Notes - Rings and Modules HT24, Correspondence theorems


Flashcards

Can you state the correspondence theorem for rings?


Suppose:

  • $\phi : R \to S$ is a surjective ring homomorphism

Then there is a bijective correspondence between the set of ideals in $R$ containing $\ker \phi$, and the set of ideals of $S$.

In particular, this means that if:

  • $I$ is an ideal of $R$ that contains $\ker \phi$
  • $J$ is an ideal of $S$

then:

  • $\phi(I)$ is an ideal of $S$
  • $\phi^{-1}(J)$ is an ideal of $R$
  • $\phi(\phi^{-1}(J)) = J$
  • $\phi^{-1}(\phi(I)) = I$
  • (and in general, if $I$ is an ideal of $R$ that doesn’t necc. contain $\ker \phi$, we still have $\phi^{-1}(\phi(I)) = I + \ker \phi$, where this is the sum of ideals, not the notation for cosets).

The correspondence theorem for rings states that if

  • $\phi : R \to S$ is a surjective ring homomorphism

then there is a bijective correspondence between the set of ideals in $R$ containing $\ker \phi$, and the set of ideals of $S$.

Using this, quickly justify that $\mathbb C[t] / \langle t^2 - 1 \rangle$ contains exactly four ideals.


Consider $\phi : \mathbb C[t] \to \mathbb C[t] / \langle t^2 - 1 \rangle$ via the natural projection $\phi\big(f(t)\big) = f(t) + \langle t^2 - 1\rangle$.

The kernel of $\phi$ is $\langle t^2 - 1\rangle$, and by properties of $\mathbb C[t]$, we know that there are just 4 ideals in $\mathbb C[t]$ containing $\langle t^2 - 1 \rangle$, corresponding to the divisors: $\langle 1 \rangle$, $\langle t - 1\rangle$, $\langle t + 1\rangle$, $\langle t^2 - 1\rangle$. By the correspondence theorem, there must be exactly 4 ideals in $\mathbb C[t] / \langle t^2 - 1 \rangle$.

The correspondence theorem for rings states that if:

  • $\phi : R \to S$ is a surjective ring homomorphism

Then there is a bijective correspondence between the set of ideals in $R$ containing $\ker \phi$, and the set of ideals of $S$.

In particular, this means that if:

  • $I$ is an ideal of $R$ that contains $\ker \phi$
  • $J$ is an ideal of $S$

then:

  1. $\phi(I)$ is an ideal of $S$
  2. $\phi^{-1}(J)$ is an ideal of $R$
  3. $\phi(\phi^{-1}(J)) = J$
  4. $\phi^{-1}(\phi(I)) = I$
  5. (and in general, if $I$ is an ideal of $R$ that doesn’t necc. contain $\ker \phi$, we still have $\phi^{-1}(\phi(I)) = I + \ker \phi$, where this is the sum of ideals, not the notation for cosets).

Quickly prove points 3 and 5.

(1 and 2 are standard results, just write out images/preimages and check the ideal property, and 4 is implied by 5, since if $I$ contains $\ker \phi$, then $I + \ker \phi = I$).


For 3: This is true given any surjective map between subsets (not just for homomorphisms).

For 4 and 5: We use double inclusion to show 5 (which implies 4):

To see $I + \ker \phi \subseteq \phi^{-1}(\phi(I))$ note that $0 \in \phi(I)$, so $\ker \phi = \phi^{-1}(0) \subseteq \phi^{-1}(\phi(I))$. Since $I \subseteq \phi^{-1}(\phi(I))$ also, it follows that $I + \ker \phi \subseteq \phi^{-1}(\phi(I))$.

To see $\phi^{-1}(\phi(I)) \subseteq I + \ker \phi$, suppose $x \in \phi^{-1}(\phi(I))$. Then by definition $\exists i \in I$ such that $\phi(i) = \phi(x)$, so $x - i \in \ker \phi$. But then $x = i + (x - i) \in I + \ker \phi$, so $\phi^{-1}(\phi(I)) \subseteq I + \ker \phi$.




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