Maximal ideals
-
- Community Team
- Posts: 426
- Joined: Mon Nov 09, 2015 1:52 pm
Maximal ideals
Let \(\displaystyle{\left(R,+,\cdot\right)}\) be a commutative ring with unity \(\displaystyle{1_{R}}\) .
Prove that, if \(\displaystyle{m}\) is a maximal ideal of this ring, then there exist a field \(\displaystyle{\mathbb{F}}\)
and an epimorphism \(\displaystyle{R\to \mathbb{F}}\) .
Conversely, if \(\displaystyle{R\to \mathbb{F}}\) is an epimorphism, where \(\displaystyle{\mathbb{F}}\)
is a field, then, find a maximal ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) .
Prove that, if \(\displaystyle{m}\) is a maximal ideal of this ring, then there exist a field \(\displaystyle{\mathbb{F}}\)
and an epimorphism \(\displaystyle{R\to \mathbb{F}}\) .
Conversely, if \(\displaystyle{R\to \mathbb{F}}\) is an epimorphism, where \(\displaystyle{\mathbb{F}}\)
is a field, then, find a maximal ideal of \(\displaystyle{\left(R,+,\cdot\right)}\) .
-
- Community Team
- Posts: 314
- Joined: Tue Nov 10, 2015 8:25 pm
Re: Maximal ideals
- Let \( \mathfrak{m} \) be a maximal ideal of \( R \). Then \( R/ \mathfrak{m} \) is a field and the canonical projection \( \pi \ \colon R \longrightarrow R / \mathfrak{m} \ , \ r \mapsto r + \mathfrak{m} \) is a ring epimorphism.
- Let \( f \ \colon R \longrightarrow \mathbb{F} \) be a ring epimorphism, where \( \mathbb{F} \) is a field. By the first isomorphism theorem, we have that \( R / \mathrm{Ker}(f) \cong \mathrm{Im}(f) = \mathbb{F} \), since \( f \) is an epimorphism. Therefore \( \mathrm{Ker}(f) \) is a maximal ideal of \( R \).
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 7 guests