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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    $\begingroup$ A field homomorphism is a ring homomorphism between fields - so yes! $\endgroup$ – Dietrich Burde Dec 11 '18 at 19:09
  • $\begingroup$ Please see my comment, which explains just what TonyK's answer means by "field structure". $\endgroup$ – user21820 Dec 12 '18 at 12:46

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 $\varphi:S\rightarrow T$ which preserves the structure. So, for instance, if $\circ$ is a binary operation in the structure, then for $x,y\in S$, we have $\varphi(x\circ y)=\varphi(x)\circ \varphi(y)$.

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.

  • $\begingroup$ 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 $(\phi(x))^{-1} = \phi(x^{-1})$.) $\endgroup$ – Daniel Schepler Dec 11 '18 at 19:34
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    $\begingroup$ 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. $\endgroup$ – hunter Dec 11 '18 at 22:59
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    $\begingroup$ @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. $\endgroup$ – rschwieb Dec 12 '18 at 1:48
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    $\begingroup$ @user21820 But the field axioms are not universal, you have an implication $x\ne0$ in it — and so fields do not form a variety in the sense of universal algebra. $\endgroup$ – Joker_vD Dec 12 '18 at 14:55
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    $\begingroup$ @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. $\endgroup$ – user21820 Dec 12 '18 at 18:08

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.


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