Establishing isomorphisms between polynomial quotient ringsIsomorphism between quotient rings of $mathbbZ[x,y]$Is this an automorphism of a polynomial ring?Non-commutative indeterminates in polynomial rings.Inducing homomorphisms on localizations of rings/modulesThe quotient of a ring by the annihilator of an idealIsomorphism of Polynomial RingsExpressing Laurent polynomials using generators and relationsShow that a ring of fractions and a quotient ring are isomorphicIsomorphism between polynomial ring and ringSome basic isomorphisms in rings and modules
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Establishing isomorphisms between polynomial quotient rings
Isomorphism between quotient rings of $mathbbZ[x,y]$Is this an automorphism of a polynomial ring?Non-commutative indeterminates in polynomial rings.Inducing homomorphisms on localizations of rings/modulesThe quotient of a ring by the annihilator of an idealIsomorphism of Polynomial RingsExpressing Laurent polynomials using generators and relationsShow that a ring of fractions and a quotient ring are isomorphicIsomorphism between polynomial ring and ringSome basic isomorphisms in rings and modules
$begingroup$
Let $A$ be a commutative ring, $I$ be an ideal of $A$ and:
$$varphi: A[X] to fracAI[X]$$
$$sum_i = 0^n a_i X^i mapsto sum_i = 0^n overlinea_i X^i$$
where $overlinea_i = a_i + I$. Prove that:
a) $varphi$ is a ring homomorphism
b) $I[X] = sum_i = 0^n a_i X^i vert a_i in I, n in mathbbN$ is an ideal of $A[X]$ and $fracA[X]I[X] cong left( fracAI right)[X]$.
a) and the first part of b) are easy. I'm having trouble with finding the last isomorphism though. I think it should be:
$$psi:fracA[X]I[X] to left( fracAI right)[X] $$
$$ left(sum_i = 0^n a_i X^iright) + I[X] mapsto sum_i = 0^n (a_i + I)X^i = sum_i = 0^n overlinea_i X^i $$
EDIT: Made the required corrections. Thanks a lot! Now I understand these things better than I did when I first asked the question.
abstract-algebra ring-theory
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
$endgroup$
add a comment |
$begingroup$
Let $A$ be a commutative ring, $I$ be an ideal of $A$ and:
$$varphi: A[X] to fracAI[X]$$
$$sum_i = 0^n a_i X^i mapsto sum_i = 0^n overlinea_i X^i$$
where $overlinea_i = a_i + I$. Prove that:
a) $varphi$ is a ring homomorphism
b) $I[X] = sum_i = 0^n a_i X^i vert a_i in I, n in mathbbN$ is an ideal of $A[X]$ and $fracA[X]I[X] cong left( fracAI right)[X]$.
a) and the first part of b) are easy. I'm having trouble with finding the last isomorphism though. I think it should be:
$$psi:fracA[X]I[X] to left( fracAI right)[X] $$
$$ left(sum_i = 0^n a_i X^iright) + I[X] mapsto sum_i = 0^n (a_i + I)X^i = sum_i = 0^n overlinea_i X^i $$
EDIT: Made the required corrections. Thanks a lot! Now I understand these things better than I did when I first asked the question.
abstract-algebra ring-theory
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
$endgroup$
1
$begingroup$
Should be addition $a_i + I[X]$, not $a_i I[X]$. (That's how quotient rings work.) But yes, that isomorphism is correct with this change throughout. You might find it easier though to define $psi$ in the other direction; your choice.
$endgroup$
– Dzoooks
4 hours ago
add a comment |
$begingroup$
Let $A$ be a commutative ring, $I$ be an ideal of $A$ and:
$$varphi: A[X] to fracAI[X]$$
$$sum_i = 0^n a_i X^i mapsto sum_i = 0^n overlinea_i X^i$$
where $overlinea_i = a_i + I$. Prove that:
a) $varphi$ is a ring homomorphism
b) $I[X] = sum_i = 0^n a_i X^i vert a_i in I, n in mathbbN$ is an ideal of $A[X]$ and $fracA[X]I[X] cong left( fracAI right)[X]$.
