Question

Let $I$ be an ideal of a commutative ring $R$ with a reduction $J=$ $\left(a_1, \ldots, a_n\right), I^{r+1}=J I^r$. If $r \geq n$ show that $\left(a_1^r, \ldots, a_n^r\right)$ is a reduction of $I^r$ and the reduction number is at most $n$. (A treatment, with additional references, of the reduction numbers of powers of ideals can be found in [Hoa93].)

   Let $I$ be an ideal of a commutative ring $R$ with a reduction $J=$ $\left(a_1, \ldots, a_n\right), I^{r+1}=J I^r$. If $r \geq n$ show that $\left(a_1^r, \ldots, a_n^r\right)$ is a reduction of $I^r$ and the reduction number is at most $n$. (A treatment, with additional references, of the reduction numbers of powers of ideals can be found in [Hoa93].)

Show more…

Integral Closure: Rees Algebras, Multiplicities, Algorithms

Integral Closure: Rees Algebras, Multiplicities, Algorithms

Wolmer Vasconcelos 1st Edition

Chapter 1, Problem 19

Instant Answer

Step 1

An ideal J is a reduction of an ideal I if J ⊆ I and there exists some integer k such that I^(k+1) = JI^k. The smallest such k is called the reduction number of I with respect to J. Step 2: We are given that J = (a₁, ..., aₙ) is a reduction of I with I^(r+1) = Show more…

Show all steps

lock

Thanks for your feedback!

Let $I$ be an ideal of a commutative ring $R$ with a reduction $J=$ $\left(a_1, \ldots, a_n\right), I^{r+1}=J I^r$. If $r \geq n$ show that $\left(a_1^r, \ldots, a_n^r\right)$ is a reduction of $I^r$ and the reduction number is at most $n$. (A treatment, with additional references, of the reduction numbers of powers of ideals can be found in [Hoa93].)

Play audio

Feedback

Powered by NumerAI

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later