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?
Best Answer
If you are trying to prove associativity, you should note that addition is only strictly defined for two elements, so you should be trying to prove the second line. You can do that from your definition so $$(a,b)+[(c,d)+(e,f)]=\\(a,b)+(c+e,d+f)=\\(a+c+e,b+d+f)=\\(a+c,b+d)+(e,f)=\\ [(a,b)+(c,d)]+(e,f)$$ You made a typo going to your fourth line. You can do the same for multiplication. The place you will fail is inverses. What is the inverse of $(1,0)$? Imagine it is $(a,b)$ and derive a contradiction.