a) and the first part of b) are easy. I'm having trouble with finding the last isomorphism though. I think it should be:
$$psi:fracA[X]I[X] to left( fracAI right)[X] $$
$$ left(sum_i = 0^n a_i X^iright) + I[X] mapsto sum_i = 0^n (a_i + I)X^i = sum_i = 0^n overlinea_i X^i $$
EDIT: Made the required corrections. Thanks a lot! Now I understand these things better than I did when I first asked the question.
abstract-algebra ring-theory
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
$endgroup$
Let $A$ be a commutative ring, $I$ be an ideal of $A$ and:
$$varphi: A[X] to fracAI[X]$$
$$sum_i = 0^n a_i X^i mapsto sum_i = 0^n overlinea_i X^i$$
where $overlinea_i = a_i + I$. Prove that:
a) $varphi$ is a ring homomorphism
b) $I[X] = sum_i = 0^n a_i X^i vert a_i in I, n in mathbbN$ is an ideal of $A[X]$ and $fracA[X]I[X] cong left( fracAI right)[X]$.
a) and the first part of b) are easy. I'm having trouble with finding the last isomorphism though. I think it should be:
$$psi:fracA[X]I[X] to left( fracAI right)[X] $$
$$ left(sum_i = 0^n a_i X^iright) + I[X] mapsto sum_i = 0^n (a_i + I)X^i = sum_i = 0^n overlinea_i X^i $$
EDIT: Made the required corrections. Thanks a lot! Now I understand these things better than I did when I first asked the question.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
edited 3 hours ago
Matheus Andrade
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
asked 4 hours ago
Matheus AndradeMatheus Andrade
1,509418
1,509418
1
$begingroup$
Should be addition $a_i + I[X]$, not $a_i I[X]$. (That's how quotient rings work.) But yes, that isomorphism is correct with this change throughout. You might find it easier though to define $psi$ in the other direction; your choice.
$endgroup$
– Dzoooks
4 hours ago
add a comment |
1
$begingroup$
Should be addition $a_i + I[X]$, not $a_i I[X]$. (That's how quotient rings work.) But yes, that isomorphism is correct with this change throughout. You might find it easier though to define $psi$ in the other direction; your choice.
$endgroup$
– Dzoooks
4 hours ago
1
1
$begingroup$
Should be addition $a_i + I[X]$, not $a_i I[X]$. (That's how quotient rings work.) But yes, that isomorphism is correct with this change throughout. You might find it easier though to define $psi$ in the other direction; your choice.
$endgroup$
– Dzoooks
4 hours ago
$begingroup$
Should be addition $a_i + I[X]$, not $a_i I[X]$. (That's how quotient rings work.) But yes, that isomorphism is correct with this change throughout. You might find it easier though to define $psi$ in the other direction; your choice.
$endgroup$
– Dzoooks
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You seem to be using the letters $R$ and $A$ to mean the same thing - better choose just one.
The left hand side of the equation preceding $mapsto$ seems to have no meaning. It can simply be deleted, and the right hand side of that equation replaced by $left(sum_i = 0^n a_i X^iright) + I[X]$.
Similarly, after the $mapsto$, the left hand side of the equation should be replaced by $sum_i = 0^n (a_i + I)X^i$.
(You also need to correct the earlier "$overlinea_i = a_i I$" to $overlinea_i = a_i + I$.)
The right hand side of the equation after $mapsto$ is then correct.
(Except: the summation sign needs at least an $i$, for clarity, and preferably also the summation limits $0$ and $n$, for consistency with the rest of your notation. Alternatively, you could drop the summation limits everywhere except in the definition of $I[X]$.)
Even with these corrections, the notation might still be hard to read, but - hint! - you can avoid this difficulty altogether, by applying a very general theorem to the result of part (a), obtaining (b).
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
$endgroup$
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
add a comment |
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$begingroup$
You seem to be using the letters $R$ and $A$ to mean the same thing - better choose just one.
The left hand side of the equation preceding $mapsto$ seems to have no meaning. It can simply be deleted, and the right hand side of that equation replaced by $left(sum_i = 0^n a_i X^iright) + I[X]$.
Similarly, after the $mapsto$, the left hand side of the equation should be replaced by $sum_i = 0^n (a_i + I)X^i$.
