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.Guest said: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?
I can't find suitable identity elements. I tried (0, 0) for addition and (1, 0) for multiplication but these don't work!Opalg said: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?
Guest said: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!
The field axioms for $G = F \times F$ are:
To check if $G = F \times F$ satisfies the field axioms, you need to verify that each axiom holds true for all elements in $F \times F$. This can be done by performing the required operations on different elements of $F \times F$ and comparing the results with the expected values based on the axioms.
If $G = F \times F$ does not satisfy the field axioms, then it cannot be considered a field. This means that some of the fundamental properties of fields, such as the existence of inverses and distributivity, do not hold true for all elements in $F \times F$.
No, $G = F \times F$ must satisfy all of the field axioms to be considered a field. The field axioms are a set of fundamental properties that define what a field is, and if any of these axioms are not satisfied, then $G = F \times F$ cannot be classified as a field.
Yes, apart from the field axioms, $G = F \times F$ must also satisfy the properties of closure, where the result of an operation on any two elements in $F \times F$ must also be an element in $F \times F$, and the property of existence of multiplicative inverses, where every element in $F \times F$ (except for 0) must have a multiplicative inverse.