Rings. Third isomorphism theorem.

Isomorphism theorem for rings (general)

Let $R$ and $S$ be a rings, and let $\phi :R\to S$ be a ring homomorphism. $\phi$ induces an ideal $\ker (\phi )$. Consider two-sided ideals contained in $\ker (\phi )$. Let $I\subseteq \ker (\phi )$ be one of these ideals. Then there is a ring homomorphism $\tilde{\phi }$ fully determined by $\phi$ and $I$ such that the following diagram commutes:

$\tilde{\phi }$ is defined as:

$$\tilde{\phi }(r+I)=\phi (r)$$

Third isomorphism theorem. Proof idea.

$J$ is the larger ideal and induces a smaller quotient ring $R/J$ through a homomorphism we will label as $\phi$. $J$ is the kernel of this homomorphism. $I$ is a subset of $J$ and is also sent to $0_{R/J}$ by $\phi$. If we consider $R/I$, we have a homomorphism $\tilde{\phi }$ that is fully determined by $\phi$ and $I$ that sends (many-to-one) cosets in $R/I$ to elements in $R/J$. It is the general isomorphism theorem that tells us this. We apply this process again to break the surjective $\tilde{\phi }$ into an isomorphism! We do this by identifying the kernel of the homomorphism, $\ker \tilde{\phi }$. It is the set of cosets in $R/I$ that are sent to $0_{R/J}$. These are isomorphic to $J/I$.

Example

$10\mathbb{Z}$ induces a quotient ring $\mathbb{Z}/10\mathbb{Z}$ with 10 elements. $30\mathbb{Z}$ is a subset of $10\mathbb{Z}$ and induces a quotient ring $\mathbb{Z}/30\mathbb{Z}$ with 30 elements. The three elements in $\mathbb{Z}/30\mathbb{Z}$ that are sent to $0_{\mathbb{Z}/10\mathbb{Z}}$ are $(0,10,30)$ and they are isomorphic to $10\mathbb{Z}/30\mathbb{Z}$.

First isomorphism theorem

In the proof above, the second application of the general isomorphism theorem is usually called the first isomorphism theorem for rings. It is recalled here:

A linear projection?

The theorem suggests a linear operation feel—that of projecting into the $R/I$ ideal:

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