(You also need to correct the earlier "$overlinea_i = a_i I$" to $overlinea_i = a_i + I$.)
The right hand side of the equation after $mapsto$ is then correct.
(Except: the summation sign needs at least an $i$, for clarity, and preferably also the summation limits $0$ and $n$, for consistency with the rest of your notation. Alternatively, you could drop the summation limits everywhere except in the definition of $I[X]$.)
Even with these corrections, the notation might still be hard to read, but - hint! - you can avoid this difficulty altogether, by applying a very general theorem to the result of part (a), obtaining (b).
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
$endgroup$
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
add a comment |
$begingroup$
You seem to be using the letters $R$ and $A$ to mean the same thing - better choose just one.
The left hand side of the equation preceding $mapsto$ seems to have no meaning. It can simply be deleted, and the right hand side of that equation replaced by $left(sum_i = 0^n a_i X^iright) + I[X]$.
Similarly, after the $mapsto$, the left hand side of the equation should be replaced by $sum_i = 0^n (a_i + I)X^i$.
(You also need to correct the earlier "$overlinea_i = a_i I$" to $overlinea_i = a_i + I$.)
The right hand side of the equation after $mapsto$ is then correct.
(Except: the summation sign needs at least an $i$, for clarity, and preferably also the summation limits $0$ and $n$, for consistency with the rest of your notation. Alternatively, you could drop the summation limits everywhere except in the definition of $I[X]$.)
Even with these corrections, the notation might still be hard to read, but - hint! - you can avoid this difficulty altogether, by applying a very general theorem to the result of part (a), obtaining (b).
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
$endgroup$
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
add a comment |
$begingroup$
You seem to be using the letters $R$ and $A$ to mean the same thing - better choose just one.
The left hand side of the equation preceding $mapsto$ seems to have no meaning. It can simply be deleted, and the right hand side of that equation replaced by $left(sum_i = 0^n a_i X^iright) + I[X]$.
Similarly, after the $mapsto$, the left hand side of the equation should be replaced by $sum_i = 0^n (a_i + I)X^i$.
(You also need to correct the earlier "$overlinea_i = a_i I$" to $overlinea_i = a_i + I$.)
The right hand side of the equation after $mapsto$ is then correct.
(Except: the summation sign needs at least an $i$, for clarity, and preferably also the summation limits $0$ and $n$, for consistency with the rest of your notation. Alternatively, you could drop the summation limits everywhere except in the definition of $I[X]$.)
Even with these corrections, the notation might still be hard to read, but - hint! - you can avoid this difficulty altogether, by applying a very general theorem to the result of part (a), obtaining (b).
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
$endgroup$
You seem to be using the letters $R$ and $A$ to mean the same thing - better choose just one.
The left hand side of the equation preceding $mapsto$ seems to have no meaning. It can simply be deleted, and the right hand side of that equation replaced by $left(sum_i = 0^n a_i X^iright) + I[X]$.
Similarly, after the $mapsto$, the left hand side of the equation should be replaced by $sum_i = 0^n (a_i + I)X^i$.
(You also need to correct the earlier "$overlinea_i = a_i I$" to $overlinea_i = a_i + I$.)
The right hand side of the equation after $mapsto$ is then correct.
(Except: the summation sign needs at least an $i$, for clarity, and preferably also the summation limits $0$ and $n$, for consistency with the rest of your notation. Alternatively, you could drop the summation limits everywhere except in the definition of $I[X]$.)
Even with these corrections, the notation might still be hard to read, but - hint! - you can avoid this difficulty altogether, by applying a very general theorem to the result of part (a), obtaining (b).
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
answered 4 hours ago
Calum GilhooleyCalum Gilhooley
5,194730
5,194730
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
add a comment |
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
$begingroup$
Thanks a lot! You got me interested, what's that very general theorem we can apply?
$endgroup$
– Matheus Andrade
3 hours ago
add a comment |
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$begingroup$
Should be addition $a_i + I[X]$, not $a_i I[X]$. (That's how quotient rings work.) But yes, that isomorphism is correct with this change throughout. You might find it easier though to define $psi$ in the other direction; your choice.
$endgroup$
– Dzoooks
4 hours ago