Commutative Algebra - Extension and Contraction

Let $f : R \to S$ be a ring homomorphism. Let $I ◃ R, J ◃ S$ . The extension $I^{e}$ of $I$ (with respect to $f$ ) is $⟨ f (I) ⟩$ . Then contraction $J^{c}$ of $J$ (with respect to $f$ ) is $f^{- 1} (J)$ i, which is an ideal in $R$ .

Note that if $J$ is prime than so is $J^{c}$ , though the same is not always true for extensions. For example, take the identity map $f : ℤ \to ℚ$ . Then for any prime $p$ , $p ℤ$ is prime in $ℤ$ but $(p ℤ)^{e} = ℚ$ which is not prime in $ℚ$ .

In general there is no simple relationship between the prime ideals of $R$ and $S$ . For example consider the identity map $f : ℤ \to ℤ [i]$ (see notes on number fields).

Other properties of extension and contraction:

$I \subset I^{e c}, J \supset J^{c e}$
$J^{c} = J^{c e c}, I = I^{e c e}$
Let $𝒞$ be the set of contracted ideals in $R$ and $ℰ$ be the set of extended ideals in $S$ . Then $𝒞 = {K ◃ R ∣ K^{e c} = K}$ , $ℰ = {L ◃ S ∣ L^{c e} = L}$ , and $K \mapsto K^{e}$ for all $K \in 𝒞$ defines a bijection $𝒞 \to ℰ$ whose inverse is $L \mapsto L^{c}$ for all $L \in ℰ$ .

Let $I_{1}, I_{2} ◃ R$ and $J_{1}, J_{2} ◃ S$ . Then

$(I_{1} + I_{2})^{e} = I_{1}^{e} + I_{2}^{e}$ , $(J_{1} + J_{2})^{c} \supset J_{1}^{c} + J_{2}^{c}$
$I_{1} \cap I_{2})^{e} \subset I_{1}^{e} \cap I_{2}^{e}$ , $(J_{1} \cap J_{2})^{c} = J_{1}^{c} \cap J_{2}^{c}$
$(I_{1} I_{2})^{e} = I_{1}^{e} I_{2}^{e}$ , $(J_{1} J_{2})^{c} \supset J_{1}^{c} J_{2}^{c}$
$(I_{1} : I_{2})^{e} \subset (I_{1}^{e} : I_{2}^{e})$ , $(J_{1} : J_{2})^{c} \subset (J_{1}^{c} : J_{2}^{c})$
$(\sqrt{I_{1}})^{e} \subset \sqrt{I_{1}^{e}}$ , $(\sqrt{J_{1}})^{c} = \sqrt{J_{1}^{c}}$

The set $ℰ$ is closed under sum and product while the set $𝒞$ is closed under intersection, forming ideal quotients and taking radicals.