Does every integral domain come from a quotient?












0












$begingroup$


Let $A$ be conmutative ring with identity and $mathfrak{p}, mathfrak{m}$ ideals. Then
$$begin{array}{ll}
mathfrak{p}text{ is a prime}iff A/mathfrak{p}text{ is an integral domain}\
mathfrak{m}text{ is maximal}iff A/mathfrak{m}text{ is a field}
end{array}$$
Does every integral domain/field come from some other ring and prime/maximal ideal? An equivalent question is can we un-quotient every ring? That is, given a ring $A$ can we find another ring $B$ and some ideal $I$ such that $B/Icong A$? If there is a way to do so is it unique (up to isomorphism)?



I want to answer in a not trivial way: every integral domain $A$ is isomorphic to $A/(0)$ and $(0)$ is prime in $A$.



Note: this is not an exercise and I will try to answer the question myself but don't have time now






abstract-algebra ring-theory maximal-and-prime-ideals







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$endgroup$








  • 2




    $begingroup$
    Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable.
    $endgroup$
    – R.C.Cowsik
    Dec 8 '18 at 17:29
















0












$begingroup$


Let $A$ be conmutative ring with identity and $mathfrak{p}, mathfrak{m}$ ideals. Then
$$begin{array}{ll}
mathfrak{p}text{ is a prime}iff A/mathfrak{p}text{ is an integral domain}\
mathfrak{m}text{ is maximal}iff A/mathfrak{m}text{ is a field}
end{array}$$
Does every integral domain/field come from some other ring and prime/maximal ideal? An equivalent question is can we un-quotient every ring? That is, given a ring $A$ can we find another ring $B$ and some ideal $I$ such that $B/Icong A$? If there is a way to do so is it unique (up to isomorphism)?



I want to answer in a not trivial way: every integral domain $A$ is isomorphic to $A/(0)$ and $(0)$ is prime in $A$.



Note: this is not an exercise and I will try to answer the question myself but don't have time now






abstract-algebra ring-theory maximal-and-prime-ideals







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable.
    $endgroup$
    – R.C.Cowsik
    Dec 8 '18 at 17:29














0












0








0





$begingroup$


Let $A$ be conmutative ring with identity and $mathfrak{p}, mathfrak{m}$ ideals. Then
$$begin{array}{ll}
mathfrak{p}text{ is a prime}iff A/mathfrak{p}text{ is an integral domain}\
mathfrak{m}text{ is maximal}iff A/mathfrak{m}text{ is a field}
end{array}$$
Does every integral domain/field come from some other ring and prime/maximal ideal? An equivalent question is can we un-quotient every ring? That is, given a ring $A$ can we find another ring $B$ and some ideal $I$ such that $B/Icong A$? If there is a way to do so is it unique (up to isomorphism)?



I want to answer in a not trivial way: every integral domain $A$ is isomorphic to $A/(0)$ and $(0)$ is prime in $A$.



Note: this is not an exercise and I will try to answer the question myself but don't have time now






abstract-algebra ring-theory maximal-and-prime-ideals







share|cite|improve this question









$endgroup$




Let $A$ be conmutative ring with identity and $mathfrak{p}, mathfrak{m}$ ideals. Then
$$begin{array}{ll}
mathfrak{p}text{ is a prime}iff A/mathfrak{p}text{ is an integral domain}\
mathfrak{m}text{ is maximal}iff A/mathfrak{m}text{ is a field}
end{array}$$
Does every integral domain/field come from some other ring and prime/maximal ideal? An equivalent question is can we un-quotient every ring? That is, given a ring $A$ can we find another ring $B$ and some ideal $I$ such that $B/Icong A$? If there is a way to do so is it unique (up to isomorphism)?



I want to answer in a not trivial way: every integral domain $A$ is isomorphic to $A/(0)$ and $(0)$ is prime in $A$.



Note: this is not an exercise and I will try to answer the question myself but don't have time now






abstract-algebra ring-theory maximal-and-prime-ideals




abstract-algebra ring-theory maximal-and-prime-ideals






share|cite|improve this question













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asked Dec 8 '18 at 17:04









PedroPedro

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517212








  • 2




    $begingroup$
    Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable.
    $endgroup$
    – R.C.Cowsik
    Dec 8 '18 at 17:29














  • 2




    $begingroup$
    Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable.
    $endgroup$
    – R.C.Cowsik
    Dec 8 '18 at 17:29








2




2




$begingroup$
Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable.
$endgroup$
– R.C.Cowsik
Dec 8 '18 at 17:29




$begingroup$
Given any ring R we can extend it to a bigger ring of which R is a quotient; e.g. the ring of polynomials in a variable.
$endgroup$
– R.C.Cowsik
Dec 8 '18 at 17:29










1 Answer
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oldest

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$begingroup$

Any ring $A$ can be written as a quotient of another ring in many many different ways. For instance, you could take any ring $B$ and consider the product ring $Atimes B$. The first projection $Atimes Bto A$ is a surjective ring-homomorphism and thus exhibits $A$ as a quotient of $Atimes B$ (by the ideal $0times B$).



