The course notes on rings have the below lemma
Let $R$ be a ring and $I$ a two sided ideal. Define $\pi : R \rightarrow R/I$ by $\pi(r)=r+I$. Then $\pi$ is a surjective ring map and ker $\pi=I$.
Proving that $\pi$ is a homomorphism is straightforward, but I don't quite understand their proof of surjectivity, which I've typed below:
If $r+I \in R/I$, then $\pi (r)=r+I$, and hence $\pi$ is surjective.
If $\pi$ is defined to be the $\pi(r)=r+I$, then why did they need to say "then $\pi(r)=r+I"$? It's almost as if they concluded that, but it was a definition in the first place!
The below is my understanding of what they were trying to say, is this correct?
$R/I$ is, by definition of the quotient, $r+I$ for all $r, I\in R$. But this is exactly the map of $\pi(r)$, so every single element under the mapping $\pi$ has a corresponding element in $R$. In other words, $\pi$ is a surjective map.
I also would like to know, does this not mean that $\pi$ is bijective? Or does the whole idea of 'modulo' when working with quotients break the uniqueness required for bijectivity?
Best Answer
Your seem to have the main idea, but let me try to clarify.
The map $\pi : R \to R/I$ is given by $\pi(r) = r + I$. In order to show $\pi$ is surjective, we need to show that given any element $ y \in R/I$, there exists an element $x \in R$ such that $\pi(x) = y$ (this is just the definition of surjectivity of a map of sets). Now, as $R/I = \{r + I \mid r \in R\}$, we can write $y$ as $a + I$ for some $a \in R$. So we need to show that there is $x \in R$ such that $\pi(x) = a + I$. As $\pi(x) = x + I$ by definition of the map $\pi$, we can rewrite the equation as $x + I = a + I$. One way to guarantee this equation holds true is to choose $x = a$; then we have $\pi(x) = \pi(a) = a + I = y$. So for any element $y \in R/I$, we have shown that there is $x \in R$ such that $\pi(x) = y$, therefore $\pi$ is surjective.
A map between sets is bijective if it is surjective and injective. The map $\pi$ is surjective, but unless $I$ is the trivial ideal (i.e. $I = \{0\}$), $\pi$ is not injective. Recall, a map $f : X \to Y$ is said to be injective if for all $x_1, x_2 \in X$, $x_1 \neq x_2$, $f(x_1) \neq f(x_2)$. To see that $\pi : R \to R/I$ is not injective if $I \neq \{0\}$, note that $I$ contains at least two elements (any ideal contains $0$ and as $I \neq \{0\}$, it must contain another element). Let $r_1, r_2 \in I$, $r_1 \neq r_2$; note that $I \subseteq R$, so $r_1, r_2 \in R$. Then $\pi(r_1) = r_1 + I = I$ (as $r_1 \in I$) and $\pi(r_2) = r_2 + I = I$ (as $r_2 \in I$); that is, $\pi(r_1) = \pi(r_2)$. Therefore, $\pi$ is not injective, so $\pi$ is not bijective.