abstract algebra question

Is there a shortcut to prove that Z, Q, R, S are pairwise nonisomorphic? I can split it into three proofs and do them individually, but I feel like there has to be some property that covers all the cases.

It can't be just the fact that Z < Q < R < S because the subset of even integers is isomorphic to the set of all integers.

Anyone care to push me in the right direction, if there is one?

what is S? and nonisomorphic as what? rings, groups, vector spaces, sets?

C not S, sorry. isomorphic as rings.

Seems to me it has to be at least three seperate ideas.

I'd start from the top down, obviously C isn't iso to any of the others because it has an element squaring to -1.

Next R isn't iso to Q or Z as R is uncountable and an isomorphism is a bijection

Q isn't isomorphic to Z because if it were, say isomorphism p:Q -> Z
p(1) = 1 by def. Also p(-1) = -1 and p(1+(-1))=0=p(1) + p(-1)
but then observe p(Z) = Z (excuse the notation), and thus the map isn't injective, as p(1/2) must also be in Z for example.

As isomorphism is an equivalence relation, this is sufficient to show pairwise non-isomorphic.

Is this simple enough? Strangely it's the last one that's the least neat, maybe there's a shorter proof than that.

Z is not divisible, Q is divisible, R have greater cardinality, C have a multiplicative elements of finite order.

>>5
Maybe simpler to just say that Q is a field and C is algebraically closed? Or just that x^2+1 splits in C.

Now prove [math]\mathbb{Q}(\sqrt{2})[math] and \mathbb{Q}(\sqrt{3}) are non-isomorphic.

>>6
(There really needs to be something here where we can edit/delete/preview posts amirite?)

Prove [math]\mathbb{Q}(\sqrt{2})[\math] and \mathbb{Q}(\sqrt{3}) are non-isomorphic.

>>7

Oh I give up.  WTF is even wrong with that?

>>8
wront slash, you need [/math] not [\math], too much texing for you.

i like >>5 best, personally.

>>8 \ --> / in [/math] ?
Prove \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt{3}) are non-isomorphic.

here's a neat one:

show (\mathbb{R},+) and (\mathbb{C},+) are isomorphic as abelian groups.

>>9
5 is retarded though.

That fact that Z is not divisible doesn't imply it's not ring isomorphic to the rationals. He's not given any form of a proof, he's merely listed a difference between the two things, if they weren't different there'd be no hope of them not being ring isomorphic.

Hell, R has multiplicative elements of finite order, take -1 for example.

I think >>4 is the best (lol samefag, but still right).

>>10
To prove that, oh look, let's use 4's idea.
If there's an isomorphism p, take p(sqrt(2)), p(sqrt(2))^2 = p(2) = 2.p(1) = 2, so that implies p(sqrt(2)) [an element inside of Q adjoined root 3] squares to 2, contradiction.

I'll do >>11 in a bit if I can be arsed

>>12
divisibility matters

say f : \mathbb{q} \longrightarrow \mathbb{Z} is an isomorphism,

then f(1) = f(q\cdot \frac{1}{q}) = f(q)\cdot f(\frac{1}{q}) = 1 in \mathbb{Z}.

but \mathbb{Z} has no multiplicative inverses.

he's listed differences that contradict the existence of an isomorphism, not just any differences will work.

>>12
divisibility is preservated by morfisms, (Q ~= Z) => (Z is divisible). But >>6 is right, being a field is a stronger reason.

In >>5 i meant to say that C have a multiplicative element of any finite order.

>>12
also, R doesn't have any finite multiplicative subgroups of order greater than 2. which results in another contradiction if you work it out.

>>11
R and C are both Q-vectorial spaces of the same dimension

>>12
>That fact that Z is not divisible doesn't imply it's not ring isomorphic to the rationals.

Divisibility, (or equivalently in this case, the property of being a field), is preserved under isomorphism.

>R has multiplicative elements of finite order, take -1 for example.

Torsion subgroup has infinite order, again something that's preserved under isomorphism.

>>11
What I want to say is that R and C are both uncountably-infinite dimensional free Z-modules and therefore isomorphic since they have the same dimension, but I'm not actually sure R is free over Z...

Hmmm.

>>16
That doesn't prove they're isomorphic as Z-modules (abelian groups).

