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Subring and ideal structure of rings in the form $Z_noplus Z_m$
Clash Royale CLAN TAG#URR8PPP
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So I was wondering how the ideals and subrings of something like this look like. I know how the structure of $Z_n$ looks like since every subring of a cyclic group is an ideal. However, this is clearly not cyclic unless $n$ is relatively prime with $m$. Any help would be helpful.
Thank you.
ring-theory
edited Jul 15 at 11:46
M.G
2,1531034
asked Jul 15 at 11:44
Sorfosh
910616
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up vote
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So I was wondering how the ideals and subrings of something like this look like. I know how the structure of $Z_n$ looks like since every subring of a cyclic group is an ideal. However, this is clearly not cyclic unless $n$ is relatively prime with $m$. Any help would be helpful.
Thank you.
ring-theory
edited Jul 15 at 11:46
M.G
2,1531034
asked Jul 15 at 11:44
Sorfosh
910616
Can you prove that if R and S are rings with unity 1, then every ideal in R x S has the form I x J, where I is an ideal in R and where J is an ideal in S?
– CJD
Jul 15 at 11:49
@CJD It is easy to prove the reverse case, every ideal in that form is, in fact, an ideal.
– Sorfosh
Jul 15 at 12:23
Here is an idea of how to prove the thing I said. Let A in R x S denote an arbitrary ideal. We want to show this ideal is of the form I x J. Let (r,s) denote an element in A. If you multiply by (1,0), you see that (r,0) is in A...
– CJD
Jul 15 at 12:49
@CJD Got it! Thanks. Let $A$ be an ideal or $Roplus S$. Then clearly the elements $(r,0)$ imply $I$ is an ideal of $R$ and $(0,r)$ imply $J$ is an ideal of $S$.
– Sorfosh
Jul 15 at 19:13
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I was wondering how the ideals and subrings of something like this look like. I know how the structure of $Z_n$ looks like since every subring of a cyclic group is an ideal. However, this is clearly not cyclic unless $n$ is relatively prime with $m$. Any help would be helpful.
Thank you.
ring-theory
edited Jul 15 at 11:46
M.G
2,1531034
asked Jul 15 at 11:44
Sorfosh
910616
So I was wondering how the ideals and subrings of something like this look like. I know how the structure of $Z_n$ looks like since every subring of a cyclic group is an ideal. However, this is clearly not cyclic unless $n$ is relatively prime with $m$. Any help would be helpful.
Thank you.
ring-theory
edited Jul 15 at 11:46
M.G
2,1531034
asked Jul 15 at 11:44
Sorfosh
910616
edited Jul 15 at 11:46
M.G
2,1531034
edited Jul 15 at 11:46
M.G
2,1531034
edited Jul 15 at 11:46
M.G
2,1531034
2,1531034
asked Jul 15 at 11:44
Sorfosh
910616
asked Jul 15 at 11:44
Sorfosh
910616
asked Jul 15 at 11:44
Sorfosh
910616
910616
Can you prove that if R and S are rings with unity 1, then every ideal in R x S has the form I x J, where I is an ideal in R and where J is an ideal in S?
– CJD
Jul 15 at 11:49
@CJD It is easy to prove the reverse case, every ideal in that form is, in fact, an ideal.
– Sorfosh
Jul 15 at 12:23
Here is an idea of how to prove the thing I said. Let A in R x S denote an arbitrary ideal. We want to show this ideal is of the form I x J. Let (r,s) denote an element in A. If you multiply by (1,0), you see that (r,0) is in A...
– CJD
Jul 15 at 12:49
@CJD Got it! Thanks. Let $A$ be an ideal or $Roplus S$. Then clearly the elements $(r,0)$ imply $I$ is an ideal of $R$ and $(0,r)$ imply $J$ is an ideal of $S$.
– Sorfosh
Jul 15 at 19:13
add a comment |Â
Can you prove that if R and S are rings with unity 1, then every ideal in R x S has the form I x J, where I is an ideal in R and where J is an ideal in S?
– CJD
Jul 15 at 11:49
@CJD It is easy to prove the reverse case, every ideal in that form is, in fact, an ideal.
– Sorfosh
Jul 15 at 12:23
Here is an idea of how to prove the thing I said. Let A in R x S denote an arbitrary ideal. We want to show this ideal is of the form I x J. Let (r,s) denote an element in A. If you multiply by (1,0), you see that (r,0) is in A...
– CJD
Jul 15 at 12:49
@CJD Got it! Thanks. Let $A$ be an ideal or $Roplus S$. Then clearly the elements $(r,0)$ imply $I$ is an ideal of $R$ and $(0,r)$ imply $J$ is an ideal of $S$.
– Sorfosh
Jul 15 at 19:13
Can you prove that if R and S are rings with unity 1, then every ideal in R x S has the form I x J, where I is an ideal in R and where J is an ideal in S?
– CJD
Jul 15 at 11:49
Can you prove that if R and S are rings with unity 1, then every ideal in R x S has the form I x J, where I is an ideal in R and where J is an ideal in S?
– CJD
Jul 15 at 11:49
@CJD It is easy to prove the reverse case, every ideal in that form is, in fact, an ideal.
– Sorfosh
Jul 15 at 12:23
@CJD It is easy to prove the reverse case, every ideal in that form is, in fact, an ideal.
– Sorfosh
Jul 15 at 12:23
Here is an idea of how to prove the thing I said. Let A in R x S denote an arbitrary ideal. We want to show this ideal is of the form I x J. Let (r,s) denote an element in A. If you multiply by (1,0), you see that (r,0) is in A...
– CJD
Jul 15 at 12:49
Here is an idea of how to prove the thing I said. Let A in R x S denote an arbitrary ideal. We want to show this ideal is of the form I x J. Let (r,s) denote an element in A. If you multiply by (1,0), you see that (r,0) is in A...
– CJD
Jul 15 at 12:49
@CJD Got it! Thanks. Let $A$ be an ideal or $Roplus S$. Then clearly the elements $(r,0)$ imply $I$ is an ideal of $R$ and $(0,r)$ imply $J$ is an ideal of $S$.
– Sorfosh
Jul 15 at 19:13
@CJD Got it! Thanks. Let $A$ be an ideal or $Roplus S$. Then clearly the elements $(r,0)$ imply $I$ is an ideal of $R$ and $(0,r)$ imply $J$ is an ideal of $S$.
– Sorfosh
Jul 15 at 19:13
add a comment |Â
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Can you prove that if R and S are rings with unity 1, then every ideal in R x S has the form I x J, where I is an ideal in R and where J is an ideal in S?
– CJD
Jul 15 at 11:49
@CJD It is easy to prove the reverse case, every ideal in that form is, in fact, an ideal.
– Sorfosh
Jul 15 at 12:23
Here is an idea of how to prove the thing I said. Let A in R x S denote an arbitrary ideal. We want to show this ideal is of the form I x J. Let (r,s) denote an element in A. If you multiply by (1,0), you see that (r,0) is in A...
– CJD
Jul 15 at 12:49
@CJD Got it! Thanks. Let $A$ be an ideal or $Roplus S$. Then clearly the elements $(r,0)$ imply $I$ is an ideal of $R$ and $(0,r)$ imply $J$ is an ideal of $S$.
– Sorfosh
Jul 15 at 19:13