Mixing
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Two non isomorphic rings with $mathbbZ_5 times mathbbZ_5 $ as their additive group.
Clash Royale CLAN TAG#URR8PPP
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I am trying to figure out a method for finding two non isomorphic rings of which their additive groups are isomorphic to $mathbbZ_5times mathbbZ_5$
abstract-algebra ring-theory
asked Jul 24 at 23:27
tmpys
462516
add a comment |Â
up vote
1
down vote
favorite
I am trying to figure out a method for finding two non isomorphic rings of which their additive groups are isomorphic to $mathbbZ_5times mathbbZ_5$
abstract-algebra ring-theory
asked Jul 24 at 23:27
tmpys
462516
1
And you found at least one???
– rschwieb
Jul 24 at 23:35
1
do rings always need identities?
– Thomas Andrews
Jul 25 at 0:05
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to figure out a method for finding two non isomorphic rings of which their additive groups are isomorphic to $mathbbZ_5times mathbbZ_5$
abstract-algebra ring-theory
asked Jul 24 at 23:27
tmpys
462516
I am trying to figure out a method for finding two non isomorphic rings of which their additive groups are isomorphic to $mathbbZ_5times mathbbZ_5$
abstract-algebra ring-theory
asked Jul 24 at 23:27
tmpys
462516
asked Jul 24 at 23:27
tmpys
462516
asked Jul 24 at 23:27
tmpys
462516
asked Jul 24 at 23:27
tmpys
462516
462516
1
And you found at least one???
– rschwieb
Jul 24 at 23:35
1
do rings always need identities?
– Thomas Andrews
Jul 25 at 0:05
add a comment |Â
1
And you found at least one???
– rschwieb
Jul 24 at 23:35
1
do rings always need identities?
– Thomas Andrews
Jul 25 at 0:05
1
1
And you found at least one???
– rschwieb
Jul 24 at 23:35
And you found at least one???
– rschwieb
Jul 24 at 23:35
1
1
do rings always need identities?
– Thomas Andrews
Jul 25 at 0:05
do rings always need identities?
– Thomas Andrews
Jul 25 at 0:05
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
- Try a cartesian product of two (obvious) rings.
- Try a finite field of an appropriate order.
To make sure those are not isomorphic, consider the number of invertible elements.
answered Jul 24 at 23:33
Alon Amit
10.2k3765
add a comment |Â
up vote
1
down vote
More generally, let $K$ be a field and consider the rings
$$
A = K times K,
qquad
B = K[x]/(x^2)
$$
Then their additive groups are both isomorphic to $K times K$ (even as $K$-vector spaces), but $A$ is not isomorphic to $B$ because $B$ has non-zero nilpotent elements while $A$ does not.
Moreover, if there is an irreducible quadratic polynomial $p in K[x]$, then $C=K[x]/(p)$ is a field and so is not isomorphic to either $A$ or $B$, which have nontrivial zero divisors.
Thus, for $K=mathbb Z_5$, we get three non-isomorphic rings with isomorphic additive groups.
answered Jul 25 at 2:42
lhf
155k9160366
add a comment |Â
up vote
0
down vote
There's a way to do this without knowledge of finite fields, but it assumes that you do NOT require rings to have a multiplicative identity. For one, take $mathbbZ_5 times mathbbZ_5$ and give it the usual multiplication of congruence classes (the direct product of the "usual" ring $mathbbZ_5$ against itself). Next, take the same underlying group but give it the trivial multiplication $ab=0$ for all $a, b in mathbbZ_5 times mathbbZ_5$. This is a ring, not iso to the first one (that one had an identity, this one doesn't).
answered Jul 25 at 1:38
Randall
7,2121825
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
- Try a cartesian product of two (obvious) rings.
- Try a finite field of an appropriate order.
To make sure those are not isomorphic, consider the number of invertible elements.
answered Jul 24 at 23:33
Alon Amit
10.2k3765
add a comment |Â
up vote
8
down vote
- Try a cartesian product of two (obvious) rings.
- Try a finite field of an appropriate order.
To make sure those are not isomorphic, consider the number of invertible elements.
answered Jul 24 at 23:33
Alon Amit
10.2k3765
add a comment |Â
up vote
8
down vote
up vote
8
down vote
- Try a cartesian product of two (obvious) rings.
- Try a finite field of an appropriate order.
To make sure those are not isomorphic, consider the number of invertible elements.
answered Jul 24 at 23:33
Alon Amit
10.2k3765
- Try a cartesian product of two (obvious) rings.
- Try a finite field of an appropriate order.
To make sure those are not isomorphic, consider the number of invertible elements.
answered Jul 24 at 23:33
Alon Amit
10.2k3765
answered Jul 24 at 23:33
Alon Amit
10.2k3765
answered Jul 24 at 23:33
Alon Amit
10.2k3765
answered Jul 24 at 23:33
Alon Amit
10.2k3765
10.2k3765
add a comment |Â
add a comment |Â
up vote
1
down vote
More generally, let $K$ be a field and consider the rings
$$
A = K times K,
qquad
B = K[x]/(x^2)
$$
Then their additive groups are both isomorphic to $K times K$ (even as $K$-vector spaces), but $A$ is not isomorphic to $B$ because $B$ has non-zero nilpotent elements while $A$ does not.
Moreover, if there is an irreducible quadratic polynomial $p in K[x]$, then $C=K[x]/(p)$ is a field and so is not isomorphic to either $A$ or $B$, which have nontrivial zero divisors.
