Rings. Third isomorphism theorem, an example.
Let \( a, b \in R \) be elements in a commutative ring \( R \). Use \( ( \bar{b} ) \) to denote the equivalence class of \( b \) in \( R/(a) \). Then firstly we have:
Recap of ring generators:
Ring generators
Let \( a \in R \) be an element of a ring. The subset \( Ra \) is a left-ideal and the subset \( aR \) is a right-ideal. If \( R \) is a commutative ring, the ideals coincide, and the syntax \( (a) \) is used to denote the ideal.
The third isomorphism theorem:
Third isomorphism theorem for rings.
Let \( R \) be a ring. Let \( I \) and \( J \) be ideals of \( R \), with \( I \) being a "smaller" ring contained within \( J \). Then:
- \( J/I \) is an ideal of \( R / I \)
- \[ R / J \cong \frac{R/I}{J/I} \]
And really, it's the general case that one should commit to intuition:
Isomorphism theorem for rings (general)
Let \( R \) and \( S \) be a rings, and let \( \varphi : R \to S \) be a ring homomorphism. \( \varphi \) induces an ideal \( \operatorname{ker}(\varphi) \). Consider two-sided ideals contained in \( \operatorname{ker}(\varphi) \). Let \( I \subseteq \operatorname{ker}(\varphi) \) be one of these ideals. Then there is a ring homomorphism \( \widetilde{\varphi} \) fully determined by \( \varphi \) and \( I \) such that the following diagram commutes:
\( \widetilde{\varphi} \) is defined as: