is the multiplication induced by C the only one that makes R2 a field?

>>8618790
yeah

any field extension of R by adding a root of a polynomial must be of degree 2 (only degree 2 polynomials are irreducible and don't give R as an extension), and any finite field extension is of this form (otherwise for a new element c there's an infinite subbasis basis c, c^2, c^3, ...). so all we need to see is how R extends by roots of quadratic equations

for an irreducible polynomial x^2 + bx + c, we add sqrt(b^2 - 4c), a root of a negative number. but sqrt(4c-b^2) is positive so dividing we get i. and R[i] = C

that means we have C inside all of them, what about other copies of C? C only has two automorphisms, the trivial one and the one that goes i -> -i. I'd say that's all

so yes, it's always C, and even if you don't consider it up to isomorphism, there are only two ways to extend it

>>8618812

I didn't fully understand the post. Could you at least tell me where I could look that stuff up?

>>8618812
OP is talking about making R2 a field, not making it complete.

So consider the operation (a,b) + (c,d) = (a+c,b+d) and (a,b)*(c,d) = (ac,bd)

Proof of associativity:
((a,b) + (c,d)) + (e,f) = (a+c,b+d) + (e,f) = (a+c+e,b+d+f)

And
(a+b) + ((c,d) + (e,f)) = (a+b) + (c+e,d+f) = (a+c+e,b+d+f) = (a+c+e,b+d+f)

The proof for multiplication is the same.

The proof of commutativity for both operations follow from the properties of operations of real numbers.

The additive identity of obviously (0,0) and the multiplicative one is (1,1).

The additive inverse of (a,b) exists and is (-a,-b). The same goes for multiplication.

The big one: Distributivity

(a,b)*( (c,d) + (e,f))) = (a,b)*(c+e,d+f) = (ac + ae, db + fb)

But also
(a,b)*(c,d) + (a,b)*(e,f) = (ac,bd) + (ae,bf) = (ac + ae,bd + bf)

This completes the proof.

Reminder that the operation induced by C is (a,b)*(c,d) = (ac - bd, ad + cb) so the operation described above is clearly different.

Exercise: Find another operation that makes R2 a field and prove that it is.

>>8618897
this is not a field as (1,0) has no multiplicative inverse and is not the zero

>>8618902
>(1,0)
Oh shit nigga. Going to kms

>>8618897
>OP is talking about making R2 a field, not making it complete.
>the same goes for multiplication
oh yeah? what's the inverse of (1,0), genius?

>>8618843
sure. this is algebra, specifically the study of fields. to study it you need to know about groups and rings first, and look for a chapter in an algebra book about "fields" or about "galois theory", which is the specific study of field extensions and automorphisms

a classic algebra book is Artin, a popular one is Dummit and Foote, my favorite one is Rotman. any is good.

>>8618907

Thanks field anon. I'm not sure if a textbook is worth it since I'll be taking field theory this semester, but I'll consider it.

>>8618790
If you don't require the multiplication to be compatible with the one in R, then yes, there are other ways to define it (but this requires the axiom of choice).

https://math.stackexchange.com/questions/1626926/is-there-only-one-way-to-make-mathbb-r2-a-field

>>8618953
If you don't require the multiplication to be compatible with the one in R then you're not reallly using R^2, are you? just a generic set of the same cardinality

>>8618972
>then you're not reallly using R^2, are you?
(a, b) x (c, d) = (e, f), where a,b,c, etc. are all real numbers. I would count this as "using" R^2

>>8618995
"making R^2 into a field without requiring the multiplication to be compatible with the one in R" is literally the same as "making a set of cardinality |R| into a field", as you can read in the math stack exchange thread

I could grab the operation of Z and a bijection to N and make N into a ring. is it N? not nearly

>>8619004
I don't really understand your objection. You can make N into a ring. So what? You seem to be confused about what a field means. The new field doesn't have to be a field extension of R.

Similar to the stuff discussed above, R^2 without a structure is just a set that can be put in bijection to R in potentially many ways. Choose such a map [math]b[/math].

[math] (x,y) \bullet_\times (u,v) := b^{-1}(\,b((x,y))\times b((u,v))\,) [/math]
[math] (x,y) \bullet_+ (u,v) := b^{-1}(\,b((x,y)) + b((u,v))\,) [/math]