When we say two fields are isomorphic, does that just mean they are isomophic as rings? [closed]












10














If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?










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closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039

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  • 8




    A field homomorphism is a ring homomorphism between fields - so yes!
    – Dietrich Burde
    Dec 11 '18 at 19:09










  • Please see my comment, which explains just what TonyK's answer means by "field structure".
    – user21820
    Dec 12 '18 at 12:46
















10














If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?










share|cite|improve this question













closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 8




    A field homomorphism is a ring homomorphism between fields - so yes!
    – Dietrich Burde
    Dec 11 '18 at 19:09










  • Please see my comment, which explains just what TonyK's answer means by "field structure".
    – user21820
    Dec 12 '18 at 12:46














10












10








10







If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?










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If we say fields $A$ and $B$ are isomorphic, does that just mean they are isomorphic as rings, or is there something else?







abstract-algebra field-theory






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asked Dec 11 '18 at 18:40









Ovi

12.4k1038111




12.4k1038111




closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Saad, Gibbs, Xander Henderson, amWhy, user 170039 Dec 14 '18 at 14:04


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Gibbs, Xander Henderson, amWhy, user 170039

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 8




    A field homomorphism is a ring homomorphism between fields - so yes!
    – Dietrich Burde
    Dec 11 '18 at 19:09










  • Please see my comment, which explains just what TonyK's answer means by "field structure".
    – user21820
    Dec 12 '18 at 12:46














  • 8




    A field homomorphism is a ring homomorphism between fields - so yes!
    – Dietrich Burde
    Dec 11 '18 at 19:09










  • Please see my comment, which explains just what TonyK's answer means by "field structure".
    – user21820
    Dec 12 '18 at 12:46








8




8




A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09




A field homomorphism is a ring homomorphism between fields - so yes!
– Dietrich Burde
Dec 11 '18 at 19:09












Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46




Please see my comment, which explains just what TonyK's answer means by "field structure".
– user21820
Dec 12 '18 at 12:46










2 Answers
2






active

oldest

votes


















25














In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.



It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.



I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.






share|cite|improve this answer





















  • So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
    – Daniel Schepler
    Dec 11 '18 at 19:34






  • 7




    I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
    – hunter
    Dec 11 '18 at 22:59






  • 8




    @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
    – rschwieb
    Dec 12 '18 at 1:48






  • 1




    @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
    – Joker_vD
    Dec 12 '18 at 14:55






  • 1




    @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
    – user21820
    Dec 12 '18 at 18:08



















15














They are just isomorphic as rings.



A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.






share|cite|improve this answer




























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    25














    In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.



    It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.



    I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.






    share|cite|improve this answer





















    • So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
      – Daniel Schepler
      Dec 11 '18 at 19:34






    • 7




      I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
      – hunter
      Dec 11 '18 at 22:59






    • 8




      @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
      – rschwieb
      Dec 12 '18 at 1:48






    • 1




      @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
      – Joker_vD
      Dec 12 '18 at 14:55






    • 1




      @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
      – user21820
      Dec 12 '18 at 18:08
















    25














    In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.



    It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.



    I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.






    share|cite|improve this answer





















    • So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
      – Daniel Schepler
      Dec 11 '18 at 19:34






    • 7




      I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
      – hunter
      Dec 11 '18 at 22:59






    • 8




      @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
      – rschwieb
      Dec 12 '18 at 1:48






    • 1




      @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
      – Joker_vD
      Dec 12 '18 at 14:55






    • 1




      @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
      – user21820
      Dec 12 '18 at 18:08














    25












    25








    25






    In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.



    It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.



    I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.






    share|cite|improve this answer












    In a sense, yes, that is what it means. But not really. When we say two structures $S$ and $T$ of a certain type are isomorphic, we mean that there is a bijection $varphi:Srightarrow T$ which preserves the structure. So, for instance, if $circ$ is a binary operation in the structure, then for $x,yin S$, we have $varphi(xcirc y)=varphi(x)circ varphi(y)$.



