# 一題代數證明題 (和「環」有關)

(⊆) R1 is a ring containing D (by i*), hence R1 contains R0 (by ii).

(⊇) R0 is a ring containing D (by i), hence R0 contains R1 (by ii*).

### 1 個解答

• Eric
Lv 6
1 0 年前
最佳解答

Consider the intersection of all rings containing D; that is,

R0 = ∩{R: R is a ring containing D} (**)

There is at least one ring being intersected, since the power set of the universal set is a ring containing D.

Claim:

(i) R0 contains D.

(ii) Any ring R containing D also contains R0.

Proof:

(i) Trivial, since every set (ring) in the intersection (**) contains D.

(ii) Trivial, since R (containing D) is a part of the intersection (**).

::

Claim: R0 is empty if and only if D is empty.

Proof:

(if) If D is empty, then the empty set is a ring containing D and thus is intersected in (**), so R0 is empty.

(only if) If R0 is empty, since we know R0 contains D, D must also be empty.

::

Claim: R0 is a ring.

Proof: If D is empty, then R0 is empty and hence trivially is a ring. Otherwise, R0 is nonempty.

If A and B are in R0, then A and B are in every ring containing D. Hence A \ B and A ∪ B are in every ring containing D, and thus A \ B and A ∪ B are both in R0. Hence R0 is a ring.

::

What remains to be proven is uniqueness:

Claim (uniqueness): If R1 is a ring satisfying

(i*) R1 contains D

(ii*) any ring R containing D also contains R1

Then R0 = R1.

Proof:

(⊆) R0 is a ring containing D (by i), hence R1 contains R0 (by ii*).

(⊇) R1 is a ring containing D (by i*), hence R0 contains R1 (by ii).

Hence, R0 = R1.

::

2006-07-15 02:02:03 補充：

Basically, R0 is the ring generated by D.

2006-07-18 02:56:56 補充：

啊，對～不好意思

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