Oh wait, maybe it does.  I guess a Q-isomorphism would be a Z-isomorphism as well, right?  Yeah, that's right.

>>17
Disregard that, I suck cocks.

I'm still curious whether R and C are free over Z, though.

>>18
Im almost sure that R or C arent free over Z. Supose they are; let f be an iso between direct sums of Z and R. take an element e from the basis of sums of Z. Then f(e) = a, for a=!0. take f^-1 (a/2). Then a = f(e) and a = 2*f(f^-1(a/2)), making a contradiction.

>>19
Is this what you're thinking of?

Let \{e_i\}_{i \in \mathbb{R}} be a basis and set a = f(e_0).  Write a/2 as a finite sum  \sum c_j e_j for c_j \in \mathbb{Z}.  Then a = 2(a/2) = \sum 2c_j e_j, which is different from e_0 since the coefficients are all even.

Yeah, that works.  Cool, thanks.

>>20
Thats the idea (Altought i didnt wrote that), show an element with 2 distinct lineal combination of the basis.

Another idea (Escentially the same): Quotient of divisible is divisible. Then ( Field(Implies divisible) isomorfic to a sum of Z )->(Z divisible)

Can any of you solve this^^

x33Hydrogen2Sulfid4+3Litium+(/10!!)×π10+(∞Nitrogen2Oxid+(Xx≠Z))33=Y
How many watt/hours is Y? if you say that little x=49 and Z=120 and big X= is constant a flame of 1gigawatt/hours.

Can any of you solve this^^
this^ sign means that it is a potense like 3^3= (3×3)×3

x3^3Hydrogen2Sulfid4+3Litium+(/10!!)×π^10+(∞Nitrogen2Oxid+(X^x≠Z))3^3=Y
How many watt/hours is Y? if you say that little x=49 and Z=120 and big X= is constant a flame of 1gigawatt/hours.

>>21
lrn2 spell

>>11
This needs a proper demonstration. I don't see how they can be isomorph, even as Z-modules. I mean, C is of dimension 2 over R as his algebraic closure (cloture ?)... And if you have example of space, set, etc that are isomorph as certain type of structure but not for other type, I'd be glad to acknowledge them.

>>25
R and C are isomorph as Q-module (As their base has the same cardinality, 2^aleph0). Every A-module morphism is a group morphism. So that imply that R and C are isomorph as groups. If you take R as an Q-module, you don't have the product between two reals, you have the structure of Q as a ring, and the structure of R as a group, and you define the "product" between elements of Q, with elements of R. In your post, you are considering R and C as fields, not as Q-vectorial spaces.

>>25
A Z-module and an abelian group are the same thing.

R and C have the same dimension as Q-vector spaces, and therefore are isomorphic as Q-vector spaces (but not as rings).  A Q-vector space isomorphism is also a Z-module isomorphism, i.e. an abelian group isomorphism.

>>26
>>27
Well, thanks a lot, and also, your moms.

Seriously speaking now, I knew for the Z-module-abelian group equivalence.  I think the main problem is my lack of knowledge of infinite cardinals and my lack of knowledge in module things. If I remember my notes, module are like a generalisation of vector space amirite ?

Ah btw, since you guys know quite a few things in algebra, I would like to ask you if what is the difference beteween a "unitary ring" and between an "algebra", aside from teh names. I mean, if I look at the definitions, ring needs two associative laws, one with a commutative group property and the other with just a group property, and is then stable for its two laws. And the definition of an algebra is the same : stability for a sum law, stability for a product law, for linear combination of elements of the algebra, but it just seems that the definition of an algebra include the action of a vector space on the algebra...

>>28
>If I remember my notes, module are like a generalisation of vector space amirite ?
Yep.  A module is like a vector space, just over an arbitrary ring instead of a field.  The fact that you don't have a field to work with makes things harder and you can't get some nice properties that you can with vector spaces (existence of a basis, etc)

>difference beteween a "unitary ring" and between an "algebra"
Basically, an algebra is like a module over a ring, except that the module also has a multiplication law.  If you know about field extensions, an algebra is essentially a ring extension.  So the difference between an algebra and a ring is basically like the difference between the field extension K/k and the field K.