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Structure of ideals in the product of two rings


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$begingroup$



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




For me is obvious why $left{ rin Rmid left(r,sright)in Jtext{ for some } sin Sright}$ 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 also an ideal.



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



Thank you for your help!










share|cite|improve this question











$endgroup$

















    24












    $begingroup$



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




    For me is obvious why $left{ rin Rmid left(r,sright)in Jtext{ for some } sin Sright}$ 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 also an ideal.



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



    Thank you for your help!










    share|cite|improve this question











    $endgroup$















      24












      24








      24


      11



      $begingroup$



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




      For me is obvious why $left{ rin Rmid left(r,sright)in Jtext{ for some } sin Sright}$ 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 also an ideal.



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



      Thank you for your help!










      share|cite|improve this question











      $endgroup$





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




      For me is obvious why $left{ rin Rmid left(r,sright)in Jtext{ for some } sin Sright}$ 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 also an ideal.



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



      Thank you for your help!







      ring-theory ideals






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 14 '16 at 21:04









      Manos

      14.1k33288




      14.1k33288










      asked Jan 22 '12 at 2:03









      IIJIIJ

      12113




      12113






















          4 Answers
          4






          active

          oldest

          votes


















          21












          $begingroup$

          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 $mathbb Z_2oplus mathbb Z_2$, then the ideal ($=$subgroup) generated by $(overline 1,overline 1)$ is a counterexample.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
            $endgroup$
            – IIJ
            Jan 22 '12 at 2:29






          • 2




            $begingroup$
            @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
            $endgroup$
            – Jonas Meyer
            Jan 22 '12 at 2:32












          • $begingroup$
            Oh yes! Thank you!. have a great day!
            $endgroup$
            – IIJ
            Jan 22 '12 at 2:44










          • $begingroup$
            Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
            $endgroup$
            – d13
            Apr 2 '14 at 10:54










          • $begingroup$
            @d13: Sorry, I don't understand your comment.
            $endgroup$
            – Jonas Meyer
            Jun 27 '14 at 7:21



















          8












          $begingroup$

          Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $Rtimes S$. 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 $I_S, I_R$ 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 $(r,s) in I$.



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






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            Wonderful remark!
            $endgroup$
            – Matemáticos Chibchas
            Oct 7 '17 at 14:35



















          5












          $begingroup$

          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 $I_1 times ldotstimes I_n$, where $I_i$ is an ideal of $A_i$ for each $iin{1,ldots,n}$.



          Proof. By induction on $n$, the cases $nin{0,1}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $Atimes B$ of two rings with $1$, and denote by $p : Atimes Bto A$ and $q : Atimes Bto B$ 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 $Ksubseteq Itimes J$. To show the inverse inclusion, let $(a,b)in Itimes J$. Then $(a,b') in K$ and $(a',b)in K$ for some $(a',b')in Atimes B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$






          share|cite|improve this answer











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            3












            $begingroup$

            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 $Rtimes S$. We say that an ideal of the product is a subproduct ideal if it is of the form $Itimes J$ 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 $Rtimes S$ are subproduct ideals.

            2. The ideals of $Rtimes R$ are subproduct ideals.

            3. $R$ is a (two-sided) $e$-ring, that is, for every element $rin R$ we have $rin Rr+rR+RrR$.


            (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 $Rtimes S$, we define
            $$I_R:={rin R | exists sin S, (r,s)in I}, I_S:={sin S | exists rin R, (r,s)in I}.$$
            That $Isubseteq I_Rtimes I_S$ is straightforward. Pick $(r,s)in I_Rtimes I_S$. Then there are $r'in R$, $s'in S$ such that $(r,s'),(r',s)in I$. By hypothesis we know that there exist $a,b,u_i,v_iin R$ 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 $I=I_Rtimes I_S$.



            1)$Rightarrow$2) Trivial.



            2)$Rightarrow$3) Pick $rin R$ and consider the ideal $I$ generated by $(r,r)$ in $Rtimes R$. By hypothesis we have $I=Jtimes K$, with $J,K$ ideals of $R$. Since $(r,r)in Jtimes K$ we have $(r,0)in Jtimes K=I=$Id$(r,r)$, what means that there exist $zinmathbb{Z}$, $a_1,a_2,b_1,b_2,u_i,v_i,x_i,y_iin R$ 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 $r=(a_1-a_2)r+r(b_1-b_2)+sum_i c_ird_iin Rr+rR+RrR$.





