Thread: Linear Algebra 1 : Fields - The Complex Field

1. Linear Algebra 1 : Fields - The Complex Field

I've just started studying about a week ago, I am extremely motivated and would do anything to succeed. Please - be as kind and tolerant as possible.

The course is Linear Algebra 1, the topic is Fields:

On this question I was asked to prove that C (the group of complex numbers) is a mathematical field. I did so, by proving that all the rules that exist on fields - exist on C.

the C field: C:=RxR={(a,b) : a,b(belongs to)R}

Then, on the second part of it, they say:

Let F be a field. Let's assume there's an a(that belongs to)F, so that a^2+1=0 . Prove that FxF isn't a field.
I can understand that it's like asking me to prove that a,b on Z=a+bi has to be that (a,b(belong to)R) on a field, because in this question it's asked to prove that it's impossible that there would be a field with (a,b) so that (a,b) can be imaginary.

Now, in order to prove that a group is NOT a field, it's enough to show that it has a,b(belong to)F so that a,b(are not equal to zero), and a*b=0.

I just couldn't find any way to prove that ^...

Would any of you mind to help me?

Thank you very much

I've just started studying about a week ago, I am extremely motivated and would do anything to succeed. Please - be as kind and tolerant as possible.

The course is Linear Algebra 1, the topic is Fields:

On this question I was asked to prove that C (the group of complex numbers) is a mathematical field. I did so, by proving that all the rules that exist on fields - exist on C.

the C field: C:=RxR={(a,b) : a,b(belongs to)R}
C is not just "RxR". You must also define (a,b)(c,d)= (ac-bd, ad+bc).

Then, on the second part of it, they say:

Let F be a field. Let's assume there's an a(that belongs to)F, so that a^2+1=0 . Prove that FxF isn't a field.

I can understand that it's like asking me to prove that a,b on Z=a+bi has to be that (a,b(belong to)R) on a field, because in this question it's asked to prove that it's impossible that there would be a field with (a,b) so that (a,b) can be imaginary.

Now, in order to prove that a group is NOT a field, it's enough to show that it has a,b(belong to)F so that a,b(are not equal to zero), and a*b=0.

I just couldn't find any way to prove that ^...

Would any of you mind to help me?

Thank you very much
Is that the whole question? Given that F is a field, there are two operations defined on the set F. But "FxF" is a set, there is NO operation automatically defined on it. I might suspect that you mean to define them "coordinate" wise: (a, b)+ (c, d)= (a+c, b+d) and (a, b)(c, d)= (ac, bd) but you just gave the complex numbers as (F,F) where that is not true. But perhaps that is what you mean (in which case C is NOT RxR as you wrote) since if RxR is NOT a field with those definitions.

One important property of a field is that every non-zero member has a multiplicative inverse. (a, 0) for a non-zero is not 0 for this field because (a, 0)+ (b, c)= (a+b, c) which is not equal to (b, c). Does (a, 0) have a multiplicative inverse?

Or, since you say "it's enough to show that it has a,b(belong to)F so that a,b(are not equal to zero), and a*b=0", what is (a,0)*(0,b)?

(Notice, again, that stating the operation is important. Using the definition of "multiplication" that makes RxR into C, (a,0)*(0,b)= a*0- b*0+ a*b+ 0*0= ab which is NOT 0.)

3. What you said about multiplying objects is mentioned in the question, I just thought it wouldn't be necessary to write it.
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The original question is:
(I just made it a little simple, or at least I thought so - I am new to academic-level math.
Let there be a field F, so that for every a(belongs to)F a^2+1(isn't equal) to zero. With the same rules of C, you can prove that FxF is a field (using the given fact).
Now, prove that the given fact is necessary - let's say that there is an a(belongs to)F so that a^2+1=0. Prove that such FxF isn't a field.
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The do define the 0, the 1, and the rest of the rules for the set C:=RxR, only I didn't copy it.

What I mean by FxF is a set of 2 objects from the field F, just like RxR means a set of 2 objects from the field R.

They same rules of complex-multiplication are right for this new set. On the first part, I proved that C is a field, with the given facts (some of which were mentioned by you or me).

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Thank you for your most generous help!

