---- Let $\phi: R \to S$ be a [[ring homomorphism]] between [[ring|rings]] $R$ and $S$. Let $A$ be a [[subring]] of $R$ and let $B$ be an [[ideal]] of $S$. > [!theorem] Theorem. ([[ring homomorphisms preserve structure|Properties of Elements under homomorphism]]) > | *Property* | Description | > | ----------- | ------------------------------------------------------------------------------------------- | > | *Multiples* | $\phi(nr)=n\phi(r)$ for all $n \in \mathbb{N}$ | > | *Powers* | $\phi(r^{n})=\phi(r)^{n}$ for all $n \in \mathbb{N}$ | > | *Units* | $S \neq (0)$, $\phi$ [[surjection\|surjective]], $u$ a [[unit]] $\implies$ $\phi(u)$ a unit | > ^theorem-1 > [!theorem] Theorem. ([[ring homomorphisms preserve structure|Properties of sets under homomorphism]]) > | *Property* | Description | > | ----------------------- | ------------------------------------------------------------------------------------------------------------------ | > | *Subrings* | $\phi(A)$ is a [[subring]] of $S$ | > | *Commutative* | $R$ is [[commutative ring\|commutative]] $\implies \phi(R)$ is [[commutative ring\|commutative]] | > | *Ideals* | $A$ an [[ideal]] & $\phi$ a [[surjection]] $\implies$ $\phi(A)$ an ideal | > | *Ideal Preimages* | $\phi ^{-1} (B)$ is an [[ideal]] of $R$ (see [[contraction of an ideal]]) | > | *Isomorphism condition* | $\phi$ is a [[ring isomorphism]] iff $\phi$ is a [[surjection]] and $\text{ker }\phi=(e)$. | > ^theorem-2 add: radical preimages: $\varphi ^{-1}( \sqrt{ I })=\sqrt{ \varphi ^{-1}(I) }$ > [!basicnonexample] Warning. > In general, the image of an [[ideal]] under a [[ring homomorphism]] may not be an [[ideal]]. Indeed, $\mathbb{Z}$ embeds into $\mathbb{Q}$ but $\mathbb{Z}$ is not an [[ideal]] of $\mathbb{Q}$ (e.g. $1 \cdot \frac{1}{59}=\frac{1}{59} \notin \mathbb{Z}$). ^nonexample > [!proof]- Proof. ([[ring homomorphisms preserve structure]]) > Most all are immediate. > ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as 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 > ``` $\to$. Let $x \in \sqrt{ \varphi ^{-1}(I) }$, say, $x^{n} \in \varphi ^{-1}(I)$. We want to show that $x \in \varphi ^{-1}(\sqrt{ I })$, i.e., find $m$ such that $\varphi(x^{})^{m} \in I$. But this is obvious. We are given that $\varphi(x^{n}) \in \varphi ^{-1}(I)$. Since $\varphi(x^{n})=\varphi(x)^{n}$, this means that $\varphi(x)^{n} \in I$. $\leftarrow$. For the other direction, let $x \in \varphi ^{-1}(\sqrt{ I })$, which is to say that $\varphi(x)^{n} \in I$ for some $n$. We want to show that $\varphi(x^{m}) \in I$ for some $m$.... clearly just take $m=n$. (this proof should just be a chain of biconditionals, not sure why i spelled it all out)