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





            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 $Rtimes S$ are a subproduct of left (resp. right) ideals.

            2. The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.

            3. $R$ is a left (resp. right) $e$-ring, that is, for every element $rin R$ exists a "local unit" $e_rin R$ such that $e_rcdot r=r$ (resp. $rcdot e_r=r$).


            The proof is analogous.





            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).






            share|cite|improve this answer











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              4 Answers
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              4 Answers
              4






              active

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              active

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              active

              oldest

              votes









              21












              $begingroup$

              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 $mathbb Z_2oplus mathbb Z_2$, then the ideal ($=$subgroup) generated by $(overline 1,overline 1)$ is a counterexample.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:29






              • 2




                $begingroup$
                @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
                $endgroup$
                – Jonas Meyer
                Jan 22 '12 at 2:32












              • $begingroup$
                Oh yes! Thank you!. have a great day!
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:44










              • $begingroup$
                Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
                $endgroup$
                – d13
                Apr 2 '14 at 10:54










              • $begingroup$
                @d13: Sorry, I don't understand your comment.
                $endgroup$
                – Jonas Meyer
                Jun 27 '14 at 7:21
















              21












              $begingroup$

              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 $mathbb Z_2oplus mathbb Z_2$, then the ideal ($=$subgroup) generated by $(overline 1,overline 1)$ is a counterexample.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:29






              • 2




                $begingroup$
                @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
                $endgroup$
                – Jonas Meyer
                Jan 22 '12 at 2:32












              • $begingroup$
                Oh yes! Thank you!. have a great day!
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:44










              • $begingroup$
                Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
                $endgroup$
                – d13
                Apr 2 '14 at 10:54










              • $begingroup$
                @d13: Sorry, I don't understand your comment.
                $endgroup$
                – Jonas Meyer
                Jun 27 '14 at 7:21














              21












              21








              21





              $begingroup$

              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 $mathbb Z_2oplus mathbb Z_2$, then the ideal ($=$subgroup) generated by $(overline 1,overline 1)$ is a counterexample.






              share|cite|improve this answer









              $endgroup$



              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 $mathbb Z_2oplus mathbb Z_2$, then the ideal ($=$subgroup) generated by $(overline 1,overline 1)$ is a counterexample.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 22 '12 at 2:15









              Jonas MeyerJonas Meyer

              41k6148257




              41k6148257












              • $begingroup$
                I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:29






              • 2




                $begingroup$
                @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
                $endgroup$
                – Jonas Meyer
                Jan 22 '12 at 2:32












              • $begingroup$
                Oh yes! Thank you!. have a great day!
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:44










              • $begingroup$
                Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
                $endgroup$
                – d13
                Apr 2 '14 at 10:54










              • $begingroup$
                @d13: Sorry, I don't understand your comment.
                $endgroup$
                – Jonas Meyer
                Jun 27 '14 at 7:21


















              • $begingroup$
                I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:29






              • 2




                $begingroup$
                @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
                $endgroup$
                – Jonas Meyer
                Jan 22 '12 at 2:32












              • $begingroup$
                Oh yes! Thank you!. have a great day!
                $endgroup$
                – IIJ
                Jan 22 '12 at 2:44










              • $begingroup$
                Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
                $endgroup$
                – d13
                Apr 2 '14 at 10:54










              • $begingroup$
                @d13: Sorry, I don't understand your comment.
                $endgroup$
                – Jonas Meyer
                Jun 27 '14 at 7:21
















              $begingroup$
              I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
              $endgroup$
              – IIJ
              Jan 22 '12 at 2:29




              $begingroup$
              I dont exactly understand what "put the zero product on the abelian group" mean. can you explain please? and Thank you for your quick answer =)
              $endgroup$
              – IIJ
              Jan 22 '12 at 2:29




              2




              2




              $begingroup$
              @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
              $endgroup$
              – Jonas Meyer
              Jan 22 '12 at 2:32






              $begingroup$
              @IIJ: If $(G,+)$ is an abelian group and you define a multiplication by $acdot b =0$ for all $a,bin G$, then $G$ becomes a ring with the zero product. E.g. $mathbb Z_2$ with the zero product has the usual addition, but $overline 0cdot overline 0 =overline 1cdot overline 0 =overline 0cdot overline 1=overline 1cdotoverline 1 = overline 0$.
              $endgroup$
              – Jonas Meyer
              Jan 22 '12 at 2:32














