Checking Field Axioms for $G = F \times F$

[Guest2]

Problem: Let $F$ be a field and let $G = F \times F$. Define multiplication and addition on $G$ by setting $(a, b)+(c, d) = (a+c, b+d)$ and $(a, b) \cdot (c, d) = (ac, bd)$. Does this define a field structure on \(\displaystyle G\)?

I know field axioms but I'm unable to apply them to this problem. How do you check associativity for example?

 

[Opalg]

Problem: Let $F$ be a field and let $G = F \times F$. Define multiplication and addition on $G$ by setting $(a, b)+(c, d) = (a+c, b+d)$ and $(a, b) \cdot (c, d) = (ac, bd)$. Does this define a field structure on \(\displaystyle G\)?

I know field axioms but I'm unable to apply them to this problem. How do you check associativity for example?

To check associativity (for multiplication – use a similar process for associativity of addition), start like this: $$\bigl((a, b) \cdot (c, d)\bigr) \cdot(e,f) = (ac,bd)\cdot (e,f) = \bigl((ac)e,(bd)f\bigr).$$ Then do a similar calculation for $(a,b)\cdot \bigl((c,d) \cdot(e,f)\bigr)$, and use associativity of $F$ to conclude that the two results are the same.

You might do better to start the problem by asking what are the zero and identity elements of $F\times F$. Does each element of $F\times F$ have a negative, and does each nonzero element have an inverse?

 

[Guest2]

Thank you.

You might do better to start the problem by asking what are the zero and identity elements of $F\times F$. Does each element of $F\times F$ have a negative, and does each nonzero element have an inverse?

I can't find suitable identity elements. I tried (0, 0) for addition and (1, 0) for multiplication but these don't work!

 

[Deveno]

Thank you.

I can't find suitable identity elements. I tried (0, 0) for addition and (1, 0) for multiplication but these don't work!

Try $(0,0)$ again for addition:

$(a,b) + (0,0) = (a+0,b+0) =?$

This might help for multiplication: suppose our identity (if it exists) is $(x,y)$.

Since we must have: $(a,b)(x,y) = (a,b)$, we obtain:

$ax = a$
$by = b$.

Note we can re-write these as:

$a(x - 1) = 0$
$b(y - 1) = 0$. Use the field properties for $F$, now.

As to your larger question, as to whether or not $F \times F$ with the operations indicated is a field, I suggest you consider whether or not it has any zero-divisors.