MATH 3231 Class Notes 02-03-2025

notering-theory

Defined Correspondence Theorem for Quotients

Take $(R,+,\cdot )$ a ring with $1_R$, and $S={}^R{/}_I$ for some ideal $I⊴R$.

Remark

Given morphism $f:R\to S$, if $a⊴S$, then $f^{-1}a⊴R$.

Remark

If $f:R\to S$ is a morphism, $I⊴R$, $f$ not necessarily surjective. Then $f(I)\subseteq S$ is an additive subgroup, but not necessarily an ideal.

Remark

If $I⊴R$, $I\subseteq J$, $J⊴R$, then ${}^J{/}_I⊴{}^R{/}_I$.

Proof

We already know from groups, $({}^J{/}_I,+)\le ({}^R{/}_I,+)$.

If $x\in J,x+I\in {}^J{/}_I$, and if $r\in R,r+I\in {}^R{/}_I$. Thus, $(r+I)(x+I)=rx+I\in {}^J{/}_I$.

Remark

$I⊴R$, $J⊴R$ that does not necessarily contain $I$. Then, ${}^J{/}_I$ does not even make sense as additive group, if $I⊈J$.

Question: What is $\phi (J)\subseteq S={}^R{/}_I$? (Where $\phi (J)=\{r+I\mid r\in J\}$, set of cosets in $R$) Answer: It is still an ideal: ${}^{J+I}{/}_I$.

Note: $J+I⊴R$, $I\subseteq J+I$, so ${}^{J+I}{/}_I⊴S={}^R{/}_I$.

Example

The image of $10\mathbb{Z}$ in ${\mathbb{Z}}_6$ is

$$(10(\mathrm{mod}6))\underset{10\equiv 4\mathrm{mod}6}{=}(4)\underset{10\equiv 4\mathrm{mod}6}{=}(-2)\underset{\text{as ideals}}{=}(2)⊴{\mathbb{Z}}_6.$$

With the previous remark: We get ${}^{10\mathbb{Z}+6\mathbb{Z}}{/}_{6\mathbb{Z}}=(2)⊴{\mathbb{Z}}_6$.

Defined Units in Rings

Given $(R,+,\cdot )$ ring with $1_R$. Assume $R$ is commutative.

Remark

$u\in R$ is a unit if and only if $((u)=Ru)=((1_R)=R).$ ($I\in R$, $I=R⟺1\in I$).

Remark

For any $u\in R$, there exists $\phi :R\to R$ given by $\phi (x)=u\cdot x$. $\phi$ is additive: $\phi (x+y)=u\cdot (x+y)=ux-uy=\phi (x)+\phi (y)$.

The following are equivalent:

• $u$ is a unit.
• $\phi$ is surjective.
• $\phi$ is bijective.

In this case, ${\phi }^{-1}=\psi$, with $\psi :R\to R$ defined by $\psi (y)=u^{-1}y$.

Defined Zero-Divisor and Prime Ideal