Unformatted text preview: 78 1. ALGEBRAIC THEMES The ﬁrst assertion is this: If x Á y .mod ab/, then x Á y .mod a/
and x Á y .mod b/. But this is actually evident, as it just says that if x y
is divisible by ab , then it is divisible by both a and b .
The second assertion is similar but slightly more subtle. We have to
show if Œxab ¤ Œyab , then .Œxa ; Œxb / ¤ .Œya ; Œyb /; that is, Œxa ¤
Œya or Œxb ¤ Œyb . Equivalently, if x y is not divisible by ab , then
x y is not divisible by a or x y is not divisible by b . The (equally
valid) contrapositive statement is: If x y is divisible by both a and b ,
then it is divisible by ab . This was shown in Exercise 1.6.11.
Now, since both Zab and Za ˚ Zb have ab elements, a one-to-one
map between them must also be onto. This means that, given integers ˛
and ˇ , there exists an integer x such that
.Œxa ; Œxb / D .Œ˛a ; Œˇb /;
or Œxa D Œ˛a and Œxb D Œˇb . But this means that x Á ˛ .mod a/ and
x Á ˇ .mod b/.
We didn’t need it here (in order to prove the Chinese remainder theorem), but the map from Zab to Za ˚ Zb deﬁned by Œxab 7! .Œxa ; Œxb /
is a ring isomorphism (See Exercise 1.11.10).
A ﬁeld is a special sort of ring:
Deﬁnition 1.11.7. A ﬁeld is a commutative ring with multiplicative identity element 1 ¤ 0 (in which every nonzero element is a unit.
(a) R, Q, and C are ﬁelds.
(b) Z is not a ﬁeld. RŒx is not a ﬁeld.
(c) If p is a prime, then Zp is a ﬁeld. This follows at once from
Proposition 1.9.9. Exercises 1.11
1.11.1. Show that the only units in the ring of integers are ˙1.
1.11.2. Let K be any ﬁeld. (If you prefer, you may take K D R.) Show
that the set KŒx of polynomials with coefﬁcients in K is a commutative
ring with the usual addition and multiplication of polynomials. Show that
the constant polynomial 1 is the multiplicative identity, and the only units
are the constant polynomials. ...
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- Fall '08
- Algebra, Prime number, multiplicative identity, zab