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How many of the following statements are false?
a) Subring of a ring is a ring.
b) Subring of commutative ring is a commutative ring.
c) Subring of a integral domain is an integral domain.
d) Subring of a field is a field.

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What have you done, what have you tried, what are your ideas, background, insights...?! –  DonAntonio Sep 7 '12 at 4:17
Are you sure you want only to know how many, now which are fals? –  Hagen von Eitzen Sep 7 '12 at 5:22
Hint for (d): is $\Bbb Z$ a subring of any field that you know? –  Brian M. Scott Sep 7 '12 at 5:53
I wonder who upvoted this? –  user641 Sep 7 '12 at 7:09
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closed as not constructive by Andres Caicedo, Steve D, William, rschwieb, LVK Sep 7 '12 at 21:33

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1 Answer

To answer these questions, carefully write down the properties of a ring/comm. ring/int. domain, and those of a subring, and then check whether or not all the properties are satisfied by the subring (or if the statement is false, find a counterexample). If you need intuition, the integers are an integral domain (and so a comm. ring, and so a ring); see whether you believe these statements are true for the integers.

For instance, for a), let R be the ring, S the subring. Because R is a ring, we have $$\forall a, b \in R, a + b \in R$$ $$\forall a, b, c \in R, (a + b) + c = a + (b + c)$$ $$\exists 0 \in R, : \forall a \in R, 0 + a = a + 0 = a$$ $$\forall a \in R, \exists b (= "-a") \in R : a + b = b + a = 0$$ $$\forall a, b \in R, a + b = b + a$$ $$\forall a, b \in R, a · b \in R$$ $$\forall a, b, c \in R, (a · b) · c = a · (b · c)$$ $$\exists 1 \in R : \forall a \in R, 1 · a = a · 1 = a$$ $$\forall a, b, c \in R, a · (b + c) = (a · b) + (a · c)$$ $$\forall a, b, c \in R, (a + b) · c = (a · c) + (b · c)$$ Because $S$ is a subring, $$\forall a, b \in S, a + b \in S$$ $$\forall a, b \in S, a · b \in S$$ $$\forall a \in S, −a \in S$$ $$1 \in S$$ Now, verify that $S$ satisfies the remaining properties of a ring (it does). It's a bit tedious, but that's how it's done. (if the text you're working from doesn't require rings to be unital, you can ignore the properties involving $1$.)

For the subsequent parts, you can build on the earlier parts; for instance, to show (b), you can start by claiming the subring is (at least) a ring by part (a), and it just remains to show it is commutative.

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Not everyone requires that a ring be unital. –  Brian M. Scott Sep 7 '12 at 5:34
Fair enough. Edited. –  BaronVT Sep 7 '12 at 5:51
The whole thing depends on how one defines "subring". One might e.g. define that $(S,+_S,\cdot_S)$ is a subring of $(R,+_R,\cdot_R)$ if it is a ring and $S\subseteq R$ and $+_S=+_R|_{S\times S}$ and $\cdot_S=\cdot_R|_{S\times S}$; then one observes that a sinmple criterion for subrings exists that allows one to avoid testing all ring axioms. Of course such a definition that a sub-foo is a foo contained in another foo is much more naturally motivated and transfers readily to most (concrete) categories. –  Hagen von Eitzen Sep 7 '12 at 6:34
You raise a good point, but since the first part of the question is "is a subring of a ring a ring?", I think it's fair to assume the working definition is something like what I posted above. –  BaronVT Sep 7 '12 at 8:23
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