---- > [!definition] Definition. ([[contraction of an ideal]]) > Let $\varphi:A \to B$ be a [[ring homomorphism]]. The **contraction** of an [[ideal]] $\mathfrak{b}$ of $B$ is the [[ideal]] $\mathfrak{b}^{c}:=\varphi ^{-1}(\mathfrak{b})$ of $A$. > > Ideals of $A$ of the form $\mathfrak{b}^{c}$ are called **contracted ideals**. ^definition > [!equivalence] > An ideal $\mathfrak{a}$ of $A$ is contracted if and only if $\mathfrak{a}^{ec}=\mathfrak{a}$. ^equivalence > [!proof] Proof of equivalence. > $\to$. Suppose $\mathfrak{a}$ is contracted: $\mathfrak{a}=\mathfrak{b}^{c}$ for some $\mathfrak{b}$. Then $\mathfrak{a}^{ec}=\mathfrak{b}^{cec}=\mathfrak{b}^{c}=\mathfrak{a}$ by the properties below. > $\leftarrow$. If $\mathfrak{a}^{ec}=\mathfrak{a}$, then $\mathfrak{a}=(\mathfrak{a}^{e})^{c}$ is obviously extended. > [!justification] > Confirming $\mathfrak{b}^{c}$ is an [[ideal]]: if $b \in \varphi ^{-1}(\mathfrak{b})$ and $r \in A$, then $\varphi(r)\varphi(a) \in \varphi(\mathfrak{b})$ since $\varphi(\mathfrak{b})$ is an [[ideal]]. This equals $\varphi(ra)$, so $\varphi(ra) \in \mathfrak{b}$. Pulling back gives us $ra \in \varphi ^{-1}(\mathfrak{b})$. ^justification > [!basicproperties] > >- $(\mathfrak{b}_{1} \cap \mathfrak{b}_{2})^{c}=\mathfrak{b}_{1}^{c} \cap \mathfrak{b}_{2}^{c}$, since [[preimages and intersections commute]] for any set-function >- $\sqrt{ \mathfrak{b} }^{c}=\sqrt{ \mathfrak{b} ^{c}}$, [[ring homomorphisms preserve structure|since preimages preserve radicals of ideals]]. >- If $\mathfrak{q}$ is a [[prime ideal]], so is $\mathfrak{q}^{c}$.[^1] >- See [[extension of an ideal]] for interactions between contraction and extension. >- Contraction is [[continuous]] wrt the [[Zariski topology on a ring spectrum|Zariski topology]] ^properties [^1]: Indeed, $1 \notin \mathfrak{q} \implies 1 \notin \varphi ^{-1}(\mathfrak{q})$, and since $\mathfrak{q}^{c}=\ker\left( A \xrightarrow{\varphi} B \to \frac{B}{\mathfrak{q}} \right)$ by [[first isomorphism theorem for rings|FIT]] one has an embedding $\frac{A}{\mathfrak{q}^{c}} \hookrightarrow \frac{B}{\mathfrak{q}}.$ $\frac{A}{\mathfrak{q}^{c}}$ is isomorphic to a [[subring]] of an [[integral domain]], hence is an [[integral domain]]. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```