Mixing
isomorphism $R oplus R / langle (a,b) rangle simeq R$
2
$begingroup$
Let $R$ be a commutative ring. Let $(a,b) in R oplus R$, where $a,b in R$ are not units. Is it possible to show that there is never an isomorphism $R to (Roplus R)/langle (a,b) rangle$ of $R$-modules?
abstract-algebra ring-theory modules
edited Jan 26 at 21:07
user26857
39.4k124183
asked Jan 26 at 11:28
zinRzinR
688
$endgroup$
|
show 2 more comments
2
$begingroup$
Let $R$ be a commutative ring. Let $(a,b) in R oplus R$, where $a,b in R$ are not units. Is it possible to show that there is never an isomorphism $R to (Roplus R)/langle (a,b) rangle$ of $R$-modules?
abstract-algebra ring-theory modules
edited Jan 26 at 21:07
user26857
39.4k124183
asked Jan 26 at 11:28
zinRzinR
688
$endgroup$
$begingroup$
Surely there can be an isomorphism?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 12:09
$begingroup$
@LordSharktheUnknown Can you think of an example?
$endgroup$
– zinR
Jan 26 at 14:18
2
$begingroup$
Take $R$ to be your favorite ring (say $k[x]$), pick $ain R$ such that neither $a$ nor $b=1-a$ is a unit.
$endgroup$
– Mohan
Jan 26 at 14:41
2
$begingroup$
$R=Bbb Z$, $(a,b)=(2,3)$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 14:59
$begingroup$
$R = mathbb{Z}/pmathbb{Z}$, $a = b = 1$. Then $|Roplus R/langle(a,b)rangle| = p$, hence $R oplus R/langle(a,b)ranglecong R$, simply because there are only two rings of order $p$, and we have non-zero multiplication in $Roplus R/langle(a,b)rangle$.
$endgroup$
– user3482749
Jan 26 at 17:14
|
show 2 more comments
2
2
2
$begingroup$
Let $R$ be a commutative ring. Let $(a,b) in R oplus R$, where $a,b in R$ are not units. Is it possible to show that there is never an isomorphism $R to (Roplus R)/langle (a,b) rangle$ of $R$-modules?
abstract-algebra ring-theory modules
edited Jan 26 at 21:07
user26857
39.4k124183
asked Jan 26 at 11:28
zinRzinR
688
$endgroup$
Let $R$ be a commutative ring. Let $(a,b) in R oplus R$, where $a,b in R$ are not units. Is it possible to show that there is never an isomorphism $R to (Roplus R)/langle (a,b) rangle$ of $R$-modules?
abstract-algebra ring-theory modules
abstract-algebra ring-theory modules
edited Jan 26 at 21:07
user26857
39.4k124183
asked Jan 26 at 11:28
zinRzinR
688
edited Jan 26 at 21:07
user26857
39.4k124183
asked Jan 26 at 11:28
zinRzinR
688
edited Jan 26 at 21:07
user26857
39.4k124183
edited Jan 26 at 21:07
user26857
39.4k124183
edited Jan 26 at 21:07
user26857
39.4k124183
39.4k124183
asked Jan 26 at 11:28
zinRzinR
688
asked Jan 26 at 11:28
zinRzinR
688
asked Jan 26 at 11:28
zinRzinR
688
688
$begingroup$
Surely there can be an isomorphism?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 12:09
$begingroup$
@LordSharktheUnknown Can you think of an example?
$endgroup$
– zinR
Jan 26 at 14:18
2
$begingroup$
Take $R$ to be your favorite ring (say $k[x]$), pick $ain R$ such that neither $a$ nor $b=1-a$ is a unit.
$endgroup$
– Mohan
Jan 26 at 14:41
2
$begingroup$
$R=Bbb Z$, $(a,b)=(2,3)$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 14:59
$begingroup$
$R = mathbb{Z}/pmathbb{Z}$, $a = b = 1$. Then $|Roplus R/langle(a,b)rangle| = p$, hence $R oplus R/langle(a,b)ranglecong R$, simply because there are only two rings of order $p$, and we have non-zero multiplication in $Roplus R/langle(a,b)rangle$.
