Fields and Groups: Proving a Set is a Field vs Non-Abelian Group

[MikeDietrich]

Homework Statement

The problem asks me to determine if the matrix [p -q ## q p] is a field with addition and multiplication. However, that is not my question.

My question is: How is proving a set is a field different from proving a set is a non-abelian group (under addition then separately under multiplication)?

 

[Mark44]

From CRC Standard Math Tables, 15th Ed.
"A field is an integral domain in which every element except z is a unit. In other words, the non-z elements form an Abelian group relative to multiplication (X).
Example 1. Therational field consiting of ordinary fractions, addition, and multiplication.
Example 2. The se of all real numbers a + b$\sqrt{2}$, a and b rational. Then
(a + b$\sqrt{2}$) + (c + d$\sqrt{2}$) = (a + c) + (b + d)$\sqrt{2}$ and
(a + b$\sqrt{2}$) X (c + d$\sqrt{2}$) = (ac + 2bd) + (ad + bc)$\sqrt{2}$."

 

[MikeDietrich]

sorry... I meant "Abelian" not "non-Abelian" (for some reason, I cannot edit my original post). Thank you Mark.

 

[HallsofIvy]

A field will never be a group under multiplication- the additive identity never has a multiplicative inverse. Also, a field requires the distributive law a(b+ c)= ab+ ac which involves both addition and multiplication.

 

[micromass]

What you DO have is that F is a field if and only if
1) F is an abelian group for addition
2) F\{0} is an abelian group for multiplication
3) the distributive laws hold.

 

[MikeDietrich]

Thank you. That is exactly what I was looking to know!