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My textbook states the following proposition: Let $f:R \rightarrow S$ be a ring homomorphism and let $s$ be in the image of $f$. Then $\{r \in R \mid f(r) = s\}$ is in one-to-one correspondence with $\ker(f)$.

What does it mean to have one to one correspondence with $\ker(f)$? Does it mean the set $\{r \in R \mid f(r) = s\}$ and the set $\ker(f)$ have the same cardinality?

Thanks!

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"One-to-one correspondence" is a poor (in my opinion, anyway) way of saying "bijection." –  Jay Kopper Apr 1 '11 at 6:06

2 Answers 2

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HINT $\rm\ $ If $\rm\:f(r)\ =\ s\ $ then $\rm\ f(r')\ =\ s\ \iff\ 0\ =\ f(r')-f(r)\ =\ f(r'-r)$

Hence $\rm\ \ f^{-1}(s)\ =\ r\ +\ ker\ f\ =\: $ particular + homogeneous solution, as in linear algebra.

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@Bill cool, it's actually the proof! I see that you are saying each $r \in R$ corresponds to (r' + k) for each $k \in ker f$, which implies the same cardinality. but what about the cases where (r'+2k), (r'+3k), and so on? since f maps them to s as well, wouldn't that make the set R have more elements than ker f? I would appreciate it if you can point out what's wrong with this reasoning... –  user8969 Apr 2 '11 at 2:19
    
@mat As above, $\rm\ z\to r+z\ $ bijects $\rm\:f^{-1}(0)$ with $\rm\:f^{-1}(s)\:,\:$ i.e. the shifting the homogeneous solution space $\rm\:f^{-1}(0)$ by any particular solution $\rm\:r\:$ puts it in bijection with the nonhomogeneous solution space $\rm\:f^{-1}(s)\:.$ –  Bill Dubuque Apr 2 '11 at 2:33
    
@Bill thanks! now I see the difference between our views. you are saying shifting the entire space by a particular solution to create a bijection, which is a nice clean way. I was thinking in terms of elements: each $r \in R$ is created by adding each $k\in$ ker f to a particular r'. after this, R is in bijection with ker f. adding 2k to r' doesn't create any new element in R, because f(2k)=f(2)*f(k)=f(2)*0=0, i.e. 2k was already in ker f, so 2k+r' is already in R. this is the fault with my previous reasoning. well, quite messy, but glad I got it. anyway, thanks a lot :) –  user8969 Apr 2 '11 at 4:30
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@mat Many times you say $\rm\: R\: $ but it should be $\rm\: f^{-1}(s)\:.\: $ $\rm\:R\:$ is partitioned into equivalence classes $\rm\:[r]\:$ where $\rm\ r'\in [r]\ \iff\ f(r') = f(r)\:.\:$ These equivalence classes are known as the kernel, or equalizer, or level-sets of $\rm\:f\:.$ –  Bill Dubuque Apr 2 '11 at 4:49
    
@Bill well, the same thing applies to your space shifting as well. one (newbie like me) might ask, after shifting r1 to the homogeneous solutions to obtain a set of non-homogeneous solutions, why not then shift the homogeneous solutions by r2 to obtain even more solutions? well, let's say f(r2) = s, and r2 = r1 + r0. then f(r2) = f(r1)+f(r0). since f(r1) = s, then f(r0) = 0, which means $r0 \in$ ker f, which means r1+r0=r2 is already obtained. so shifting by r2 won't lead to any new solutions. sorry for the long comments, since they are not immediately clear to me xP –  user8969 Apr 2 '11 at 4:50

It means that there is a map $g: \{r\in R | f(r) = s\} \rightarrow \operatorname{ker}f$ such that every element of $\operatorname{ker}f$ is the image of one and only one element of $\{r\in R | f(r) = s\}$. This implies that the cardinality is the same.

In this case, if $s'\in R$ is such that $f(s') = s$, then one such map is $g(r) = r - s'$.

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thanks! just quick question: since R is the domain of f, wouldn't that mean {r in R | f(r)=s} is just the entire R? but this doesn't seem right, because this would mean the entire R is mapped to 0 due to the one-to-one correspondence. I would really appreciate it if you can point out what I am missing. –  user8969 Apr 1 '11 at 5:46
    
@mathcat: No, what they are doing is fixing a single value for $s$ and defining the set in question as the set of elements mapped to $s$. –  Alex Becker Apr 1 '11 at 5:48
    
think of a group homomorphism $f:G\to H$, then $G/\text{ker}f\cong\text{im}f$, ie everything in a coset gets mapped to the same element... –  yoyo Apr 1 '11 at 15:30

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