Showing that three rings are isomorphic
0
Let $C_3 = langle a | a^3 = e rangle$ and let $R=(mathbb{Z}/2)[C_3]$ be the group ring of $C_3$ with $mathbb{Z}/2$ coefficients.
Let $S = (mathbb{Z}/2)[y]/(y^3-[1])$ and let $T = mathbb{Z}[x]/(2,x^3-1)$. Prove that $S,T$, and $R$ are pairwise isomorphic rings.
ring-theory group-rings ring-isomorphism
asked Nov 28 '18 at 3:14
Wesley
525313
add a comment |
0
Let $C_3 = langle a | a^3 = e rangle$ and let $R=(mathbb{Z}/2)[C_3]$ be the group ring of $C_3$ with $mathbb{Z}/2$ coefficients.
Let $S = (mathbb{Z}/2)[y]/(y^3-[1])$ and let $T = mathbb{Z}[x]/(2,x^3-1)$. Prove that $S,T$, and $R$ are pairwise isomorphic rings.
ring-theory group-rings ring-isomorphism
asked Nov 28 '18 at 3:14
Wesley
525313
1
Can you think of an isomorphism between $R$ and $S$ at least? If this is not possible, then try to write down the elements of $R$ and $S$ explicitly, and find a pattern that you can exploit. A similar sort of procedure should be adopted if you cannot figure out anything between $S$ and $T$.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 3:23
For $S$ and $T,$ try to construct a surjective homomorphism from $mathbb{Z}[x]$ to $S$ with kernel $(2, x^3-1).$ Again, the same for $R$ and $S.$
– dhk628
Nov 28 '18 at 5:45
From $(mathbb{Z}/2)[x]$ to R it seems that an isomorphism would be the evaluation homomorphism at $a$. Would that work?
– Wesley
Nov 28 '18 at 16:35
add a comment |
0
0
0
Let $C_3 = langle a | a^3 = e rangle$ and let $R=(mathbb{Z}/2)[C_3]$ be the group ring of $C_3$ with $mathbb{Z}/2$ coefficients.
Let $S = (mathbb{Z}/2)[y]/(y^3-[1])$ and let $T = mathbb{Z}[x]/(2,x^3-1)$. Prove that $S,T$, and $R$ are pairwise isomorphic rings.
ring-theory group-rings ring-isomorphism
asked Nov 28 '18 at 3:14
Wesley
525313
Let $C_3 = langle a | a^3 = e rangle$ and let $R=(mathbb{Z}/2)[C_3]$ be the group ring of $C_3$ with $mathbb{Z}/2$ coefficients.
Let $S = (mathbb{Z}/2)[y]/(y^3-[1])$ and let $T = mathbb{Z}[x]/(2,x^3-1)$. Prove that $S,T$, and $R$ are pairwise isomorphic rings.
ring-theory group-rings ring-isomorphism
ring-theory group-rings ring-isomorphism
asked Nov 28 '18 at 3:14
Wesley
525313
asked Nov 28 '18 at 3:14
Wesley
525313
asked Nov 28 '18 at 3:14
Wesley
525313
asked Nov 28 '18 at 3:14
Wesley
525313
asked Nov 28 '18 at 3:14
Wesley
525313
525313
1
Can you think of an isomorphism between $R$ and $S$ at least? If this is not possible, then try to write down the elements of $R$ and $S$ explicitly, and find a pattern that you can exploit. A similar sort of procedure should be adopted if you cannot figure out anything between $S$ and $T$.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 3:23
For $S$ and $T,$ try to construct a surjective homomorphism from $mathbb{Z}[x]$ to $S$ with kernel $(2, x^3-1).$ Again, the same for $R$ and $S.$
– dhk628
Nov 28 '18 at 5:45
From $(mathbb{Z}/2)[x]$ to R it seems that an isomorphism would be the evaluation homomorphism at $a$. Would that work?
– Wesley
Nov 28 '18 at 16:35
add a comment |
1
Can you think of an isomorphism between $R$ and $S$ at least? If this is not possible, then try to write down the elements of $R$ and $S$ explicitly, and find a pattern that you can exploit. A similar sort of procedure should be adopted if you cannot figure out anything between $S$ and $T$.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 3:23
For $S$ and $T,$ try to construct a surjective homomorphism from $mathbb{Z}[x]$ to $S$ with kernel $(2, x^3-1).$ Again, the same for $R$ and $S.$
– dhk628
Nov 28 '18 at 5:45
From $(mathbb{Z}/2)[x]$ to R it seems that an isomorphism would be the evaluation homomorphism at $a$. Would that work?
– Wesley
Nov 28 '18 at 16:35
1
1
Can you think of an isomorphism between $R$ and $S$ at least? If this is not possible, then try to write down the elements of $R$ and $S$ explicitly, and find a pattern that you can exploit. A similar sort of procedure should be adopted if you cannot figure out anything between $S$ and $T$.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 3:23
Can you think of an isomorphism between $R$ and $S$ at least? If this is not possible, then try to write down the elements of $R$ and $S$ explicitly, and find a pattern that you can exploit. A similar sort of procedure should be adopted if you cannot figure out anything between $S$ and $T$.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 3:23
For $S$ and $T,$ try to construct a surjective homomorphism from $mathbb{Z}[x]$ to $S$ with kernel $(2, x^3-1).$ Again, the same for $R$ and $S.$
– dhk628
Nov 28 '18 at 5:45
For $S$ and $T,$ try to construct a surjective homomorphism from $mathbb{Z}[x]$ to $S$ with kernel $(2, x^3-1).$ Again, the same for $R$ and $S.$
– dhk628
Nov 28 '18 at 5:45
From $(mathbb{Z}/2)[x]$ to R it seems that an isomorphism would be the evaluation homomorphism at $a$. Would that work?
– Wesley
Nov 28 '18 at 16:35
From $(mathbb{Z}/2)[x]$ to R it seems that an isomorphism would be the evaluation homomorphism at $a$. Would that work?
– Wesley
Nov 28 '18 at 16:35
add a comment |
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1
Can you think of an isomorphism between $R$ and $S$ at least? If this is not possible, then try to write down the elements of $R$ and $S$ explicitly, and find a pattern that you can exploit. A similar sort of procedure should be adopted if you cannot figure out anything between $S$ and $T$.
– астон вілла олоф мэллбэрг
Nov 28 '18 at 3:23
For $S$ and $T,$ try to construct a surjective homomorphism from $mathbb{Z}[x]$ to $S$ with kernel $(2, x^3-1).$ Again, the same for $R$ and $S.$
– dhk628
Nov 28 '18 at 5:45
From $(mathbb{Z}/2)[x]$ to R it seems that an isomorphism would be the evaluation homomorphism at $a$. Would that work?
– Wesley
Nov 28 '18 at 16:35