              $begingroup$
              Oh yes! Thank you!. have a great day!
              $endgroup$
              – IIJ
              Jan 22 '12 at 2:44




              $begingroup$
              Oh yes! Thank you!. have a great day!
              $endgroup$
              – IIJ
              Jan 22 '12 at 2:44












              $begingroup$
              Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
              $endgroup$
              – d13
              Apr 2 '14 at 10:54




              $begingroup$
              Hello Sir, I have a question, what if J in this case is prime ideal? how do we prove that its a product of one prime ideal of R and S itself or product of R and prime ideal of S? I came up with the R*S/J being integral domain.. But I'm stuck. Thankyou
              $endgroup$
              – d13
              Apr 2 '14 at 10:54












              $begingroup$
              @d13: Sorry, I don't understand your comment.
              $endgroup$
              – Jonas Meyer
              Jun 27 '14 at 7:21




              $begingroup$
              @d13: Sorry, I don't understand your comment.
              $endgroup$
              – Jonas Meyer
              Jun 27 '14 at 7:21











              8












              $begingroup$

              Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $Rtimes S$. 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 $I_S, I_R$ 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 $(r,s) in I$.



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






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Wonderful remark!
                $endgroup$
                – Matemáticos Chibchas
                Oct 7 '17 at 14:35
















              8












              $begingroup$

              Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $Rtimes S$. 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 $I_S, I_R$ 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 $(r,s) in I$.



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






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Wonderful remark!
                $endgroup$
                – Matemáticos Chibchas
                Oct 7 '17 at 14:35














              8












              8








              8





              $begingroup$

              Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $Rtimes S$. 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 $I_S, I_R$ 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 $(r,s) in I$.



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






              share|cite|improve this answer











              $endgroup$



              Let us assume that $R$ has an identity element $1_R$. Let $I$ be a two-sided ideal of $Rtimes S$. 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 $I_S, I_R$ 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 $(r,s) in I$.



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







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Oct 7 '17 at 14:33









              Learnmore

              17.8k325103




              17.8k325103










              answered Feb 11 '16 at 20:00









              ManosManos

              14.1k33288




              14.1k33288








              • 1




                $begingroup$
                Wonderful remark!
                $endgroup$
                – Matemáticos Chibchas
                Oct 7 '17 at 14:35














              • 1




                $begingroup$
                Wonderful remark!
                $endgroup$
                – Matemáticos Chibchas
                Oct 7 '17 at 14:35








              1




              1




              $begingroup$
              Wonderful remark!
              $endgroup$
              – Matemáticos Chibchas
              Oct 7 '17 at 14:35




              $begingroup$
              Wonderful remark!
              $endgroup$
              – Matemáticos Chibchas
              Oct 7 '17 at 14:35











              5












              $begingroup$

              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 $I_1 times ldotstimes I_n$, where $I_i$ is an ideal of $A_i$ for each $iin{1,ldots,n}$.



              Proof. By induction on $n$, the cases $nin{0,1}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $Atimes B$ of two rings with $1$, and denote by $p : Atimes Bto A$ and $q : Atimes Bto B$ 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 $Ksubseteq Itimes J$. To show the inverse inclusion, let $(a,b)in Itimes J$. Then $(a,b') in K$ and $(a',b)in K$ for some $(a',b')in Atimes B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                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 $I_1 times ldotstimes I_n$, where $I_i$ is an ideal of $A_i$ for each $iin{1,ldots,n}$.



                Proof. By induction on $n$, the cases $nin{0,1}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $Atimes B$ of two rings with $1$, and denote by $p : Atimes Bto A$ and $q : Atimes Bto B$ 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 $Ksubseteq Itimes J$. To show the inverse inclusion, let $(a,b)in Itimes J$. Then $(a,b') in K$ and $(a',b)in K$ for some $(a',b')in Atimes B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  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 $I_1 times ldotstimes I_n$, where $I_i$ is an ideal of $A_i$ for each $iin{1,ldots,n}$.



                  Proof. By induction on $n$, the cases $nin{0,1}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $Atimes B$ of two rings with $1$, and denote by $p : Atimes Bto A$ and $q : Atimes Bto B$ 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 $Ksubseteq Itimes J$. To show the inverse inclusion, let $(a,b)in Itimes J$. Then $(a,b') in K$ and $(a',b)in K$ for some $(a',b')in Atimes B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$






                  share|cite|improve this answer











                  $endgroup$



                  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 $I_1 times ldotstimes I_n$, where $I_i$ is an ideal of $A_i$ for each $iin{1,ldots,n}$.



