Structure of ideals in the product of two rings

Question

$R$ and $S$ are two rings. Let $J$ be an ideal in $R\times S$. Then there are $I_{1}$, ideal of $R$, and $I_{2}$, ideal of $S$ such that $J=I_{1}\times I_{2}$.

For me it's obvious why $\left\{ r\in R\mid \left(r,s\right)\in J\text{ for some } s\in S\right\}$ is an ideal of $R$ (and the same for $S$) so I can prove $J$ is a subset of the product of two ideals and also that the product of these two ideals is an ideal.

But I can't see how to prove equality without assuming existence of unity or commutativity. Trying basically to show that if $\left(r,s\right)\in J$ then also $\left(r,0\right),\left(0,s\right)\in J$.

Thank you for your help!

Answer 1

If you have an identity, multiplying by $(1_R,0)$ and $(0,1_S)$ will help. If you don't, then the result need not be true. E.g. if you put the zero product on the abelian group $\mathbb Z_2\oplus \mathbb Z_2$, then the ideal ($=$subgroup) generated by $(\overline 1,\overline 1)$ is a counterexample.

Answer 2

Let us assume that $R$ has an identity element $1_R$.

Let $I$ be a two-sided ideal of $R\times S$. Define $I_R = \left\{r \in R: \, \exists s \in S \, \, \, \text{s.t.} \, \, (r,s) \in I \right\}$ and similarly $I_S = \left\{s \in S: \, \exists r \in R \, \, \, \text{s.t.} \, \, (r,s) \in I \right\}$. Then $I_R, I_S$ are ideals of $R,S$ and $I \subset I_R \times I_S$.

Now take $r \in I_R$ and $s \in I_S$. We will show that $(r,s) \in I$. By definition there is some $s' \in S$ such that $(r,s') \in I$. Similarly there is some $r' \in R$ such that $(r',s) \in I$. Since $I$ is a two-sided ideal, we have that $(1_R,s)(r,s') = (r,ss') \in I$. Similarly, $(r',s)(1_R,s') = (r',ss') \in I$. Hence $(r-r',0) \in I$ and so $(r', s) + (r - r', 0) = (r,s) \in I$.

Remark: As we see, all we need is that only one of $R,S$ has an identity element.

Answer 3

Lemma. Let $(A_i)_{i=1,\ldots,n}$ be a (finite) family of rings with $1$. Then ideals of $A = \prod_{i=1}^n A_i$ are of the form $I_1 \times \ldots\times I_n$, where $I_i$ is an ideal of $A_i$ for each $i\in\{1,\ldots,n\}$.

Proof. By induction on $n$, the cases $n\in\{0,1\}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $A\times B$ of two rings with $1$, and denote by $p : A\times B\to A$ and $q : A\times B\to B$ the two canonical projections. Then $I = p(K)$ (resp. $J = q(K)$) is an ideal of $A$ (resp. of $B$.), because the image of an ideal under a surjective ring homomorphism is an ideal. Obviously $K\subseteq I\times J$. To show the inverse inclusion, let $(a,b)\in I\times J$. Then $(a,b') \in K$ and $(a',b)\in K$ for some $(a',b')\in A\times B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) \in K$. $\square$

Answer 4

The answer of Manos can be generalized to a characterization (references at the end):

Let $R,S$ be rings (not necessarily with identity element). Consider the product $R\times S$. We say that an ideal of the product is a subproduct ideal if it is of the form $I\times J$ with $I$ ideal of $R$, $J$ ideal of $S$. Now fix $R$. The following assertions are equivalent:

1. For all rings $S$, the ideals of $R\times S$ are subproduct ideals.
2. The ideals of $R\times R$ are subproduct ideals.
3. $R$ is a (two-sided) $e$-ring, that is, for every element $r\in R$ we have $r\in Rr+rR+RrR$.

(Observe that $Rr+rR+RrR$ may fell short of being the ideal generated by $r$, as it doesn't necessarily include $\mathbb{Z}r$).

Let us prove the theorem:

3)$\Rightarrow$1) This is essentially Manos' proof. Given $I$ ideal of $R\times S$, we define $$I_R:=\{r\in R \ | \ \exists s\in S, (r,s)\in I\}, \ I_S:=\{s\in S \ | \ \exists r\in R, (r,s)\in I\}.$$ That $I\subseteq I_R\times I_S$ is straightforward. Pick $(r,s)\in I_R\times I_S$. Then there are $r'\in R$, $s'\in S$ such that $(r,s'),(r',s)\in I$. By hypothesis we know that there exist $a,b,u_i,v_i\in R$ such that $r=ar+rb+\sum_iu_irv_i$. Since $I$ is an ideal, $(ar,0)=(a,0)(r,s')$, $(rb,0)=(r,s')(b,0)$ and $(u_irv_i,0)=(u_i,0)(r,s')(v_i,0)$ are in $I$, and this implies $(r,0)\in I$. Similarly $(r',0)\in I$, and thus $(r,s)=(r-r',0)+(r',s)\in I$. Therefore $I=I_R\times I_S$.

