07-03. Ring homomorphisms and quotient rings

Definiiton. ring homomorphisim

• $R$, $S$,
  • ring
• $\phi: R \rightarrow S$,

$\phi$ is said to be ring homomorphism if

• (i) $\phi(a + b) = \phi(a) + \phi(b)$
• (ii) $\phi(ab) = \phi(a)\phi(b)$

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Definiiton. kernel of ring homomorphisim

• $R$, $S$,
  • ring
• $\phi: R \rightarrow S$,
  • ring homomorphisim

The kernel of the $\phi$, denoted $\mathrm{ker}\phi$, is

\[\mathrm{ker}\phi := \{ x \in R \mid \phi(x) = 0 \} .\]

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Definiiton. isomorphisim

• $R$, $S$,
  • ring
• $\phi: R \rightarrow S$,
  • ring homomorphisim

A bijective ring homomorphisim is called an isomorphisim

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Examples

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Proposition 5.

• $R, S$,
  • ring
• $\phi:R \rightarrow S$,
  • homomorphisimo
• (1) $\mathrm{Im}\phi$ is a subring of $S$,
• (2) $\mathrm{ker}\phi$ is a subring of $R$,
  • $\forall a \in R$, $b \in \mathrm{ker}\phi$, $ab \in \mathrm{ker}\phi$,

proof.

(1)

$s_{1}, s_{2} \in \mathrm{im}\phi$, then $s_{1} := \phi(r_{1}), s_{2} := \phi(s_{2})$ for some $r_{1}, r_{2} \in R$. Then

\[\begin{eqnarray} \phi(r_{1} - r_{2}) & = & s_{1} - s_{2} \in \nonumber \\ \phi(r_{1}r_{2}) & = & s_{1}s_{2} \nonumber \end{eqnarray}\]

$\mathrm{im}\phi$ is closed under subtraction and multiplication. Thus, $\mathrm{im}\phi$ is subring.

(2)

If $\alpha, \beta \in \mathrm{ker}\phi$, $\phi(\alpha) = \phi(\beta) = 0$.

\[\begin{eqnarray} \phi(\alpha - \beta) & = & 0 \in \nonumber \\ \phi(\alpha\beta) & = & 0 \nonumber \end{eqnarray}\]

$\mathrm{ker}\phi$ is closed under subtraction and multiplication. Thus, $\mathrm{ker}\phi$ is subring. Similary,

\[\begin{eqnarray} \forall r \in \mathbb{R}, \ \phi(r\alpha) & = & \phi(r) 0 \nonumber \\ & = & 0 \nonumber \\ \ \phi(\alpha r) & = & 0 \phi(r) \nonumber \\ & = & 0 \nonumber \end{eqnarray}\]

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Definition. ideal

• $R$,
  • ring
• $I \subset I$,
• $r \in R$,

(1) \(rI := \{ra \mid a \in I\}\), \(Ir := \{a r \mid a \in I\}\),

(2) $I$ is said to be left ideal if

• (i) $I$ is a subring of $R$,
• (ii) $r \in R$, $rI \subseteq I$,

(2) $I$ is said to be right ideal if

• (i)
• (ii) $r \in R$, $Ir \subseteq I$,

(3) $I$ is said to be ideal if $I$ is left and right ideal.

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Theorem 8. isomorphism theory

• $R$,
  • ring
• $A \subseteq R$,
  • subring
• $B \subseteq R$,
  • ideal

(1) The second Isomorphism Theorem for Rings

\[A + B := \{ a + b \mid a \in A, \ b \in B \}\]

is a subring.

$A \cap B$ is an ideal of $A$.

\[(A + B)/B \cong A/(A \cap B) .\]

(2) The third Isomorphism Theorem for Rings

• $I \subseteq J \subseteq R$,
  • ideals

Then

$R/J$ is an ideal of $R/I$ and $(R/I)/(J/I) \cong R/J$.

(3) The fourth Isomorphism Theorem for Rings, Lattice Isomorphism Theorem for Rings

• $I \subseteq R$,
  • ideal
• $\pi: R \rightarrow R/I$,
  • natural surjection

Then

• $\pi$ is bijection.
• $\pi(A) = A/I$ is a subring of $R/I$.
• If $I \subset A$, $A$ is ideal if and only if $A/I$ is ideal of $R/I$.

proof.

(1)

Let $x := a + b, y := a^{\prime} + b^{\prime} \in A + B$.

\[\begin{eqnarray} x + y & = & a + b + a^{\prime} + b^{\prime} \nonumber \\ & = & (a + a^{\prime}) + (b + b^{\prime}) \nonumber \\ xy & = & aa^{\prime} + ( ab^{\prime} + ba^{\prime} + bb^{\prime} ) \in A + B \quad (\because B \text{ is a ideal}) \nonumber \end{eqnarray} .\]

Let $[x]{B} := [a + b]{B} \in (A + B)/B$.

\[(A + B)\]

$\Box$

Definition. arithmetrics of ideals

• $R$,
  • commutative ring
• $I, J \subseteq R$,
  • ideals
• (1) \(I + J := \{a + b \mid a \in I, b \in J\}\),
• (2) \(IJ := \{ \sum_{i} a_{i}b_{i} \mid k \in \mathbb{N}, a_{i} \in I, b_{i} \in J\}\),
• (3) For any $n \in \mathbb{N}$, $n$th power of $I$ is defined inductively

\[I^{1} := I, \quad I^{n} := II^{n-1}.\]

This is equivalent to

\[I^{n} = \left\{ \sum_{i = 1}^{k} a_{1}^{i} \cdots a_{n}^{i} \mid a_{j}^{i} \in I, \ k \in \mathbb{N}, \ j = 1, \ldots, n, \ i = 1, \ldots, k \right\}\]

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Proposition.

• (1) $I + J$ is the smallest ideal containing both $I$ and $J$.
• (2) $IJ$ is a ideal contained in $I \cap J$.

proof.

(1)

(2) By definition, $IJ$ is closed under addition.

\[\forall r \in R, \ ab \in IJ, \ r(ab) = (ra)b \in IJ .\]

Thus, $IJ$ is ideal Let $ab \in IJ$. Since $J$ is (right) ideal, $a \in R$ and $b \in J$, $ab \in J$. Similarly, $ab \in I$. Hence $ab \in I \cap J$. $I \cap J$ is ideal so that $IJ \subseteq I \cap J$.

$\Box$

Reference