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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













2












$begingroup$



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.










share|cite|improve this question











$endgroup$







  • 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
















2












$begingroup$



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.










share|cite|improve this question











$endgroup$







  • 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














2












2








2





$begingroup$



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.










share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 3 hours ago







Matheus Andrade

















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













  • 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











1 Answer
1






active

oldest

votes


















4












$begingroup$

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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks a lot! You got me interested, what's that very general theorem we can apply?
    $endgroup$
    – Matheus Andrade
    3 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









4












$begingroup$

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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks a lot! You got me interested, what's that very general theorem we can apply?
    $endgroup$
    – Matheus Andrade
    3 hours ago















4












$begingroup$

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).






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Thanks a lot! You got me interested, what's that very general theorem we can apply?
    $endgroup$
    – Matheus Andrade
    3 hours ago













4












4








4





$begingroup$

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).






share|cite|improve this answer









$endgroup$



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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • $begingroup$
    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

















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