1)$\Rightarrow$2) Trivial.

2)$\Rightarrow$3) Pick $r\in R$ and consider the ideal $I$ generated by $(r,r)$ in $R\times R$. By hypothesis we have $I=J\times K$, with $J,K$ ideals of $R$. Since $(r,r)\in J\times K$ we have $(r,0)\in J\times K=I=$Id$(r,r)$, what means that there exist $z\in\mathbb{Z}$, $a_1,a_2,b_1,b_2,u_i,v_i,x_i,y_i\in R$ such that $$r=zr+a_1r+rb_1+\sum_iu_irv_i, \ 0=zr+a_2r+rb_2+\sum_ix_iry_i.$$ The key here is that $z$ is the same in both equations, so we can equate by $zr$ to get $r=(a_1-a_2)r+r(b_1-b_2)+\sum_i c_ird_i\in Rr+rR+RrR$.

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EDITED: Of course, rings with identity are $e$-rings. Another interesting class of $e$-rings is that of von Neumann regular rings.

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Observe that we have a similar (less convoluted, more restrictive) version for one-sided ideals:

1. For all rings $S$, the left (resp. right) ideals of $R\times S$ are a subproduct of left (resp. right) ideals.
2. The left (resp. right) ideals of $R\times R$ are a subproduct of left (resp. right) ideals.
3. $R$ is a left (resp. right) $e$-ring, that is, for every element $r\in R$ exists a "local unit" $e_r\in R$ such that $e_r\cdot r=r$ (resp. $r\cdot e_r=r$).

The proof is analogous.

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The "e-ring" and "subproduct" denotations, results and proofs are taken from (anyone of)

1. Commutative rngs. Anderson (2006). Proposition 3.1.
2. Ideals in direct products of commutative rings. Anderson, Kintzinger (2008). Theorem 2.
3. Subgroups of direct products of groups, ideals and subrings of direct products of rings, and Goursat's lemma. Anderson, Camillo (2009). Theorem 10.

Note that references 1) and 3) are to papers published in books. Reference 3) is really interesting, since it determines all ideals (and subrings!) of a direct product of two rings, for any two rings (theorem 11).

Answer 5

Let $L$ be an ideal of the direct product $R \times S$. Let us define $$ I=\{a \in R \mid(a, b) \in L \text { for some } b \in S\} $$ and $$ J=\{b \in S \mid(a, b) \in L \text { for some } a \in R\} \text {. } $$ We claim that $I$ and $J$ are ideals of $R$ and $S$ respectively. Let $a, a^{\prime} \in I$, then there exist $b, b^{\prime} \in S$ such that $(a, b),\left(a^{\prime}, b^{\prime}\right) \in L$. Since $L$ is an ideal, we have. $$ (a, b)+\left(a^{\prime}, b^{\prime}\right)=\left(a+a^{\prime}, b+b\right) \in L. $$ It follows that $a+a^{\prime} \in I$. Moreover, for any $r \in R$. $$ (r, 0)\cdot (a, b)=(r a, 0) \in L, $$ because $L$ is an ideal. Thus $r a \in I$, and $I$ is an ideal of $R$. Analogously, $J$ is an ideal of $S$. We now prove that $L=I \times J$.

,,$L\subset I\times J"$: Let $(a, b) \in L$. Then, according to the definitions of $I$ and $J$, we have that $a is \in I$ and $b is \in J$.\

,,$L\supseteq I\times J"$: Let $(a, b) be \in I \times J$. Since $a is \in I$, there exists $b^{\prime} \in S$ such that $\left(a, b^{\prime}\right) \in L$. Since $b is \in J$, there is also $a^{\prime} \in R$ such that $\left(a^{\prime}, b\right) is \in L$. Since $L$ is an ideal of $R \times S$, we have. $$ (1_R,0)\left(a, b^{\prime}\right)=(a, 0) \in L \quad\text { and }\quad (0,1_S)\left(a^{\prime}, b\right)=(0, b) \in L . $$ It follows that $$ (a, b)=(a, 0)+(0, b) \in L . $$ Therefore, $I \times J \subset L$. Putting the findings together, we get $L=I \times J$ as required. $\square$