The ideals tag has no usage guidance.

**-2**

votes

**0**answers

20 views

### Show that $(X_{k_1}^{a_1},…,X_{k_s}^{a_s})$ is $(X_{k_1},…,X_{k_s})$-primary [migrated]

Show that $I=(X_{k_1}^{a_1},...,X_{k_s}^{a_s})$ is $(X_{k_1},...,X_{k_s})$-primary, where $I$ is the ideal generated by monomials $X_{k_1}^{a_1},...,X_{k_s}^{a_s}$ .$\qquad$ ...

**2**

votes

**0**answers

80 views

### $T$-nilpotent ideals

Recall that a subset $I$ of a ring $R$ is left (resp.,
right) $T$-nilpotent in case for every sequence $$a_1,a_2,\cdots $$
in $I$ there is an $n$ such that $a_1\cdots a_n=0$ (resp.,
$a_n\cdots ...

**2**

votes

**0**answers

55 views

### some sort of 'saturation' of module quotients

Let $R$ be a local Noetherian ring over a field, with the maximal ideal $\mathfrak{m}$. (e.g. $R=k[[x_1,\dots,x_{p>1}]]$) Given two $R$-modules, $N\subset M$, of the same (finite, non-zero) rank. ...

**0**

votes

**0**answers

101 views

### Can it occur that $q^{ce}$ is a prime ideal (of $S$), while $q^{ce}\neq q $?

Let $R$ and $S$ be commutative rings (with $1$) and $f : R\to S$ be a ring homomorphism. For an ideal $I$ of $R$, set $I^e:=\langle f(I)S\rangle$ (called the extension of $I$ to $S$) and for an ideal ...

**0**

votes

**1**answer

128 views

### In what conditions every ideal is an extension ideal? Is every prime ideal extension of prime ideal?

Let $R$ and $S$ be commutative rings (with $1$), and $f : R\to S$ be a ring homomorphism. For an ideal $I$ of $R$, set $I^e:=\langle f(I)S\rangle$ (called the extension of $I$ to $S$). When $f$ is ...