This is far from the only possibility. For instance, you could take a polynomial ring $B$ over $A$ in any number of variables, and get a surjective ring homomorphism $Bto A$ by mapping all the variables to $0$. Or, you could map the variables to any elements of $A$ instead of $0$ (assuming $A$ is commutative; if $A$ is noncommutative you would need to instead use a ring of noncommuting polynomials). Or, you could take a polynomial ring $B$ over $mathbb{Z}$ in lots of variables, and map those variables to elements of $A$. As long as the chosen elements of $A$ generate $A$ as a ring, the homomorphism $Bto A$ will be surjective.



Another way to get more examples: given any ring $B$ with an ideal $I$ such that $B/Icong A$, you can take any ideal $Jsubseteq B$ such that $Jsubseteq I$ and then $A$ is also a quotient of $B/J$ (by the ideal $I/J$).






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    $begingroup$

    Any ring $A$ can be written as a quotient of another ring in many many different ways. For instance, you could take any ring $B$ and consider the product ring $Atimes B$. The first projection $Atimes Bto A$ is a surjective ring-homomorphism and thus exhibits $A$ as a quotient of $Atimes B$ (by the ideal $0times B$).



    This is far from the only possibility. For instance, you could take a polynomial ring $B$ over $A$ in any number of variables, and get a surjective ring homomorphism $Bto A$ by mapping all the variables to $0$. Or, you could map the variables to any elements of $A$ instead of $0$ (assuming $A$ is commutative; if $A$ is noncommutative you would need to instead use a ring of noncommuting polynomials). Or, you could take a polynomial ring $B$ over $mathbb{Z}$ in lots of variables, and map those variables to elements of $A$. As long as the chosen elements of $A$ generate $A$ as a ring, the homomorphism $Bto A$ will be surjective.



    Another way to get more examples: given any ring $B$ with an ideal $I$ such that $B/Icong A$, you can take any ideal $Jsubseteq B$ such that $Jsubseteq I$ and then $A$ is also a quotient of $B/J$ (by the ideal $I/J$).






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Any ring $A$ can be written as a quotient of another ring in many many different ways. For instance, you could take any ring $B$ and consider the product ring $Atimes B$. The first projection $Atimes Bto A$ is a surjective ring-homomorphism and thus exhibits $A$ as a quotient of $Atimes B$ (by the ideal $0times B$).



      This is far from the only possibility. For instance, you could take a polynomial ring $B$ over $A$ in any number of variables, and get a surjective ring homomorphism $Bto A$ by mapping all the variables to $0$. Or, you could map the variables to any elements of $A$ instead of $0$ (assuming $A$ is commutative; if $A$ is noncommutative you would need to instead use a ring of noncommuting polynomials). Or, you could take a polynomial ring $B$ over $mathbb{Z}$ in lots of variables, and map those variables to elements of $A$. As long as the chosen elements of $A$ generate $A$ as a ring, the homomorphism $Bto A$ will be surjective.



      Another way to get more examples: given any ring $B$ with an ideal $I$ such that $B/Icong A$, you can take any ideal $Jsubseteq B$ such that $Jsubseteq I$ and then $A$ is also a quotient of $B/J$ (by the ideal $I/J$).






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Any ring $A$ can be written as a quotient of another ring in many many different ways. For instance, you could take any ring $B$ and consider the product ring $Atimes B$. The first projection $Atimes Bto A$ is a surjective ring-homomorphism and thus exhibits $A$ as a quotient of $Atimes B$ (by the ideal $0times B$).



        This is far from the only possibility. For instance, you could take a polynomial ring $B$ over $A$ in any number of variables, and get a surjective ring homomorphism $Bto A$ by mapping all the variables to $0$. Or, you could map the variables to any elements of $A$ instead of $0$ (assuming $A$ is commutative; if $A$ is noncommutative you would need to instead use a ring of noncommuting polynomials). Or, you could take a polynomial ring $B$ over $mathbb{Z}$ in lots of variables, and map those variables to elements of $A$. As long as the chosen elements of $A$ generate $A$ as a ring, the homomorphism $Bto A$ will be surjective.



        Another way to get more examples: given any ring $B$ with an ideal $I$ such that $B/Icong A$, you can take any ideal $Jsubseteq B$ such that $Jsubseteq I$ and then $A$ is also a quotient of $B/J$ (by the ideal $I/J$).






        share|cite|improve this answer









        $endgroup$



        Any ring $A$ can be written as a quotient of another ring in many many different ways. For instance, you could take any ring $B$ and consider the product ring $Atimes B$. The first projection $Atimes Bto A$ is a surjective ring-homomorphism and thus exhibits $A$ as a quotient of $Atimes B$ (by the ideal $0times B$).



        This is far from the only possibility. For instance, you could take a polynomial ring $B$ over $A$ in any number of variables, and get a surjective ring homomorphism $Bto A$ by mapping all the variables to $0$. Or, you could map the variables to any elements of $A$ instead of $0$ (assuming $A$ is commutative; if $A$ is noncommutative you would need to instead use a ring of noncommuting polynomials). Or, you could take a polynomial ring $B$ over $mathbb{Z}$ in lots of variables, and map those variables to elements of $A$. As long as the chosen elements of $A$ generate $A$ as a ring, the homomorphism $Bto A$ will be surjective.



        Another way to get more examples: given any ring $B$ with an ideal $I$ such that $B/Icong A$, you can take any ideal $Jsubseteq B$ such that $Jsubseteq I$ and then $A$ is also a quotient of $B/J$ (by the ideal $I/J$).







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 20:54









        Eric WofseyEric Wofsey

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        186k14215342






























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