                  Proof. By induction on $n$, the cases $nin{0,1}$ being trivial, one sees that it suffices to prove the assertion for $n=2$. Let $K$ be an ideal of a product $Atimes B$ of two rings with $1$, and denote by $p : Atimes Bto A$ and $q : Atimes Bto B$ 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 $Ksubseteq Itimes J$. To show the inverse inclusion, let $(a,b)in Itimes J$. Then $(a,b') in K$ and $(a',b)in K$ for some $(a',b')in Atimes B$. Then $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 '16 at 17:20







                  user363520

















                  answered Feb 26 '15 at 17:31









                  ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj

                  9,25711638




                  9,25711638























                      3












                      $begingroup$

                      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 $Rtimes S$. We say that an ideal of the product is a subproduct ideal if it is of the form $Itimes J$ 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 $Rtimes S$ are subproduct ideals.

                      2. The ideals of $Rtimes R$ are subproduct ideals.

                      3. $R$ is a (two-sided) $e$-ring, that is, for every element $rin R$ we have $rin Rr+rR+RrR$.


                      (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 $Rtimes S$, we define
                      $$I_R:={rin R | exists sin S, (r,s)in I}, I_S:={sin S | exists rin R, (r,s)in I}.$$
                      That $Isubseteq I_Rtimes I_S$ is straightforward. Pick $(r,s)in I_Rtimes I_S$. Then there are $r'in R$, $s'in S$ such that $(r,s'),(r',s)in I$. By hypothesis we know that there exist $a,b,u_i,v_iin R$ 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 $I=I_Rtimes I_S$.



                      1)$Rightarrow$2) Trivial.



                      2)$Rightarrow$3) Pick $rin R$ and consider the ideal $I$ generated by $(r,r)$ in $Rtimes R$. By hypothesis we have $I=Jtimes K$, with $J,K$ ideals of $R$. Since $(r,r)in Jtimes K$ we have $(r,0)in Jtimes K=I=$Id$(r,r)$, what means that there exist $zinmathbb{Z}$, $a_1,a_2,b_1,b_2,u_i,v_i,x_i,y_iin R$ 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 $r=(a_1-a_2)r+r(b_1-b_2)+sum_i c_ird_iin Rr+rR+RrR$.





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





                      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 $Rtimes S$ are a subproduct of left (resp. right) ideals.

                      2. The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.

                      3. $R$ is a left (resp. right) $e$-ring, that is, for every element $rin R$ exists a "local unit" $e_rin R$ such that $e_rcdot r=r$ (resp. $rcdot e_r=r$).


                      The proof is analogous.





                      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).






                      share|cite|improve this answer











                      $endgroup$


















                        3












                        $begingroup$

                        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 $Rtimes S$. We say that an ideal of the product is a subproduct ideal if it is of the form $Itimes J$ 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 $Rtimes S$ are subproduct ideals.

                        2. The ideals of $Rtimes R$ are subproduct ideals.

                        3. $R$ is a (two-sided) $e$-ring, that is, for every element $rin R$ we have $rin Rr+rR+RrR$.


                        (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 $Rtimes S$, we define
                        $$I_R:={rin R | exists sin S, (r,s)in I}, I_S:={sin S | exists rin R, (r,s)in I}.$$
                        That $Isubseteq I_Rtimes I_S$ is straightforward. Pick $(r,s)in I_Rtimes I_S$. Then there are $r'in R$, $s'in S$ such that $(r,s'),(r',s)in I$. By hypothesis we know that there exist $a,b,u_i,v_iin R$ 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 $I=I_Rtimes I_S$.



                        1)$Rightarrow$2) Trivial.



                        2)$Rightarrow$3) Pick $rin R$ and consider the ideal $I$ generated by $(r,r)$ in $Rtimes R$. By hypothesis we have $I=Jtimes K$, with $J,K$ ideals of $R$. Since $(r,r)in Jtimes K$ we have $(r,0)in Jtimes K=I=$Id$(r,r)$, what means that there exist $zinmathbb{Z}$, $a_1,a_2,b_1,b_2,u_i,v_i,x_i,y_iin R$ 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 $r=(a_1-a_2)r+r(b_1-b_2)+sum_i c_ird_iin Rr+rR+RrR$.





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





                        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 $Rtimes S$ are a subproduct of left (resp. right) ideals.

