A result from ring theory, which is a subfield of abstract algebra (no pun intended; go look at the mathematical definition of field if you don't know why I'm apologizing). Basically, it says that for two rings that correspond according to a particular kind of mapping, there exists a one-to-one correspondence between the ideals of one ring that contain the kernel of the mapping, and the ideals of the other ring. If you haven't had any ring theory, that probably means nothing to you, but you might be able to get something out of it by reading slowly through the theorem and the proof and following hardlinks to definitions where necessary. Good luck!

Statement of Theorem: If R and R' are rings, and φ : R -> R' is a surjective (onto) ring homomorphism with kernel K, let I' be an ideal in R' and let I = {r in R : φ(r) is in I'} = φ-1(I'). Then I is an ideal in R, and I/K and I' are isomorphic.

Proof: First, note that I' is a subring of R' because it's an ideal in R'. We can use this to prove that I is an ideal in R, as follows:

  • I is an additive subgroup of R
    1. Closure under addition: Given arbitrary elements a, b in I, a+b is in R because φ(a) is in I' and φ(b) is in I', and since φ is a homomorphism, φ(a) + φ(b) = φ(a+b). Now, since I' is an ideal, it is closed under addition, so φ(a) + φ(b) is in I', and so φ(a+b) is in I', so a+b is in I.
    2. Associativity is inherited from the ring R
    3. Identity elements: I' is an ideal in R', so it contains the identity element for R', which I'll denote 0'. Since R is a ring, it must contain an identity element, which I'll denote 0, and since φ is a homomorphism, φ(0) = 0'.
    4. Inverses: Given an arbitrary element a in I, we know a-1 is in the ring R. We also know φ(a) is in I' by definition, and since I' is an ideal, -φ(a) is in I'. Recall from group theory that -φ(a) = φ(-a). Therefore, φ(-a)I', and -a is in I.
  • For any a in R, b in I, the products ab and ba are both in I.
    1. By definition, φ(b) is in I', and since φ is a homomorphism onto R', φ(a) is in R'. Since I' is an ideal in R', φ(a)φ(b) is in I'. Since φ is a homomorphism, φ(a)φ(b)=φ(ab). So ab is in I.
    2. Likewise, φ(b)φ(a) is in I', and since φ is a homomorphism, φ(b)φ(a)=φ(ba), so φ(ba) is in I', and ba is in I.

So I is an ideal in R. It remains to be shown that there is a correspondence (hence the name of this theorem) between the ideals in R' and the ideals in R containing K: namely, that I' is isomorphic to I/K, or that for every ideal I' in R', there exists an ideal I in R that corresponds to I'.

Let I' and J' be ideals in R' that are not identical (i.e. there exists some element x in I' that is not in J', or vice versa.) Without loss of generality, say x is in I', but not J'. Thus, φ-1(I') contains at least one point, φ-1(x) that is not in φ-1(J'). Now, since φ is a surjective (onto) homomorphism, there exists some y in R such that φ(y)=x. This means that φ(y) is in I' but not in J'. Take the reverse images of I' and J' and you get two corresponding non-identical ideals in R (call them φ-1(I')=I and φ-1(J')=J).

(Note that any ideal I of the form φ-1(I') must contain K, the kernel of the mapping φ by definition: I', as an ideal in R', is a subring and therefore must contain the additive identity element 0'. But K=ker φ={x in R | φ(x)=0'}, and since I={x in R | φ(x) is in I'}, I contains K. This was not at all obvious to me the first time I looked at this theorem, although it seems trivial in retrospect.)

Now to confirm that any arbitrary ideal in R that contains K corresponds (via φ) to an ideal in R'. Let I be an ideal in R, where I contains K. Define φ(I)={φ(a) | a is in I}, a subset of R'. φ(I) is an ideal in R'.

  • φ(I) is an additive subgroup of R'
    1. Closure: Given arbitrary elements φ(a) and φ(b) in φ(I), we know because φ is a surjective homomorphism (onto) that these correspond to elements a, b in I. Because I is an ideal and hence an additive subgroup, a+b is in I. By the definition of φ(I), φ(a+b) is in I. But φ is a homomorphism, so φ(a+b)=φ(a)+φ(b), so φ(a)+φ(b) is in φ(I).
    2. Associativity is inherited from R, and R', both rings.
    3. Identity elements: Since I is an ideal in R, it contains the additive identity element 0. Since φ is a homomorphism, φ(0)=0', the additive identity element of the ring R'. So φ(I) contains 0'.
    4. Inverses Given an arbitrary element φ(a) in phi;(I), we know (because φ is a surjective homomorpism) that it corresponds to some element a in the ideal I in R. Since I is an ideal, it contains a's inverse, -a. And by the definition of φ(I), the image of that inverse φ(-a)=-φ(a) is in φ(I) as well.
  • For any elements φ(a) in φ(I) and φ(b) in R', the products φ(a)φ(b) and φ(b)φ(a) are in φ(I)
    1. Note that φ(a) corresponds to an element a in the ideal I in R, and that φ(b) corresponds to some element b in the ring R, because φ is a surjective homomorphism. Also, by the definition of ring homomorphism, φ(a)φ(b)=φ(ab). Upon closer examination, it turns out that since a is in the ideal I, ab is in I as well, so φ(ab) and hence φ(a)φ(b) are in φ(I).
    2. Similarly, the definition of φ as a ring homomorphism says that φ(b)φ(a)=φ(ba) and since a is in the ideal I, ba is in I as well, and so φ(ba)=φ(b)φ(a) is in φ(I).

Thus, φ(I) is an ideal in R'.

Conclusion: Long story short, there exists a one-to-one correspondence between the ideals in R that contain K, and the ideals in R'. In other words, the sets {I/K | I is an ideal in R and I contains K} and {I' | I' is an ideal in R'} are isomorphic. QED. (phew!)

Disclaimer: I apologize if this proof or any explanation was a little sketchy; it is based on my lecture notes from a Rings and Fields class at UC Irvine, and I decided to node my homework in an effort to improve my understanding. I think it helped me, but since the lecture I based my noding on was more than a week ago and involved considerable proof by handwaving, I wouldn't be surprised if there were still plenty of errors. I welcome corrections, suggestions, comments, and suchforth.

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