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Let $R$ and $S$ be rings. If $f : R \rightarrow S$ is a surjective ring homomorphism, then $$\bar{f} : R / \ker(f) \longrightarrow S, x + \ker(f) \longmapsto f(x)$$ is an isomorphism.

  1. First, I wanted to show that $\ker(f)$ is an ideal. I let $r \in R$ and $x \in \ker(f)$. Since $f$ is a homomorphism, we have $$f(rx)=f(r)f(x)=f(r)\cdot 0 = 0$$which means $rx \in \ker(f)$. Similarly, we have$$f(xr)=f(x)f(r)=0 \cdot f(r) = 0$$ which means $xr \in \ker(f)$ as well. Hence, $\ker(f)$ is an ideal, which means $$(x+\ker(f))(y+\ker(f))=xy+\ker(f)$$ and $$(x+\ker(f))+(y+\ker(f))=x+y+\ker(f)$$

  2. We have, for multiplication, \begin{align} \bar{f}((x+\ker(f))(y+\ker(f)))&=\bar{f}(xy+\ker(f))\\ &=f(xy)\\ &=f(x)f(y)\\ &=\bar{f}(x+\ker(f))\bar{f}(y+\ker(f)) \end{align} and, for addition, \begin{align} \bar{f}((x+\ker(f))(y+\ker(f)))&=\bar{f}(x+y+\ker(f))\\ &=f(x+y)\\ &=f(x)+f(y) \\ &=\bar{f}(x+\ker(f))+\bar{f}(y+\ker(f)) \end{align} so $\bar{f}$ is a ring homormorphism.

  3. Do I also show that $\bar{f}$ is surjective, so that I have everything I need to establish that $\bar{f}$ is an isomorphism?

Cookie
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  • You also need to show $\ker(f)$ is closed under addition. 2. You need to show that $\overline f$ is well-defined, and respects multiplication and addition (and the multiplicative identity). The computation you have displayed accomplishes none of these things; you need to show that $\overline f(xy+\ker(f)) = \overline f(x+\ker(f)) \overline f(y+\ker(f))$. 3. You need to show that $\overline f$ is one-to-one and onto. None of these tasks are particularly difficult, but as you can see, it is a somewhat lengthy laundry list of things to do.
    • – Dustan Levenstein Apr 08 '14 at 20:46
  • On the homomorphism part, can I say that $\bar{f}(xy+ker(f)) = f(xy)=f(x)f(y)$? – Cookie Apr 08 '14 at 20:52
  • Yes, that's essentially the proof that it respects multiplication (maybe with one last equality $=f(x+\ker(f))f(y+\ker(f))$). Well-definedness is important here, though! – Dustan Levenstein Apr 08 '14 at 20:55
  • Regarding well-definedness, I have fundamental trouble applying that concept to rings here (and groups as a whole for that matter). All I was informed about a function being well-defined is that for every, say, $x \in D$ (domain $D$) there exists a $y \in f(D)$ such that $f(x)=y$. – Cookie Apr 08 '14 at 20:59
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    almost - well-definedness is existence and uniqueness of such a $y$. Here there are multiple ways of expressing the same input to $f$, so you have to show that if $x+\ker(f)=x'+\ker(f)$, then the definitions for $\overline f(x+\ker(f))$ and $\overline f(x'+\ker(f))$ give the same thing, i.e., that $f(x)=f(x')$. (Here, $x'$ is just another element of R.) – Dustan Levenstein Apr 08 '14 at 23:11
  • Okay, so I wrote $$x+\ker(f)=y+\ker(f) \implies \bar{f}(x+\ker(f))=\bar{f}(y+\ker(f)) \implies f(x)=f(y).$$ Does this suffice to show well-definedness? – Cookie Apr 09 '14 at 05:06
  • The implication "$x+\ker(f)=y+\ker(f) \implies \overline f(x+\ker(f)) = \overline f(y+\ker(f))$ is exactly what you need to show. And no, you haven't shown it here. Hint: use the fact that $x+\ker(f)=y+\ker(f)$ implies $x-y\in \ker(f)$. – Dustan Levenstein Apr 09 '14 at 12:27