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Abstract Algebra
Notes We will now discuss some properties of divisibility in PIDs. If R is a ring and a,b R, with a 0,
then a divides b if there exists c R such that b = ac.
Definition: Given two elements a and b in a ring. R, we say that c R is a common divisor of a
and b if c | a and c | b.
An element d R is a greatest common divisor (g.c.d, in short) of a, b R if
(i) d | a and d | b, and
(ii) for any common divisor c of a and b, c | d.
For example, in Z a g.c.d of 5 and 15 is 5 , and a g.c.d of 5 and 7 is 1.
We will show you that if the g.c.d of two elements exists, it is unique up to units, i.e., if d and d
are two g.c.ds of a and b, then d=ud , for some unit u.
So now let us prove the following result.
Theorem 3: Let R be an integral domain and a, b R. If a g.c.d of a and b exists, then it is unique
up to units.
Proof: So, let d and d be two g.c.ds of a and b. Since d is a common divisor and d is a g.c.d, we
get d | d . Similarly, we get d|d. Thus, we see that d and d are associates in R. Thus, the g.c.d of
a and b is unique up to units.
Theorem 3 allows us to say the g.c.d instead of a g.c.d. We denote the g.c.d of a and b by (a,b).
(This notation is also used for elements of R × R. But there should be no cause for confusion. The
context will clarify what we are using the notation for.
How do we obtain the g.c.d of two elements in practice? How did we do it in Z? We looked at the
common factors of the two elements and their product turned out to be the required g.c.d.
We will use the same method in the following example.
Example: In Q[x] find the g.c.d of
p(x) = x + 3x 10 and
2
q(x) = 6x 10x 4
2
Solution: By the quadratic formula, we know that the roots of p(x) are 2 and 5, and the roots of
q(x) are 2 and 1/3.
Therefore, p(x) = (x 2) (x + 5) and q(x) = 2(x 2) (3x + 1).
The g.c.d of p(x) and q(x) is the product of the common factors of p(x) and q(x), which is (x 2).
Let us consider the g.c.d of elements in a PD.
Theorem 4: Let R be a PID and a, b R. Then (a, b) exists and is of the form ax + by for some x,y
R.
Proof: Consider the ideal <a, b>. Since R is a PID, this ideal must be principal also. Let d R such
that <a, b> = <d>. We will show that the g.c,d of a and b is d.
Since a <d>, d | a, Similarly, d | b.
Now suppose c R such that c | a and c | b.
Since d E <a,b>, 3 x, y R such that d = ax+by.
Since c | a and c | b, c | (ax+by), i.e., c | d.
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