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Timeline for Monomial order and prime ideals

Current License: CC BY-SA 4.0

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Mar 22, 2022 at 22:45 comment added pinaki Well, it is possible that $I$ is prime, but $in_<(I)$ is radical, but not prime, e.g. when $I$ is the ideal generated by $x_1x_2 - x_3^2$ and $<$ is a monomial order such that $x_1x_2 < x_3^2$.
Mar 22, 2022 at 20:56 comment added Wágner Badilla Yes, see Lemma 1.2.5. Book A singular introduction to commutaitva algebra.
Mar 22, 2022 at 20:48 comment added markvs Is that proved somewhere?
Mar 22, 2022 at 20:45 comment added Wágner Badilla The global monomial order implies well order.
Mar 22, 2022 at 20:44 comment added Wágner Badilla $x^{\alpha}=x_1^{\alpha_1}\cdots x_n^{\alpha_n}$ with $\alpha=(\alpha_1,\ldots,\alpha_n) \in \mathbb{Z}_{\geq 0}^n$, $\alpha \not=0$.
Mar 22, 2022 at 20:17 comment added markvs I do not know what is $x^\alpha$ (or even what is $\alpha$)? And why does it imply "well order".Total order just means linear order.
Mar 22, 2022 at 19:40 comment added Wágner Badilla Is that $<$ is a total order, $1<x^{\alpha}$, and $<$ respects multiplication by monomials.
Mar 22, 2022 at 19:37 comment added markvs What is a "global monomial order"?
Mar 22, 2022 at 19:28 history edited Wágner Badilla CC BY-SA 4.0
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Mar 22, 2022 at 19:26 comment added Wágner Badilla I assume that $<$ is a global monomial order. Thus, $<$ is a well order.
Mar 22, 2022 at 19:23 comment added markvs There are two different definitions of "monomial order". Do you assume that $<$ is a well order?
Mar 22, 2022 at 19:18 comment added Wágner Badilla To be specific. Suppose $in_<(I)$ is a radical ideal. Let $P_1, \ldots, P_t$ be the minimal prime ideals of $I$. Then, $in_<(P_1), \ldots,in_<(P_t)$ are the minimal prime ideals of $in_<(i)$. Is this true?
Mar 22, 2022 at 4:55 comment added markvs Depends on "<"?
S Mar 22, 2022 at 2:54 review First questions
Mar 22, 2022 at 7:31
S Mar 22, 2022 at 2:54 history asked Wágner Badilla CC BY-SA 4.0