Algebra (Unknown 27). - Index of
Algebra (Unknown 27). - Index of
Algebra (Unknown 27). - Index of
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
4 <strong>Algebra</strong> 2<br />
Conversely, let A/I be an integral domain. If ab ∈ I then (a+I)(b+I) = I = 0+I,<br />
hence either a + I = I and so a ∈ I, or b + I = I and so b ∈ I. Thus, I is a prime<br />
ideal <strong>of</strong> A.<br />
Example: for a prime number p the ideal pZ is a prime ideal <strong>of</strong> Z. The zero ideal<br />
(0) is a prime ideal <strong>of</strong> Z.<br />
Corollary. Every maximal ideal is prime.<br />
Pro<strong>of</strong>. Every field is an integral domain.<br />
Remark. In general, not every prime ideal is maximal. For instance, (0) is a prime<br />
ideal <strong>of</strong> Z which is not maximal.<br />
1.6. For rings A i define their product A 1 × · · · × A n as the set theoretical product<br />
endowed with the componentwise addition and multiplication.<br />
Chinese Remainder Theorem. Let I 1 , . . . , I n be ideals <strong>of</strong> A such that I i + I j = A<br />
for every i ≠ j. Then<br />
A/(I 1 . . . I n ) ≃<br />
∏<br />
A/I k , a + I 1 . . . I n ↦→ (a + I k ) 1kn .<br />
Pro<strong>of</strong>. First let n = 2. Then<br />
1kn<br />
I 1 I 2 ⊂ I 1 ∩ I 2 = (I 1 ∩ I 2 )A = (I 1 ∩ I 2 )(I 1 + I 2 ) ⊂ (I 1 ∩ I 2 )I 1 + (I 1 ∩ I 2 )I 2 ⊂ I 1 I 2 .<br />
So I 1 I 2 = I 1 ∩ I 2 . The kernel <strong>of</strong> the homomorphism<br />
A →<br />
∏<br />
A/I k , a ↦→ (a + I 1 , a + I 2 )<br />
1k2<br />
is I 1 ∩ I 2 = I 1 I 2 . It is surjective: since I 1 + I 2 = A, there are elements x ∈ I 1 , y ∈ I 2<br />
such that x + y = 1 and hence bx + ay = a + (b − a)x ∈ a + I 1 and similarly<br />
bx + ay ∈ b + I 2 .<br />
Now proceed by induction on n. Denote J 1 = I 1 , J 2 = I 2 . . . I n , so J 1 J 2 =<br />
I 1 . . . I n . Since I 1 + I k = A for all k > 1, we deduce using 1.2 that<br />
J 1 + J 2 = I 1 + I 2 . . . I n ⊃ (I 1 + I 2 ) . . . (I 1 + I n ) = A,<br />
so<br />
∏<br />
J 1 +J 2 = A. Now in the same way as in the previous paragraph one gets A/(J 1 J 2 ) ≃<br />
1k2 A/J k. By the induction hypothesis A/J 2 ≃ ∏ 2kn A/I k. Thus,<br />
A/(I 1 . . . I n ) ≃<br />
∏<br />
A/I k .<br />
1kn<br />
Example.<br />
Let p i be distinct primes and r i positive integers. Then<br />
Z/(p r 1<br />
1 . . . pr n<br />
n Z) ≃ ∏ Z/p r i<br />
i Z.