Are there any commutative rings in which no nonzero prime ideal is finitely generated? I feel like the example (or proof of impossibility) ought to be obvious, but I’m not seeing it.

Let $R$ be a commutative ring with unity , $M$ be a finitely generated module over $R$ , let $N,P$ be submodules of $M$ such that $P\subseteq N \subseteq M$ and $M\cong P$ , then is it true that $M\cong N$ ? If not true , then what happens if we also assume that $M$ […]

I was looking for a proper subgroup of $\mathbb Q$ which is not finitely generated under the addition operation. We know every finitely generated subgroup of $\mathbb Q$ is cyclic. For a proper subgroup I am just thinking about the subgroup $H$ generated by $\{\frac{1}{p} : p \text{ prime }\}$ may work. It seems $1/4$ […]

I need some help on this exercise from A Course in Ring Theory by Donald S. Passman Find all finitely generated graded $K[x]$-modules up to abstract isomorphism. Remember, $K[x]$ is a principal ideal domain. The result is supposedly similar to the well-known structure theorem in the non-graded case. So let $M$ be a finitely generated […]

This question already has an answer here: $A\oplus C \cong B \oplus C$. Is $A \cong B$ when $C$ is finite, A and B infinite. 1 answer

Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$. Let $M$ be a finitely generated module. Then there exists an element $x\in M$ such that $\frac{x}{1}\not\in\mathfrak m_iM_{{\mathfrak m}_i}$ for every $i=1,\dots,n$. I cannot prove that such an element there exists. I was trying to prove it by induction on $n$. […]

Suppose $G$ is any group and $X$ is the set of all elements of order $p$ in that group where $p$ is a prime. Prove that if $X$ is finite, then $\langle X\rangle$ is finite, where $\langle X\rangle$ is group generated by the elements of $X$. If possible, I’m interested in a solution that assumes […]

$\fbox{1}$ Prove that any finitely generated subgroup of $(\mathbb{Q},+)$ is cyclic. $\fbox{2}$ Prove that $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}\times \mathbb{Q}$. Any hints would be appreciated.

If over a commutative ring $R$ we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ are finitely generated projective? We have finite generation, because if $M\otimes N$ is generated by $\sum_i a_{ij} x_i^j\otimes y_i^j$, then we have $x_{i_1}^{j_1}\otimes y_{i_2}^{j_2}\otimes x^{j_3}_{i_3}$ generating $M^n$. Projecting onto $M$, we see that $x_i^j$ generate […]

Let $G$ a finitely generated group, $\mathrm{Aut}(G)$ is its automorphism group, then it is necessary that $\mathrm{Aut}(G)$ is a finitely generated group? Thanks in advance.

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