- If \( \mathfrak{a} \) is a homogeneous ideal of \( B \), then \( \sqrt{\mathfrak{a}} \) is also a homogeneous ideal of \( B \).
- Let \( \mathfrak{a} \) be a homogeneous ideal of \( B \). Then \( \mathfrak{a} \) is prime if and only if for all homogeneous elements \( a,b \in B \) the following implication holds: \( ab \in \mathfrak{a} \implies a \in \mathfrak{a} \text{ or } b \in \mathfrak{a} \)
Basic Ring Theory - 10
-
- Community Team
- Posts: 314
- Joined: Tue Nov 10, 2015 8:25 pm
Basic Ring Theory - 10
Let \( B \) be a \( \mathbb{Z} \)-graded ring. Show that:
Create an account or sign in to join the discussion
You need to be a member in order to post a reply
Create an account
Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute
Sign in
Who is online
Users browsing this forum: No registered users and 3 guests