Proving a structure is a field? Please help with what I am doing wrong here. It has been awhile since Ive been in school and need some help. The question is:
Let $F$ be a field and let $G=F\times F$. Define operations of addition and multiplication on $G$ by setting $(a,b)+(c,d)=(a+c,b+d)$ and $(a,b)*(c,d)=(ac,db)$. Do these operations define the structure of a field on $G$?
In order to be a field, the following conditions must apply:


*

*Associativity of addition and multiplication

*commutativity of addition and mulitplication

*distributivity of multiplication over addition

*existence of identy elements for addition and multiplication

*existence of additive inverses

*existence of multiplicative inverse 0 cannot equala, a-1*a=1


I started with 1. saying
$(a,b)+(c,d)+(e,f)=(a+c+e,b+d+f)$
$(a,b)+[(c,d)+(e,f)]=[(a,b)+(c,d)]+(e,f)$
$(a,b)+(c+e,d+f)=(a+c,b+d)+(e,f)$
$(a+c+d,b+e+f)=(a+c+e,b+d+f)$
which is not correct but I'm not sure where I went wrong. Is my logic incorrect?

It can help proving some more general result; let $A$ be any non empty set with an operation $\%$ on it (I use a generic symbol). Consider now the operation $?$ on $A\times A$ defined by
$$
(a,b)\mathbin{?}(c,d)=(a\mathbin{\%}c,b\mathbin{\%}d)
$$


*

*If $\%$ is associative, then also $?$ is associative

*$?$ is commutative if and only if $\%$ is commutative

*If $\%$ has a neutral element $e$, then $(e,e)$ is a neutral element for $?$

*An element $(a,b)$ has an inverse (with respect to $?$) if and only if both $a$ and $b$ have an inverse (with respect to $\%$)


Once you have verified these facts, you immediately have 1, 2, 4 and 5. Moreover, the neutral element for addition in $G$ is $(0,0)$. Distributivity of multiplication in $G$ can be verified directly.
However, condition 6 fails, because $(1,0)$ is not the neutral element for addition, but it has no multiplicative inverse.
