Page 217 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes where p , p , ..., p are distinct irreducible elements of R and r and s are non-negative integers 
1
i = 12, ..., n. 2 n i i
(If some p does not occur in the factorisation of a, then the corresponding r = 0. Similarly, if
i
i
some p is not a factor of b, then the corresponding s = 0. For example, take 20 and 15 in Z. Then
i
i
20 = 2 × 3 × 5 and 15 = 2 × 3 × 5 .)
2
0
1
0
1
Now, let t, = min (r , s )  i = 1, 2 ,....,n .
i
i
2 t
1 t
Then d  p p ...p n n t divides a as well as b, since t  r and t  s  i = 1, 2, ...., n.
1
2
i
i
i
i
Now, let c | a and c | b. Then every irreducible factor of c must be an irreducible factor of a and
of b, because of the unique factorisation property.
Thus, c  p 1 m 1 p 2 m 2 ...p n m n , where m  r and m  s  i = 1,2, ...,n . Thus, m  t  i = 1,2 ,..., n.
i
i
i
i
i
i
Therefore, c | d.
Hence, d = (a, b).
Over there we found a non-UFD in which an irreducible element need not be a prime element.
The following result says that this distinction between irreducible and prime elements can only
occur in a domain that is not a UFD.
Theorem 4: Let R be a UFD. An element of R is prime iff it is irreducible.
Proof: We know that every prime in R is irreducible. So let us prove the converse.
Let a  R be irreducible and let a | bc, where b, c  R.
Consider (a,b). Since a is irreducible, (a,b) = 1 or (a,b) = a.
If (a,b) = a, a| b.
If (a,b) = I, then d| b . Let bc = ad, where d  R.
1 r
1 s
2 r
Let b  p p ...p m m r , and c  q q 2 2 s ...q n n s , be irreducible factorisations of b and c. Since bc = ad
1
2
1
and a is irreducible, a must be one of the pis or one of the qs. Since a| b, a  p for any i.
i
j
Therefore, a = q for some j. That is, a | c.
j
Thus, if (a,b) = 1 then alc.
So, we have shown that a/ bc  a | b or a | c.
Hence, a is prime.
Theorem 5: Let R be a UFD. Then R[x] is a UFD.
We will not prove this result here, even though it is very useful to mathematicians. But let us
apply it. You can use it to solve the following exercises.
Lemma: Let D be a unique factorization domain, and let p be an irreducible element of D. If a,b
are in D and p|ab, then p|a or p|b.

Definition: Let D be a unique factorization domain. A non-constant polynomial
f(x) = a x + a x + · · · + a x + a in D[x]
n-1
n
n
n-1
1
0
is called primitive if there is no irreducible element p in D such that p | a for all i.
i

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