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# System prepended metadata

title: Quotient Rings and Homomorphism

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# Quotient Rings and Homomorphism 

## Congruence modulo I
### Definition
Congruence modulo I 
R: ring, I: ideal of R, $a,b \in R$, then (congruence class containing a) = $\{b \in R| b \equiv a(mod I)\}$ = $\{b \in R|b-a = i \in I$ | for some $i\in I\}$ = $\{a+i|i \in I\}$


a의 congruence class 는 본질적으로 $Z_n$ 과 같고, a와 congruence인 모든 원소를 모아놓은 집합이다. 
$b-a = i \in I$ 인 모든 원소들은 $a+i$로 표현되고, 따라서, 우리는 a+I 를 coset of I로 표기한다. 

### Theorem 6.6 $a \equiv c$ (mod I) $\leftrightarrow a+I = c+I$

We skip the proof

#### Examples 
1. R = $\mathbb{z}$, I = (3) 

$a+I = c+I \rightarrow a \equiv c$ (mod I) $\rightarrow a-c = I = (3) \rightarrow a \equiv c$ (mod 3)

2. This example illustrates the quotient ring. 

R = $\mathbb{Z}[x]$, I : {polynomials whose constant term is even} = (x,2)
R/I is {0+I, 1+I} because this means that 1. 0+I $\neq$ 1+I and all $f(x) + I \in R/I$ must be in 0+I or 1+I. To prove this second statement, if we assume the existence of $f(x)$, if we let the constant term be $a_0$, then $a_0$ has to be even or odd. If $a_0$ is even, then $0+I$ is in $f(x)$ and if it is odd, then $f(x)$ is in $1+I$. Therefore, $R/I = \{0+I, 1+I \}$. 

**New Perspective**
Ideal의 원소가 어떤 집합에서 0 이라고 생각해보게 되면 어떠한가? 0 이 된다는 것은 다시 ideal 에 속하는 것이다. 


## Quotient Rings 

### Definition of Addition and Multiplication 

> R : ring, I : ideal of R, 
> Let a + I, c+ I $\in$ R/I
> Define + and * on R/I by 
> $(a+I) + (c+I) = (a+c) + I$ 
> $(a+I)(c+I) = (a*c) + I$
>  