# Notes - Rings and Modules HT24, Isomorphism theorems

### Flashcards

State the first isomorphism theorem for rings.

Suppose:

- $\phi : R \to S$ is a ring homomorphism

Then:

- $\exists \overline\phi : R / \ker(\phi) \to \text{Im } \phi$ isomorphism

So

\[R/\ker\phi \cong \text{Im }\phi\]Quickly prove, by appealing to the universal property of quotients, the first isomorphism theorem, i.e. that if:

- $\phi : R \to S$ is a ring homomorphism

then:

- $\exists \overline\phi : R / \ker(\phi) \to \text{Im } \phi$ isomorphism

For reference, the universal property of quotients states that

- $R$ is a ring
- $\phi : R \to S$ any ring homomorphism
- $I \subseteq \ker \phi$
- $I \trianglelefteq R$
- $q : R \to R/I$ quotient homomorphism

then:

- $\exists ! \overline \phi : R/I \to S$ such that $\phi = \overline \phi \circ q$
- $\ker \overline \phi = \ker \phi / I = \{ m + I \mid m \in \ker \phi \}$

Apply universal property of quotients to $I = \ker\phi$. Then $\overline \phi : R / \ker(\phi) \to \text{Im } \phi$ follows, and it’s an isomorphism since $\ker \overline \phi = \ker \phi / I$, so we are done.

State the second isomorphism theorem, which relates $A + I$ and $A \cap I$.

Suppose:

- $R$ is a ring
- $A$ is a subring
- $I$ is an ideal of $R$

Then:

\[\frac{A}{A\cap I} \cong \frac{A + I}{I}\]Quickly prove, by appealing to the first isomorphism theorem, the second isomorphism theorem, i.e. that if

- $R$ is a ring
- $A$ is a subring
- $I$ is an ideal of $R$

then:

\[\frac{A}{A\cap I} \cong \frac{A + I}{I}\]

Overall idea: Consider the restriction of the projection map $q : R \to R/I$ to $A$.

Note that $A + I$ is a subring that includes $I$. Then consider the quotient map $q: R \to R/I$, and consider its restriction $p : A \to R/I$. Then $\text{Im } p = (A+I)/I$, so if we can show $\ker p = A \cap I$, then by the first isomorphism theorem, we are done. $\ker p = \{r \in A \mid r + I = 0\}$, but $r + I = 0 \iff r \in I$. Then $\ker p = A \cap I$.

State the third isomorphism theorem, which lets you sometimes “cancel” in quotients.

Suppose:

- $R$ is a ring
- $I _ 1 \subseteq I _ 2$ are ideals in $R$

Then:

\[\frac{R/I_1}{I_2/I_1} \cong R/I_2\]Quickly prove, by appealing to both the universal property of quotients and the first isomorphism theorem, the third isomorphism theorem, i.e. that if

- $R$ is a ring
- $I _ 1 \subseteq I _ 2$ are ideals in $R$

then:

\[\frac{R/I_1}{I_2/I_1} \cong R/I_2\]

For reference, the universal property of quotients states that

- $R$ is a ring
- $\phi : R \to S$ any ring homomorphism
- $I \subseteq \ker \phi$
- $I \trianglelefteq R$
- $q : R \to R/I$ quotient homomorphism

then:

- $\exists ! \overline \phi : R/I \to S$ such that $\phi = \overline \phi \circ q$
- $\ker \overline \phi = \ker \phi / I = \{ m + I \mid m \in \ker \phi \}$

We aim to find a homomorphism $p : R/I _ 1 \to R/I _ 2$ where $\ker p = I _ 2/I _ 1$. Let $q _ i : R \to R/I _ i$ be the quotient homomorphisms. Then by the universal property for $q _ 2$ applied to the homomorphism $q _ 1$, there exists a homomorphism $\overline q _ 2 : R/I _ 1 \to R/I _ 2$ with $\ker \overline q _ 2 = \ker q _ 2 / \ker q _ 1 = I _ 2 / I _ 1$, and it is surjective since $\overline q _ 2 \circ q _ 1 = q _ 2$ and $q _ 2$ is surjective.