# Extension and Contraction

Let $$f: R\rightarrow S$$ be a ring homomorphism. Let $$I \triangleleft R, J \triangleleft S$$. The extension $$I^e$$ of $$I$$ (with respect to $$f$$) is $$\langle f(I) \rangle$$. Then contraction $$J^c$$ of $$J$$ (with respect to $$f$$) is $$f^{-1}(J)$$, 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 $$f:\mathbb{Z} \rightarrow \mathbb{Q}$$. Then for any prime $$p$$, $$p\mathbb{Z}$$ is prime in $$\mathbb{Z}$$ but $$(p\mathbb{Z})^e = \mathbb{Q}$$ which is not prime in $$\mathbb{Q}$$.

In general there is no simple relationship between the prime ideals of $$R$$ and $$S$$. For example consider the identity map $$f:\mathbb{Z}\rightarrow\mathbb{Z}[i]$$ (see notes on <a href="../numberfield"> number fields</a>).

Other properties of extension and contraction:

1. $$I\subset I^{e c}, J\supset J^{c e}$$

2. $$J^{c} = J^{c e c}, I = I^{e c e}$$

3. Let $$\mathcal{C}$$ be the set of contracted ideals in $$R$$ and $$\mathcal{E}$$ be the set of extended ideals in $$S$$. Then $$\mathcal{C} = \{ K\triangleleft R | K^{e c} = K\}$$, $$\mathcal{E} = \{ L\triangleleft S| L^{c e} = L\}$$, and $$K \mapsto K^{e}$$ for all $$K \in \mathcal{C}$$ defines a bijection $$\mathcal{C} \rightarrow \mathcal{E}$$ whose inverse is $$L \mapsto L^{c}$$ for all $$L \in \mathcal{E}$$.

Let $$I_1,I_2\triangleleft R$$ and $$J_1,J_2 \triangleleft S$$. Then

1. $$(I_1 + I_2)^e = I_1^e + I_2^e$$, $$(J_1 + J_2)^c \supset J_1^c + J_2^c$$

2. $$I_1 \cap I_2)^e \subset I_1^e \cap I_2^e$$, $$(J_1 \cap J_2)^c = J_1^c \cap J_2^c$$

3. $$(I_1 I_2)^e = I_1^e I_2^e$$, $$(J_1 J_2)^c \supset J_1^c J_2^c$$

4. $$(I_1 :I_2)^e \subset (I_1^e :I_2^e)$$, $$(J_1 :J_2)^c \subset (J_1^c :J_2^c)$$

5. $$(\sqrt{I_1})^e \subset \sqrt{I_1^e}$$, $$(\sqrt{J_1})^c = \sqrt{J_1^c}$$

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

Ben Lynn blynn@cs.stanford.edu 💡