Thus, for $K=mathbb Z_5$, we get three non-isomorphic rings with isomorphic additive groups.
answered Jul 25 at 2:42
lhf
155k9160366
add a comment |Â
up vote
1
down vote
More generally, let $K$ be a field and consider the rings
$$
A = K times K,
qquad
B = K[x]/(x^2)
$$
Then their additive groups are both isomorphic to $K times K$ (even as $K$-vector spaces), but $A$ is not isomorphic to $B$ because $B$ has non-zero nilpotent elements while $A$ does not.
Moreover, if there is an irreducible quadratic polynomial $p in K[x]$, then $C=K[x]/(p)$ is a field and so is not isomorphic to either $A$ or $B$, which have nontrivial zero divisors.
Thus, for $K=mathbb Z_5$, we get three non-isomorphic rings with isomorphic additive groups.
answered Jul 25 at 2:42
lhf
155k9160366
add a comment |Â
up vote
1
down vote
up vote
1
down vote
More generally, let $K$ be a field and consider the rings
$$
A = K times K,
qquad
B = K[x]/(x^2)
$$
Then their additive groups are both isomorphic to $K times K$ (even as $K$-vector spaces), but $A$ is not isomorphic to $B$ because $B$ has non-zero nilpotent elements while $A$ does not.
Moreover, if there is an irreducible quadratic polynomial $p in K[x]$, then $C=K[x]/(p)$ is a field and so is not isomorphic to either $A$ or $B$, which have nontrivial zero divisors.
Thus, for $K=mathbb Z_5$, we get three non-isomorphic rings with isomorphic additive groups.
answered Jul 25 at 2:42
lhf
155k9160366
More generally, let $K$ be a field and consider the rings
$$
A = K times K,
qquad
B = K[x]/(x^2)
$$
Then their additive groups are both isomorphic to $K times K$ (even as $K$-vector spaces), but $A$ is not isomorphic to $B$ because $B$ has non-zero nilpotent elements while $A$ does not.
Moreover, if there is an irreducible quadratic polynomial $p in K[x]$, then $C=K[x]/(p)$ is a field and so is not isomorphic to either $A$ or $B$, which have nontrivial zero divisors.
Thus, for $K=mathbb Z_5$, we get three non-isomorphic rings with isomorphic additive groups.
answered Jul 25 at 2:42
lhf
155k9160366
edited Jul 31 at 11:15
answered Jul 25 at 2:42
lhf
155k9160366
answered Jul 25 at 2:42
lhf
155k9160366
answered Jul 25 at 2:42
lhf
155k9160366
155k9160366
add a comment |Â
add a comment |Â
up vote
0
down vote
There's a way to do this without knowledge of finite fields, but it assumes that you do NOT require rings to have a multiplicative identity. For one, take $mathbbZ_5 times mathbbZ_5$ and give it the usual multiplication of congruence classes (the direct product of the "usual" ring $mathbbZ_5$ against itself). Next, take the same underlying group but give it the trivial multiplication $ab=0$ for all $a, b in mathbbZ_5 times mathbbZ_5$. This is a ring, not iso to the first one (that one had an identity, this one doesn't).
answered Jul 25 at 1:38
Randall
7,2121825
add a comment |Â
up vote
0
down vote
There's a way to do this without knowledge of finite fields, but it assumes that you do NOT require rings to have a multiplicative identity. For one, take $mathbbZ_5 times mathbbZ_5$ and give it the usual multiplication of congruence classes (the direct product of the "usual" ring $mathbbZ_5$ against itself). Next, take the same underlying group but give it the trivial multiplication $ab=0$ for all $a, b in mathbbZ_5 times mathbbZ_5$. This is a ring, not iso to the first one (that one had an identity, this one doesn't).
answered Jul 25 at 1:38
Randall
7,2121825
add a comment |Â
up vote
0
down vote
up vote
0
down vote
There's a way to do this without knowledge of finite fields, but it assumes that you do NOT require rings to have a multiplicative identity. For one, take $mathbbZ_5 times mathbbZ_5$ and give it the usual multiplication of congruence classes (the direct product of the "usual" ring $mathbbZ_5$ against itself). Next, take the same underlying group but give it the trivial multiplication $ab=0$ for all $a, b in mathbbZ_5 times mathbbZ_5$. This is a ring, not iso to the first one (that one had an identity, this one doesn't).
answered Jul 25 at 1:38
Randall
7,2121825
There's a way to do this without knowledge of finite fields, but it assumes that you do NOT require rings to have a multiplicative identity. For one, take $mathbbZ_5 times mathbbZ_5$ and give it the usual multiplication of congruence classes (the direct product of the "usual" ring $mathbbZ_5$ against itself). Next, take the same underlying group but give it the trivial multiplication $ab=0$ for all $a, b in mathbbZ_5 times mathbbZ_5$. This is a ring, not iso to the first one (that one had an identity, this one doesn't).
answered Jul 25 at 1:38
Randall
7,2121825
answered Jul 25 at 1:38
Randall
7,2121825
answered Jul 25 at 1:38
Randall
7,2121825
answered Jul 25 at 1:38
Randall
7,2121825
7,2121825
add a comment |Â
add a comment |Â
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1
And you found at least one???
– rschwieb
Jul 24 at 23:35
1
do rings always need identities?
– Thomas Andrews
Jul 25 at 0:05