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?

If you know a little about Ring Theory, this explanation might help.
Consider the projection maps $\pi_1,\pi_2:R\times R\rightarrow R$. Then from your definitions, we have that


*

*$\pi_1(\alpha+\beta)=\pi_1(\alpha)+\pi_1(\beta)$

*$\pi_2(\alpha+\beta)=\pi_2(\alpha)+\pi_2(\beta)$

*$\pi_1(\alpha\beta)=\pi_1(\alpha)\pi_1(\beta)$

*$\pi_2(\alpha\beta)=\pi_2(\alpha)\pi_2(\beta)$

*$\pi_1((1,1))=1$

*$\pi_2((1,1))=1$

*$\pi_1(R\times R)=\pi_2(R\times R)=R$


where $\alpha,\beta\in R\times R$. Thus $\pi_1$ and $\pi_2$ are surjective ring homomorphisms from $R\times R$ to $R$. If $R\times R$ were a field, it would have exactly two ideals, namely itself and $\{0\}$. But we can find at least 4 distinct ideals in $R\times R$, namely itself, $\pi_1^{-1}(R)$, $\pi_2^{-1}(R)$ and $\{0\}$. 
