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I am currently reading A First Course in Rings and Ideals and the theorem for subring states:

Let $R$ be a ring and $S$ be a nonempty subset of $R$. Then, $S$ is a subring of $R$ if and only if

  1. $a, b \in S$ implies $a - b \in S$ (closure under differences),
  2. $a, b \in S$ implies $ab \in S$ (closure under multiplication).

Why did theorem use "closure under differences" instead of "closure under addition? Also, does this theorem also imply that subrings are closed under addition?

Thanks!

emacs drives me nuts
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2 Answers2

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Yes, if $S \subseteq R$ is closed under subtraction, then it is also closed under addition.

Proof. Let $a, b \in S$ be arbitrary. Then $0 = b-b \in S$, so $-b = 0-b \in S$. Now $a + b = a - (-b) \in S$. $\square$

However, the converse is not true! There can exist subsets of $R$ which are closed under addition but not subtraction. For example, $\mathbb{N} \subseteq \mathbb{Z}$. So, if the theorem said "closed under addition" instead of "closed under subtraction", it would be wrong.

Let me also make a more philosophical point:

does this theorem also imply that subrings are closed under addition?

This question doesn't really make sense to ask. Subrings are closed under addition, by definition (Definition 1-4 in your book, if I'm not mistaken). The theorem does not tell you anything new about subrings -- instead, it gives you an efficient way to check if a subset of a ring is a subring. If this doesn't quite make sense yet, that's ok! Reread this page of your book and then go talk to your professor about it.

diracdeltafunk
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Just recall the definition of subring (or def of ring if you've forgotten): let $(R, +, \cdot )$ be ring, $A\subset R$ is called a subring of $R$ if $(A, +, \cdot)$ is also a ring.

If $A$ is subring, it is clear that 1 and 2 are correct.

on the other hand, 1 implies $A$ is a subgroup of $R$ with respect to addtion. 2 implies $A$ is closed under multiplication. Hence you can check that $(A, +, \cdot)$ is a ring straight from definition.

I guess your confusion mostly comes from the following theorome:

Let $(G, \cdot)$ be group, then $H\subset G$ is subgroup iff $\forall x, y\in A, xy^{-1}\in H$

come back to this thm and you'll figure out why it's subtraction not addtion.

Westlifer
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