First Isomorphism Theorem

If $\varphi : R \rightarrow S$ is a ring homomorphism with $ker\varphi=I$, then there is an isomorphism $R/I \rightarrow im\varphi$ given by $r+I\rightarrow\varphi(r)$.

I have trouble to understand the second part of the theorem

then there is an isomorphism $R/I \rightarrow im\varphi$ given by $r+I\rightarrow\varphi(r)$.

I can't image how it looks like. Can anyone explain it more easily? Thank you.

  • $\begingroup$ I am writing an answer. I request you to wait. $\endgroup$ – астон вілла олоф мэллбэрг Jun 15 '18 at 8:04
  • $\begingroup$ Thank you! I'm moving on to the next Chapter. $\endgroup$ – user8314628 Jun 15 '18 at 8:05
  • $\begingroup$ Just as a comment, homomorphism is an epimorphism onto its image(trivially). İt is a monomorphism if kernel is 0, and R/I identifies kernel as equivalent to 0. Hence on R/I, it is an isomorphism. $\endgroup$ – Atbey Jun 15 '18 at 8:21
  • $\begingroup$ The terms make me more confused... $\endgroup$ – user8314628 Jun 15 '18 at 8:23

See, $R/I$ is a ring, whose elements are equivalence classes under the relation $x \sim y$ if $x - y \in I$. That is, if I ask you "name an element of $R / I$", then you would tell me the name of some element in $R$, say $r \in R$, and I will take the equivalence class of $r$, which is denoted $r+I$, as the element of $\frac RI$ which you are referring to.

Let's take an example. Take the ring $\mathbb Z $ under usual addition and multiplication, along with the ideal $3\mathbb Z = \{3n : n \in \mathbb Z\}$, the multiples of $3$.

Then, what are the elements of $\frac{\mathbb Z}{3\mathbb Z}$? You name any element of $\mathbb Z$, and I will interpret the equivalence class it belongs to, as the element you are referring to. For example, if you say $73$, then $73 + 3\mathbb Z$ is an element of $\frac{\mathbb Z}{3 \mathbb Z}$. Similarly, so is $65 + {3 \mathbb Z}$.

Now, every element of $\frac{\mathbb Z}{3 \mathbb Z}$ is of the form $r + {3 \mathbb Z}$ where $r \in \mathbb Z$.

So what is the isomorphism given to you actually doing? Given an element $E$ of $\frac RI$, it is an equivalence class, therefore there is some $r$ such that $E = r + I$. Now, find where that $r$ goes under $\phi$, and send $E$ to $\phi(r)$.

For example, if $\phi : Z \to S$ (where $S$ is some ring) is a map whose kernel is exactly $3 \mathbb Z$, then the isomorphism between $\frac{\mathbb Z}{\mathbb 3Z}$ and $S$ is like this :

The isomorphism would take $73 + {3\mathbb Z}$ to $\phi(73)$. It would take $65 +{3 \mathbb Z}$ to $\phi(65)$. It would take $1 + {3\mathbb Z}$ to $\phi(1)$. And so on.

Now, I hope at least you know what the isomorphism is in the general case. But there is a deeper issue.

If I give you an equivalence class, $E$, in $\frac RI$, I told you there is an element such that $E = r+I$. But $r$ need not be unique! For example, $1 +3 \mathbb Z$ is equal to $73 + \mathbb 3Z$, because $73 - 1 = 72 = 3 \times 24 \in \mathbb 3Z$, so $1 \sim 73$, and related elements belong to the same equivalence class, so $1 + 3 \mathbb Z = 73 + 3\mathbb Z$.

But then, what will $\phi$ do when confronted with this? If I give it $1 + 3\mathbb Z$, then it will return $\phi(1)$, but if I give it $73 + \mathbb Z$, then it will return $\phi(73)$.

How can $\phi $ return two elements, $\phi(1)$ and $\phi(73)$, while taking the same equivalence class $1 + 3\mathbb Z = 73 + \mathbb Z$ as input? Every function has a single input and returns a single output, but it seems there could be two outputs here, $\phi(1)$ and $\phi(73)$!

This is not the case : actually, $\phi(1) = \phi(73)$, by the fact that $\phi(72) = 0$(what is the kernel of $\phi$?) and $73 = 72 +1$ so $\phi(73) = \phi(72 + 1) = \phi(72) + \phi(1) = 0 + \phi(1) = \phi(1)$. So, $\phi$ is not doing anything wrong. More formally, we say it is well-defined in this situation.

Putting the deeper issue to rest, I hope you now see :

  • what the isomorphism from $\mathbb R/I$ to the image of $\phi$ is doing.

  • why it is well-defined (I gave an example, but you can generalize and work it out).

As to why it is an isomorphism, well, that I shall explain only if you require it to be done.


Define $ \psi: R/I \to im \varphi$ by

$$ \psi(r+I):=\varphi(r).$$

Then show that $ \psi$ is well defined (this means that $r+I=s+I $ implies $\varphi(r)= \varphi(s)$) and show that $\psi$ is a ring isomorphism.


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