Structure of ideals in the product of two ringsIdeals in direct product of ringsIdeals of the real...
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Structure of ideals in the product of two rings
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$R$ 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
$endgroup$
add a comment |
$begingroup$
$R$ 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!
ring-theory ideals
$endgroup$
add a comment |
$begingroup$
$R$ 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!
ring-theory ideals
$endgroup$
$R$ 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
ring-theory ideals
edited Feb 14 '16 at 21:04
Manos
14.1k33288
14.1k33288
asked Jan 22 '12 at 2:03
IIJIIJ
12113
12113
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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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.
$endgroup$
$begingroup$
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
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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
add a comment |
$begingroup$
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.
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1
$begingroup$
Wonderful remark!
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– Matemáticos Chibchas
Oct 7 '17 at 14:35
add a comment |
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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 $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$
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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:
- For all rings $S$, the ideals of $Rtimes S$ are subproduct ideals.
- The ideals of $Rtimes R$ are subproduct ideals.
- $R$ 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:
- For all rings $S$, the left (resp. right) ideals of $Rtimes S$ are a subproduct of left (resp. right) ideals.
- The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.
- $R$ 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)
Commutative rngs. Anderson (2006). Proposition 3.1.
Ideals in direct products of commutative rings. Anderson, Kintzinger (2008). Theorem 2.
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).
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
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active
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$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
Wonderful remark!
$endgroup$
– Matemáticos Chibchas
Oct 7 '17 at 14:35
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
Wonderful remark!
$endgroup$
– Matemáticos Chibchas
Oct 7 '17 at 14:35
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
$begingroup$
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 $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$
$endgroup$
add a comment |
$begingroup$
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 $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$
$endgroup$
add a comment |
$begingroup$
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 $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$
$endgroup$
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 $(a,b) = (1,0)(a,b') + (0,1)(a',b) in K$. $square$
edited Nov 28 '16 at 17:20
user363520
answered Feb 26 '15 at 17:31
ujsgeyrr1f0d0d0r0h1h0j0j_jujujsgeyrr1f0d0d0r0h1h0j0j_juj
9,25711638
9,25711638
add a comment |
add a comment |
$begingroup$
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:
- For all rings $S$, the ideals of $Rtimes S$ are subproduct ideals.
- The ideals of $Rtimes R$ are subproduct ideals.
- $R$ 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:
- For all rings $S$, the left (resp. right) ideals of $Rtimes S$ are a subproduct of left (resp. right) ideals.
- The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.
- $R$ 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)
Commutative rngs. Anderson (2006). Proposition 3.1.
Ideals in direct products of commutative rings. Anderson, Kintzinger (2008). Theorem 2.
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).
$endgroup$
add a comment |
$begingroup$
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:
- For all rings $S$, the ideals of $Rtimes S$ are subproduct ideals.
- The ideals of $Rtimes R$ are subproduct ideals.
- $R$ 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:
- For all rings $S$, the left (resp. right) ideals of $Rtimes S$ are a subproduct of left (resp. right) ideals.
- The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.
- $R$ 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)
Commutative rngs. Anderson (2006). Proposition 3.1.
Ideals in direct products of commutative rings. Anderson, Kintzinger (2008). Theorem 2.
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).
$endgroup$
add a comment |
$begingroup$
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:
- For all rings $S$, the ideals of $Rtimes S$ are subproduct ideals.
- The ideals of $Rtimes R$ are subproduct ideals.
- $R$ 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:
- For all rings $S$, the left (resp. right) ideals of $Rtimes S$ are a subproduct of left (resp. right) ideals.
- The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.
- $R$ 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)
Commutative rngs. Anderson (2006). Proposition 3.1.
Ideals in direct products of commutative rings. Anderson, Kintzinger (2008). Theorem 2.
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).
$endgroup$
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:
- For all rings $S$, the ideals of $Rtimes S$ are subproduct ideals.
- The ideals of $Rtimes R$ are subproduct ideals.
- $R$ 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:
- For all rings $S$, the left (resp. right) ideals of $Rtimes S$ are a subproduct of left (resp. right) ideals.
- The left (resp. right) ideals of $Rtimes R$ are a subproduct of left (resp. right) ideals.
- $R$ 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)
Commutative rngs. Anderson (2006). Proposition 3.1.
Ideals in direct products of commutative rings. Anderson, Kintzinger (2008). Theorem 2.
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).
edited Nov 12 '17 at 18:50
answered Nov 12 '17 at 18:45
Jose BroxJose Brox
3,42211129
3,42211129
add a comment |
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