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Here is a restatement of the Theorem 3.6d.
"The only ideals in Z are the obvious ones."
Note: Our text calls Thm 3.3 the "division algorithm". It
certainly is a theorem. Neither the text nor I have explained
what the *algorithm* is; it turns out that this theorem can be
proven algorithmically. Indeed, the "long division" algorithm
that you learned in school can be used to prove Thm 3.3, if the
axioms are set up appropriately.
So maybe I should refer to (3.3) as "the Division Algorithm
Theorem"...(?)
Best Wishes, "Prof. Jonathan"
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Use Z for the set of integers.
For b an integer, let b·Z denote the set of multiples of b. Our
book also calls this J_b, i.e, "J sub b".
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Lemma 1: Consider an ideal J. Suppose we have integers x, b, q and r
such that
2: x = q·b + r .
If x and b are in the ideal, then so is r.
Proof. Certainly q·b is in the ideal. Thus x - qb is in J too. QED.
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Problem B3: More generally, here is a proof of Theorem 3.6d:
Suppose that J is an ideal in Z. Then there exists a
natnum b such that
3: J = b·Z.
Sketch of Proof:
WLOGenerality J has a non-zero element.
WLOG J has a positive element. Thus the set
P := {n in J | n is positive}
of natnums is non-void. So we can define a natnum
b := min(P).
Our goal now is to show (3). Courtesy (Thm3.6c) we have the inclusion
J superset b·Z.
To show the opposite inclusion, J subset b·Z, fix an element x in J.
Our goal is to establish that x is a multiple of b. So I want to
exhibit an integer q for which
4: x = q·b .
STEP TWO: Use the Division Algorithm thm to provide *integers*
q and r such that
MOM: x = q·b + r
where
MOM': 0 <= r < b .
Courtesy lemma 1, this r is in J. But r is less-than b, and b was
the MINIMUM positive integer in J. Hence r is not positive. Thanks
to (MOM'), then, r must be zero. Consequently [thanks (MOM)!] we
have (4), as desired. QED
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