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Subring proof

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].

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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