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維正 發問時間: 科學數學 · 1 0 年前

一題代數證明題 (和「環」有關)

試證:If D is any class of sets, then there exists a unique ring R0 such that (i) R0 包含 D, and (ii) any ring R containing D contains also R0.

已更新項目:

我覺得最後一個proof的前兩行應該是這樣才對:

(⊆) 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

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  • 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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