x)=0$ for all $x\in I$ and $f(y)\neq 0$ for all $y\in S$.   [You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0\notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]  Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.   For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that?
55902
Let $d \in \mathbb{Z}$ be a square free integer.  $R=\mathbb{Z}[\sqrt d]$ =$\{a+b\sqrt d | a,b \in \mathbb{Z} \}$. Overall, I'm trying to show that every prime ideal $P \subset R$ is a maximal ideal.     So far I showed that $I \subset R$ is finitely generated.  $I=\{(x,s+y \sqrt d)\}$ And now I'm trying to show that R/P, for some prime ideal is a finite ring with no zero divisors.  From there it would follow that R/P is a field and any prime ideal is maximal.   I know R could also be written as $R= \mathbb{Z}[x]/(x^2-d)$.  How could I show that R/P is a quotient of $\mathbb{Z}/n\mathbb{Z}[x]/(x^2-d)$?
54424
Maybe this question is too obvious to everyone.      In Noetherian ring $R$, ideal $I\subset R$, $P\in\mathrm{Ass}(I)$, then $P=I:a$ for some $a\in R$ where $\mathrm{Ass}(I)$ denotes associated primes of $I$.   This is an exercise 8.29 in Gathmann's Commutative Algebra notes. I am supposed to find this statement obvious. However, up to now, I still did not find this statement obvious. Can someone explain intuitively why noetherianness here allows me removing radical $\sqrt{I:a}$ without going into details?  If $Q$ is a primary ideal of $I$, then certainly $\sqrt{Q:a}=P$ for $a\not\in P$. It seems taking $(p_1,\dots, p_n)=P$, $p_i^{n_i}a\in Q$. Since $Q$ primary, I must have $p_i\in Q$ and $p_i^{n_i}\in Q$. (This argument seems very wrong as this deduces the same conclusion without noetherian property.)
118290
I know that if (p) is maximal then Z/(p) is a field which implies its an integral domain which then implies that (p) is a prime ideal and p is prime. I'm having trouble going the forward direction assuming that p is prime and then showing it's a prime ideal.
8897
Let $K$ be a field. Then $K[x,y]$ is not a PID because $x$ is irreducible but the quotient $K[x,y]/(x)$ is isomorphic to $K[y]$, which is not a field.  So there must be an ideal in there which is not principal. What is an example of such an ideal?
39458
I'd like to show that for $p,q$ distinct primes, the extensions $\mathbb Q(\sqrt p),\mathbb Q(\sqrt q)$ are not isomorphic. I don't really have knowledge of the "high-level language" of algebraic number theory, so my approach is going to be very bit-wise, if you will.  Here's my try:  If there were such an isomorphism, call it $\varphi$. I know that we must have: $$\left\{\begin{align}&amp;\varphi\mid_{\mathbb Q} = \text{id} \\ &amp;\varphi(\sqrt p) =\pm\sqrt p \end{align}\right.$$  So essentially the existence of $\varphi$ means that for any $a+b'\sqrt p$ there exist $c,d$ such that $\varphi(a+b'\sqrt p) = a+b\sqrt p = c+d\sqrt q$ (where $b$ might be a sign change of $b'$). Take, to simplify $a=0, b=1$. Let $\alpha = \sqrt p - d\sqrt q = c$, from above. We can rearrange to obtain $$\begin{align}&amp;(\alpha-\sqrt p)^2 = d^2q \Rightarrow \\ \Rightarrow&amp; \ \alpha^2 - 2\alpha\sqrt p + p - d^2q = 0 \\ \Rightarrow&amp; \ \sqrt p -\frac{c^2+p-d^2q}{2c} = \sqrt p - \lambda = 0\end{align}$$  But then it has to be that $(x^2-p) \mid (x-\lambda)$, which cannot be.  This proof however doesn't satisfy me for two reasons. First of all, I'd like to have stopped at the polynomial I found in $\alpha$,  shown that it was irreducible in $\mathbb Q(\sqrt2)$, and argued in terms of extension degrees. However I don't know how to show that. The second reason is that I'm not using the primality of $p,q$ other than having it imply the irreducibility of $x^2-p$. It seems to me from the way the exercise is worded that if they were compound and $x^2-p,x^2-q$ were still irreducible, we might have a homomorphism.  Can someone clarify?
21154
I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is principal. I already proved that if R and S are ring every ideal in R x S is I x J with ideals in the original ring. But I cant follow from that that $\mathbb Z\times\mathbb Z$ has only principal ideals.      Explicitly, if $I$ is an ideal $(a)\times(b)$ which would be its generator $(c,d)$ in $\mathbb Z\times\mathbb Z$?    Thanks.
105287
I am trying to prove $(a)$ is a maximal ideal in $\mathbb{Z}$,  if and only if $a$ is prime number.  Now I wrote: assume $a \in\mathbb{Z}$, while it's not prime number we can write as $a=xy$, for some integers $x$ and $y$. then $(a)\subset (x)$ and $(a)\subset (y)$  while if $(a)$ is maximal ideal, it's not exist $(k)$ such that $(a)\subset (k)\subset\mathbb{Z}$. so $(a)$ cannot be maximal ideal. we can say by contradictory if $a$ is prime, $(a)$ is maximal ideal.  is this correct?  and how to prove the other way.
73545
I'm giving a seminar talk on the ideal class group, and am looking for an example of a fractional ideal that is not invertible. Does know of a simple example in say $k[x,y]$ or $\mathbb{Z}[\sqrt{-3}]$. Such examples must exist but computation there is very hard.   Thanks.