----- > [!proposition] Proposition. ([[ideals of product ring are products of ideals]]) > The [[ideal|ideals]] of a [[direct product of rings|product ring]] $R \times S$ are precisely of the form $I \times J$ with $I$ an [[ideal]] of $R$ and $J$ an [[ideal]] of $S$. > > The *[[prime ideal|prime]]* [[ideal|ideals]] of $R \times S$ take the form $\mathfrak{p} \times (1)$ for $\mathfrak{p}$ a [[prime ideal]] of $R$ and $(1) \times \mathfrak{q}$ for $\mathfrak{q}$ a [[prime ideal]] of $S$. > > ^proposition > [!proof]- Proof. ([[ideals of product ring are products of ideals]]) > $\to.$ Let $I$ be an [[ideal]] of $R$, $J$ of $S$, and consider $I \times J \subset R \times S$. We have $(I \times J)(r,s)=\{ (ir, js): i \in I , j \in J\}=Ir \times Js \subset I \times J$ as required for any $(r,s) \in R \times S$. > > $\leftarrow.$ Conversely, suppose $K$ is an [[ideal]] of $R \times S$. Define two 'candidates' $I:=\{ r \in R: (r, 0) \in K\} \subset R$ and $J:= \{ s \in S : (0, s) \in K \} \subset S.$ > It is clear that $I$ and $J$ are ideals of $R$ and $S$ respectively. > > **Claim:** $K=I \times J$. Take an arbitrary $(k_{R}, k_{S}) \in K$. It decomposes as $(k_{R},k_{S})=(k_{R}, 0)+(0, k_{S})$, showing $k_{R} \in I$ and $k_{S} \in J$. Thus $K \subset I \times J$. For the reverse inclusion, take $(r,s) \in I \times J$. This equals $(r,0)+(0,s)$ with $(r,0) \in K$ and $(0,s) \in K$; since $K$ is an [[abelian group]] this implies $(r,s) \in K$. > > > **Prime ideals.** Now let $K=I \times J$ be an arbitrary ideal of $R \times S$. We claim that $K$ is prime if and only if either $I=\mathfrak{p}$ for some $\mathfrak{p} \in \text{Spec }R$ and $J=(1)$, or $J=\mathfrak{q}$ for some $\mathfrak{q} \in \text{Spec }J$ and $I=(1)$. > > Indeed, $K$ is prime if and only if $\frac{R \times S}{K}=\frac{R \times S}{I \times J} \cong \frac{R}{I} \times \frac{S}{J}$ is an [[integral domain]]. A product ring (with coordinate-wise multiplication) is an integral domain if and only if one of the factors is the zero ring, hence either $\frac{R}{I}=(0)$ or $\frac{S}{J}=(0)$, and correspondingly either $I=R$ or $J=S$. ----- #### ---- #### 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 > ```