                        2. The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.

                        3. $R$ is a left (resp. right) $e$-ring, that is, for every element $rin R$ exists a "local unit" $e_rin R$ such that $e_rcdot r=r$ (resp. $rcdot e_r=r$).


                        The proof is analogous.





                        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).






                        share|cite|improve this answer











                        $endgroup$
















                          3












                          3








                          3





                          $begingroup$

                          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 $Rtimes S$. We say that an ideal of the product is a subproduct ideal if it is of the form $Itimes J$ 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 $Rtimes S$ are subproduct ideals.

                          2. The ideals of $Rtimes R$ are subproduct ideals.

                          3. $R$ is a (two-sided) $e$-ring, that is, for every element $rin R$ we have $rin Rr+rR+RrR$.


                          (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 $Rtimes S$, we define
                          $$I_R:={rin R | exists sin S, (r,s)in I}, I_S:={sin S | exists rin R, (r,s)in I}.$$
                          That $Isubseteq I_Rtimes I_S$ is straightforward. Pick $(r,s)in I_Rtimes I_S$. Then there are $r'in R$, $s'in S$ such that $(r,s'),(r',s)in I$. By hypothesis we know that there exist $a,b,u_i,v_iin R$ 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 $I=I_Rtimes I_S$.



                          1)$Rightarrow$2) Trivial.



                          2)$Rightarrow$3) Pick $rin R$ and consider the ideal $I$ generated by $(r,r)$ in $Rtimes R$. By hypothesis we have $I=Jtimes K$, with $J,K$ ideals of $R$. Since $(r,r)in Jtimes K$ we have $(r,0)in Jtimes K=I=$Id$(r,r)$, what means that there exist $zinmathbb{Z}$, $a_1,a_2,b_1,b_2,u_i,v_i,x_i,y_iin R$ 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 $r=(a_1-a_2)r+r(b_1-b_2)+sum_i c_ird_iin Rr+rR+RrR$.





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





                          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 $Rtimes S$ are a subproduct of left (resp. right) ideals.

                          2. The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.

                          3. $R$ is a left (resp. right) $e$-ring, that is, for every element $rin R$ exists a "local unit" $e_rin R$ such that $e_rcdot r=r$ (resp. $rcdot e_r=r$).


                          The proof is analogous.





                          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).






                          share|cite|improve this answer











                          $endgroup$



                          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 $Rtimes S$. We say that an ideal of the product is a subproduct ideal if it is of the form $Itimes J$ 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 $Rtimes S$ are subproduct ideals.

                          2. The ideals of $Rtimes R$ are subproduct ideals.

                          3. $R$ is a (two-sided) $e$-ring, that is, for every element $rin R$ we have $rin Rr+rR+RrR$.


                          (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 $Rtimes S$, we define
                          $$I_R:={rin R | exists sin S, (r,s)in I}, I_S:={sin S | exists rin R, (r,s)in I}.$$
                          That $Isubseteq I_Rtimes I_S$ is straightforward. Pick $(r,s)in I_Rtimes I_S$. Then there are $r'in R$, $s'in S$ such that $(r,s'),(r',s)in I$. By hypothesis we know that there exist $a,b,u_i,v_iin R$ 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 $I=I_Rtimes I_S$.



                          1)$Rightarrow$2) Trivial.



                          2)$Rightarrow$3) Pick $rin R$ and consider the ideal $I$ generated by $(r,r)$ in $Rtimes R$. By hypothesis we have $I=Jtimes K$, with $J,K$ ideals of $R$. Since $(r,r)in Jtimes K$ we have $(r,0)in Jtimes K=I=$Id$(r,r)$, what means that there exist $zinmathbb{Z}$, $a_1,a_2,b_1,b_2,u_i,v_i,x_i,y_iin R$ 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 $r=(a_1-a_2)r+r(b_1-b_2)+sum_i c_ird_iin Rr+rR+RrR$.





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





                          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 $Rtimes S$ are a subproduct of left (resp. right) ideals.

                          2. The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.

                          3. $R$ is a left (resp. right) $e$-ring, that is, for every element $rin R$ exists a "local unit" $e_rin R$ such that $e_rcdot r=r$ (resp. $rcdot e_r=r$).


                          The proof is analogous.





                          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).







                          share|cite|improve this answer














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                          edited Nov 12 '17 at 18:50

























                          answered Nov 12 '17 at 18:45









                          Jose BroxJose Brox

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