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Theorem 25.2: Modern Algebra an Introduction

A subset S of a Ring R is a subring of R iff S is nonempty, S is closed under both addition and multiplication of R, and S contains the negative of each of its elements.

Can anyone prove this? (note it is if and only if)

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What is your definition of a subring? – Alex Wertheim Dec 10 '13 at 1:40
What is your definition of a ring? – dfeuer Dec 15 '13 at 6:05

I'll give you some advice on how to do proofs.

When you want to prove something satisfies a definition, then you need to be familiar with that definition. In this case, you need to know the definition of a subring. From what I vaguely recall, S is a subring if it is also a ring and it has the same multiplicative identity as R (I'm not totally sure on this though).

So how do you start the proof? Well, start with the assumptions in the theorem. In the first direction you have "S is a subring of R" implies $\Rightarrow$ "S is nonempty, S is closed, etc."

This part is easy. HINT: Look at the definition of a subring.

Next, since this is an "if and only if" ($\Leftrightarrow$) statement then you have to also show that it works from the other direction:

S is nonempty, S is closed, S contains negatives" implies $\Rightarrow$ "S is a subring of R"

So work with your assumptions (S is nonempty, S is closed, etc) and figure out how these assumptions alone are enough to show that S satisfies all the other requirements of being a subring (which you will find in the definition of subring).

I would provide more detail but, like I said, I don't remember the definitions of rings or subrings off the top of my head.

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It depends on the definition of ring. Some texts require a ring to have a multiplicative identity and some don't. – dfeuer Dec 15 '13 at 6:02
Ah ok, good to know. I thought I'd just give it a broad shot since OP still hasn't replied to the first comment. – andraiamatrix Dec 15 '13 at 6:04
Some have ring and ring with unity. Some have rng and ring. Blah blah. It appears that rings with unity are in some ways nicer to study, but I don't really know anything about that. They seem to make modules work out better—if the ring doesn't have an identity, then there is no identity matrix. – dfeuer Dec 15 '13 at 6:08

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