$endgroup$
– user3482749
Jan 26 at 17:14
|
show 2 more comments
$begingroup$
Surely there can be an isomorphism?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 12:09
$begingroup$
@LordSharktheUnknown Can you think of an example?
$endgroup$
– zinR
Jan 26 at 14:18
2
$begingroup$
Take $R$ to be your favorite ring (say $k[x]$), pick $ain R$ such that neither $a$ nor $b=1-a$ is a unit.
$endgroup$
– Mohan
Jan 26 at 14:41
2
$begingroup$
$R=Bbb Z$, $(a,b)=(2,3)$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 14:59
$begingroup$
$R = mathbb{Z}/pmathbb{Z}$, $a = b = 1$. Then $|Roplus R/langle(a,b)rangle| = p$, hence $R oplus R/langle(a,b)ranglecong R$, simply because there are only two rings of order $p$, and we have non-zero multiplication in $Roplus R/langle(a,b)rangle$.
$endgroup$
– user3482749
Jan 26 at 17:14
$begingroup$
Surely there can be an isomorphism?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 12:09
$begingroup$
Surely there can be an isomorphism?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 12:09
$begingroup$
@LordSharktheUnknown Can you think of an example?
$endgroup$
– zinR
Jan 26 at 14:18
$begingroup$
@LordSharktheUnknown Can you think of an example?
$endgroup$
– zinR
Jan 26 at 14:18
2
2
$begingroup$
Take $R$ to be your favorite ring (say $k[x]$), pick $ain R$ such that neither $a$ nor $b=1-a$ is a unit.
$endgroup$
– Mohan
Jan 26 at 14:41
$begingroup$
Take $R$ to be your favorite ring (say $k[x]$), pick $ain R$ such that neither $a$ nor $b=1-a$ is a unit.
$endgroup$
– Mohan
Jan 26 at 14:41
2
2
$begingroup$
$R=Bbb Z$, $(a,b)=(2,3)$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 14:59
$begingroup$
$R=Bbb Z$, $(a,b)=(2,3)$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 14:59
$begingroup$
$R = mathbb{Z}/pmathbb{Z}$, $a = b = 1$. Then $|Roplus R/langle(a,b)rangle| = p$, hence $R oplus R/langle(a,b)ranglecong R$, simply because there are only two rings of order $p$, and we have non-zero multiplication in $Roplus R/langle(a,b)rangle$.
$endgroup$
– user3482749
Jan 26 at 17:14
$begingroup$
$R = mathbb{Z}/pmathbb{Z}$, $a = b = 1$. Then $|Roplus R/langle(a,b)rangle| = p$, hence $R oplus R/langle(a,b)ranglecong R$, simply because there are only two rings of order $p$, and we have non-zero multiplication in $Roplus R/langle(a,b)rangle$.
$endgroup$
– user3482749
Jan 26 at 17:14
|
show 2 more comments
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$begingroup$
Surely there can be an isomorphism?
$endgroup$
– Lord Shark the Unknown
Jan 26 at 12:09
$begingroup$
@LordSharktheUnknown Can you think of an example?
$endgroup$
– zinR
Jan 26 at 14:18
2
$begingroup$
Take $R$ to be your favorite ring (say $k[x]$), pick $ain R$ such that neither $a$ nor $b=1-a$ is a unit.
$endgroup$
– Mohan
Jan 26 at 14:41
2
$begingroup$
$R=Bbb Z$, $(a,b)=(2,3)$.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 14:59
$begingroup$
$R = mathbb{Z}/pmathbb{Z}$, $a = b = 1$. Then $|Roplus R/langle(a,b)rangle| = p$, hence $R oplus R/langle(a,b)ranglecong R$, simply because there are only two rings of order $p$, and we have non-zero multiplication in $Roplus R/langle(a,b)rangle$.
$endgroup$
– user3482749
Jan 26 at 17:14