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# Subring proof

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Let R={f(x) included in Q[x]: the coefficient of x in f(x) is 0}
Prove that R is a subring of Q[x].

https://brainmass.com/math/ring-theory/subring-proof-140711

#### Solution Preview

subrings aren't (always) required to have a 1 (the identity of the ambient ring, R), but they might have an identity of their own

consider, for example, the subring S of R = Z x Z, the direct product of two copies of ring of integers, consisting
of pairs (x, 0), where x is in Z, i.e.

S = {(x, 0) : x in Z}

addition and multiplication in Z x Z are defined component-wise,

so (x, y) + (x', y') = (x + x', y + y')
and (x, y).(x', y') = (xx', yy')

the additive identity of Z x Z is (0, 0), and the multiplicative identity is (1, 1)
but S is a subring, whose own multiplicative identity is (1, 0), which is not the same as (1, 1), the identity of the larger ring R

indeed, for any (x, 0) in S,

(x, 0).(1, 0) = (x1, 0) = (x, 0)
(1, 0).(x, 0) = (1x, 0) = (x, 0)

formally speaking, all you ...

#### Solution Summary

This show how to prove that a ring is a subring.

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## Commutative Ring Proofs

See the attached file.
Let R be any commutative ring and S a subset of R n f0g containing no zero divisors.
Let X be the Cartesian product R  S and denote a relation  on X where (a; b)  (c; d).
(a) Show that  is an equivalence relation on X.
(b) Denote the equivalence class of (a; b) by a=b and the set of equivalence classes by RS (called the
localization of R at S). Show that RS is a commutative ring with 1.
(c) If a 2 S show that fra=a: r 2 Rg is a subring of RS and that r 7! ra=a is a monomorphism, so that
R can be identified with a subring with RS.
(d) Show that every s 2 S is a unit in RS.
(e) Give a universal" definition for the ring RS and show that RS is unique up to isomorphism.