4. You were on the right track in the original post. As I understand it, the question is to show that the "ring" direct product (as in the direct product with componentwise operations) of two fields is not a field? Then since $\displaystyle \mathbb{F}$ is a field, then $\displaystyle \mathbb{F}$ certainly contains at least a zero element and an identity element which are distinct right? Call them 0 and 1. Then Consider $\displaystyle a=(1,0), b=(0,1)$.

5. In this case, I need to prove that if there's an a, any a, in the field F, that for this a: a^2+1=0, then FxF can't be a field.

The field FxF means it's sets of numbers that each of them belongs to F.

The a and b you've (siclar) mentioned are sets (2 sets..), while a is supposed to be a sigle object. (the set is defined by FxF)

Anyway, (a)*(b)=(0,1), mainly because (1,0) is the identity element on FxF, so every object from the field FxF you multiply it by - stays the same.

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The solution should be something like this:

*There is an a^2+1=0, a(is in the field)F
*I need to prove that FxF is not a field.

So, I can show that if a=(m,n) are in the field FxF - then, it's not a field.
But how? How can I use the information given?

It looks so simple at first sight, but I just can't seem to find the absolute answer for that!

Thank you all for your help!!

6. If I understand things correctly you are being asked to prove that the set F*F is not a field if you have the same multiplication as the complex numbers i.e. (a,b)*(c,d) = (ac-bd,ad+bc).
Now as you say in you just need to find two non-zero elements a,b such that a*b=0.
If you know anything about complex numbers you already know that (a,b)*(a,-b) is (aa+bb,0).
This should lead you to consider what might happen when the "real" part is indeed the special a that solves a^2+1=0. Surely that is a big enough hint for you to quickly find two elements from the set F*F that have 0 product!
I'm sure that the fact that the multiplication is the same as the complex numbers is important, because for example you could probably define other multiplications whereby F*F would end up being the skew field like the quaternions or another field isomorpic to F.

7. In this case, I need to prove that if there's an a, any a, in the field F, that for this a: a^2+1=0, then FxF can't be a field.

The field FxF means it's sets of numbers that each of them belongs to F.
Right, $\displaystyle \mathbb{F}\times\mathbb{F}=\{(x,y): x\in\mathbb{F}, y\in\mathbb{F}\}$. I cannot tell what the operations are defined to be on this set. I was assuming you wanted to show this is not a field under the usual componentwise operations, where $\displaystyle (w,x)+(y,z)=(w+y,x+z)$ and $\displaystyle (w,x)*(y,z)=(w*y,x*z)$. In this case, the additive identity is $\displaystyle (0,0)$ and multiplicative identity is $\displaystyle (1,1)$. However, I think that you mean to have the operations the same as when constructing the complex numbers from the direct products of the reals.

The a and b you've (siclar) mentioned are sets (2 sets..), while a is supposed to be a sigle object. (the set is defined by FxF)
This was just a conflict between my notation and yours, so let $\displaystyle s=(1,0)$ and $\displaystyle t=(0,1)$. Then $\displaystyle s*t=(1*0,0*1)=(0,0)$ so the existence of zero divisors implies this is not a field under componentwise operations, which is what my original understanding of the problem was.

Anyway, (a)*(b)=(0,1), mainly because (1,0) is the identity element on FxF, so every object from the field FxF you multiply it by - stays the same.
See above.

I understand the question correctly now. See the post above for a good explanation of the problem.

8. Originally Posted by alunw
If I understand things correctly you are being asked to prove that the set F*F is not a field if you have the same multiplication as the complex numbers i.e. (a,b)*(c,d) = (ac-bd,ad+bc).
Now as you say in you just need to find two non-zero elements a,b such that a*b=0.
If you know anything about complex numbers you already know that (a,b)*(a,-b) is (aa+bb,0).
This should lead you to consider what might happen when the "real" part is indeed the special a that solves a^2+1=0. Surely that is a big enough hint for you to quickly find two elements from the set F*F that have 0 product!
I'm sure that the fact that the multiplication is the same as the complex numbers is important, because for example you could probably define other multiplications whereby F*F would end up being the skew field like the quaternions or another field isomorpic to F.
Thank you very much! I now understand how to solve this problem!

Thanks to everyone else for trying to help, it was indeed very useful to reach the solution!