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Unit 21: Unique Factorization Domains
This function is the norm function, and is usually denoted by N. Notes
You can check that this function has the property that
5
f() = f() f() , , Z [ ].
Now, 9 has two factorisations in Z[ 5], namely,
9 = 3.3 = ( 2 + 5 ) ( 2 5 ).
5
You have already shown that the only units of Z [ ] are 1 and 1. Thus, no two of 3, 2+ 5
and 2 are associates of each other.
5
5
Also, each of them is irreducible. For suppose any one of them, say 2 + , is reducible. Then
2+ = for some non-invertible a, Z[ ].
5
5
Applying the function f we see that
f ( 2 + 5 ) = f() f(),
i.e., 9 = f() f().
Since f(), f() N and a, are not units, the only possibilities are f() = 3 = f().
So, if a = a + b , then a + 5b = 3.
5
2
2
But, if b 0, then a + 5b 5; and if b = 0, then a = 3 is not possible in Z. So we reach a
2
2
2
5
contradiction. Therefore, our assumption that 2 + is reducible is wrong. That is, 2 + is
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irreducible.
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Similarly, we can show that 3 and 2 are irreducible. Thus, the factorisation of 9 as a product
of irreducible elements is not unique. Therefore, Z[ ] is not a UFD.
5
From this example you can also see that an irreducible element need not be a prime element.
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For example, 2 + is irreducible and 2+ |3.3, but 2+ | 3 . Thus, 2 + is not a
5
5
5
prime element.
Now let us discuss some properties of a UFD. The first property says that any two elements of a
UFD have a g.c.d. and their g.c.d. is the product of all their common factors. Here we will use the
fact that any element a in a UFD R can be written as
1 r
2 r
a p p ...p n n r
1
2
where the pis are distinct irreducible elements of R. For example, in Z[x] we have
x x x + 1 = (x 1) (x + l) (x 1 ) = (x 1) (x + 1).
2
2
3
So, let us prove the following result.
Theorem 3: Any two elements of a UFD have a g.c.d.
Proof: Let R be a UFD and a.b R.
2 r
1 s
1 r
Let a p p ...p n n r and b p p 2 2 s ...p n n s
1
2
1
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