Prove that a set of operations define a field

[elic307]

Hi,
I am trying to find a solution for the following problem:

F is given as follows:

F = {(a,b) where a,b are rational numbers}

Let's define:
(a, b) + (c, d) = (a + c, b + d)
(a, b) * (c, d) = (ac+2bd, ad + bc)

Prove that F is an algebraic field

I was able to prove that all the axioms hold here, till I needed to find the
zero, the one, and the inverse the "a in the power of -1". Please forgive my
English, I am not a native speaker.

For the Zero, I found it must be (0, 0)

For the "One" (assume (c,d) is the one) I get
ac + 2bd = a
ad + bc = b

Which can happen for arrbitrary (a,b) if c=1 and d=0.
That is the one is (1, 0)

Again, assume (c,d) is the inverse.
So I need to find c and d such that

ac+2bd = 1
ad+bc = 0

for any a,b

I can't find any because a and b are in both terms.

I suspect my choice of the one is not so successful or maybe this can't be
proved at all?

Thanks for you help

 

[blamocur]

You seem to be on the right way with your equation, but you statement that "I can't find any because a and b are in both terms." baffles me: all you need to do express c and d in terms of a and b. You might find your equation more readable if you use x and y instead of c and d.

 

[Steven G]

For the "One" (assume (c,d) is the one) I get
ac + 2bd = a
ad + bc = b

Which can happen for arrbitrary (a,b) if c=1 and d=0.
That is the one is (1, 0)

Is (1,0) the only one or are there others? Can you have more than one One in a field?

 

[elic307]

You seem to be on the right way with your equation, but you statement that "I can't find any because a and b are in both terms." baffles me: all you need to do express c and d in terms of a and b. You might find your equation more readable if you use x and y instead of c and d.

I meant that the solution must not depend on either a or b since it should be correct for any a,b.

 

[elic307]

For the "One" (assume (c,d) is the one) I get
ac + 2bd = a
ad + bc = b

Which can happen for arrbitrary (a,b) if c=1 and d=0.
That is the one is (1, 0)

Is (1,0) the only one or are there others? Can you have more than one One in a field?

I don't know. This is the only item I could found that can act as a "one". So the solution for the "inverse" takes that into account.

 

[blamocur]

I meant that the solution must not depend on either a or b since it should be correct for any a,b.

There is no single inverse in a field (except for the field with only 2 elements {0,1}). Unlike identity ("One"), an inverse is defined for every non-zero element of the field. It is unique once you choose which element to compute the inverse for ((a,b) in your case).
To summarize: inverse definitely has to depend on a,b (https://en.wikipedia.org/wiki/Field_(mathematics)).

 

[elic307]

There is no single inverse in a field (except for the field with only 2 elements {0,1}). Unlike identity ("One"), an inverse is defined for every non-zero element of the field. It is unique once you choose which element to compute the inverse for ((a,b) in your case).
To summarize: inverse definitely has to depend on a,b (https://en.wikipedia.org/wiki/Field_(mathematics)).

Now that you say this, you're definitely correct. I got carried away with the definition of the "zero" and the identity which do not depend on a or b.
I will revisit my work and see.
Thanks for the clarification.