    It turns out that preserving the ring structure is enough to preserve the field structure; a field is just a commutative ring with inverses, so the property of being a field is preserved if the operations $+$ and $times$ are preserved. Thus two fields are isomorphic if and only if they are isomorphic when considered as rings. But this is a contingent fact, and it's not really what we mean when we say that two fields are isomorphic.



    I realise that this view verges on philosophy, and I wouldn't defend it to the death. I am just trying to give an idea of what mathematicians are thinking of when they say isomorphic.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 11 '18 at 19:03









    TonyK

    41.5k353132




    41.5k353132












    • So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
      – Daniel Schepler
      Dec 11 '18 at 19:34






    • 7




      I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
      – hunter
      Dec 11 '18 at 22:59






    • 8




      @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
      – rschwieb
      Dec 12 '18 at 1:48






    • 1




      @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
      – Joker_vD
      Dec 12 '18 at 14:55






    • 1




      @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
      – user21820
      Dec 12 '18 at 18:08


















    • So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
      – Daniel Schepler
      Dec 11 '18 at 19:34






    • 7




      I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
      – hunter
      Dec 11 '18 at 22:59






    • 8




      @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
      – rschwieb
      Dec 12 '18 at 1:48






    • 1




      @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
      – Joker_vD
      Dec 12 '18 at 14:55






    • 1




      @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
      – user21820
      Dec 12 '18 at 18:08
















    So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
    – Daniel Schepler
    Dec 11 '18 at 19:34




    So in other words, you would want the field isomorphism to include a condition like: if $x ne 0$ then $phi(x) ne 0$ and $(phi(x))^{-1} = phi(x^{-1})$, right? (Or possibly, if $x ne 0$ and $phi(x) ne 0$ then $(phi(x))^{-1} = phi(x^{-1})$.)
    – Daniel Schepler
    Dec 11 '18 at 19:34




    7




    7




    I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
    – hunter
    Dec 11 '18 at 22:59




    I'm not sure I agree with this answer. There is no such thing as a "field structure": a ring is a set with additional structure, and a field is a ring with the additional property that the multiplication operation is invertible away from zero. From this perspective, it's automatic that a homomorphism of fields is just a homomorphism of rings.
    – hunter
    Dec 11 '18 at 22:59




    8




    8




    @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
    – rschwieb
    Dec 12 '18 at 1:48




    @hunter I think some might call that “property” a “unary operation with a special compatibility condition with multiplication” so that it is sort of a structure.
    – rschwieb
    Dec 12 '18 at 1:48




    1




    1




    @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
    – Joker_vD
    Dec 12 '18 at 14:55




    @user21820 But the field axioms are not universal, you have an implication $xne0$ in it — and so fields do not form a variety in the sense of universal algebra.
    – Joker_vD
    Dec 12 '18 at 14:55




    1




    1




    @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
    – user21820
    Dec 12 '18 at 18:08




    @Joker_vD: I'm using "universal sentence" as defined in standard logic texts (e.g. Rautenberg) and on Math SE. I didn't say we get a universal algebra, but universal theories do have some nice properties. With the original field axioms, not every substructure (ring) of a field is a field. But with the Skolemized field axioms (and if you stipulate $i(0) = 0$), every substructure (closed under $+,-,·,i$) is a field. That is how we think of subfields too. Every universal theory has a term model. What you're referring to are universal Horn formulae.
    – user21820
    Dec 12 '18 at 18:08











    15














    They are just isomorphic as rings.



    A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.






    share|cite|improve this answer


























      15














      They are just isomorphic as rings.



      A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.






      share|cite|improve this answer
























        15












        15








        15






        They are just isomorphic as rings.



        A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.






        share|cite|improve this answer












        They are just isomorphic as rings.



        A ring isomorphism already preserves both operations of the field, and it's trivial to prove that a ring isomorphism "preserves inverses," so there's nothing else you could ask of an isomorphism between fields that isn't already there.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 11 '18 at 18:44









        rschwieb

        105k1299244




        105k1299244















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