Is R={0, 2, 4, 6, 8} a Field under Addition and Multiplication Modulo 10?

cmajor47
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Homework Statement


Let R={0, 2, 4, 6, 8} under addition and multiplication modulo 10. Prove that R is a field.


Homework Equations


A field is a commutative ring with unity in which every nonzero element is a unit.


The Attempt at a Solution


I know that the unity of R is 6, and that the multiplicative inverse of 2 is 8, of 3 is 2, of 4 is 4, and of 6 is 6.
So I've shown that each nonzero element is a unit.
I'm not sure how to go about showing that R is a commutative ring.
To show the commutative part, I'm guessing that I can just do out the Cayley table of R.
I don't know how to show that R is a ring without going through all of the characteristics of a ring though. I don't think I need to do that but I'm not sure.
 
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Which properties are non-obvious?
 
Well I can clearly see that a+b=b+a, (a+b)+c=a+(b+c), a(bc)=(ab)c, a(b+c)=ab+ac, and (b+c)a=ba+ca, and that there is an additive identity 0 st a+0=a for all a in R
So, I guess the non-obvious property is that there is an element -a in R such that a+(-a)=0, but is this property even true?
 
I claim that what -[/color]a needs to be is obvious, and thus that property holds depending on whether or not that is actually an element of R.


(I use red to indicate it's the putative negation operation on R, rather than the negation operation on the integers modulo 10 which I will express in black)
 
Then I'm not sure what the non-obvious property is.
 
I am hoping that while you found the existence of additive inverses non-obvious earlier, that from my hint you now find it obvious -- or can at least prove it even if it isn't obvious.

So, if that was the only field property that wasn't clear to you, does that mean you are now happy that it's a field?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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