I is a prime ideal iff R/I is an integral domain
Dependencies:
Let $R$ be a commutative ring with unity. Let $I$ be a proper ideal of $R$.
$I$ is defined to be a prime ideal iff $ab \in I \Rightarrow (a \in I \vee b \in I)$.
$I$ is a prime ideal iff $R/I$ is an integral domain.
Proof
Proof of 'only-if' part
Let $I$ be a prime ideal.
\begin{align} & (a+I)(b+I) = 0+I \\ &\Rightarrow ab + I = I \\ &\Rightarrow ab \in I \\ &\Rightarrow a \in I \vee b \in I \tag{$I$ is prime} \\ &\Rightarrow a+I = 0+I \vee b+I = 0+I \end{align}
Therefore, $R/I$ is an integral domain.
Proof of 'if' part
Let $R/I$ be an integral domain.
\begin{align} & ab \in I \\ &\Rightarrow ab + I = I \\ &\Rightarrow (a+I)(b+I) = 0 + I \\ &\Rightarrow a+I = 0+I \vee b+I = 0+I \tag{$R/I$ is an integral domain} \\ &\Rightarrow a \in I \vee b \in I \end{align}
Therefore, $I$ is a prime ideal.
Dependency for: None
Info:
- Depth: 6
- Number of transitive dependencies: 14