Page 205 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes So, if b  0, then (a + 5b ) | b, which is not possible.
2
2
 b = 0.
Thus, the only units of R are the invertible elements of Z.

Theorem 1: Let R be a Euclidean domain with Euclidean valuation d. Then, for any a  R \ {0},
d(a) = d(l) iff a is a unit in K.
Proof: Let us first assume that a  R\ {0] with d(a) = d(1).
By the division algorithm in R,  q, r  R such that 1 = aq+r,
where r = 0 or d(r) < d(a) = d(1).

Now, if r  0, d(r) = d(r.1)  d(1). Thus, d(r) < d(1) can’t happen.
Thus, the only possibility for r is r = 0,
Therefore, 1 = aq, so that a is a unit.
Conversely, assume that a is a unit in R. Let b  R such that ab = 1. Then d(a)  d(ab) = d(1). But
we know that d(a) = d(a.1)  d(1). So, we must have d(a) = d(1).

Using this theorem, we can immediately solve Example, since f(x) is a unit in F[x] iff deg f(x) =
deg (1 ) = 0.
Now let us look at the ideals of a Euclidean domain.
Theorem 2: Let R be a Euclidean domain with Euclidean valuation d. Then every ideal I % of R
is of the form I = Ra for some a  R.
Proof: If I = (01, then I = Ka, where a = 0. So let us assume that I  {0}. Then I\ {0} is non-empty.
Consider the set {d(a) | a  I \{0}). The well ordering principle this set has a minimal element.
Let this be d(b), where b e I \ {0}. We will show that I = Rb.
Since b  1 and I is an ideal of R,
Rb  I. ...(1)
Now take any a  I. Since I  R and R is a Euclidean domain, we can find q, r  R such that
a = bq + r, where r = 0 or d(r) < d(b).
Now, b  I  bq  I. Also, a  I. Therefore, r = a – bq  I.
But r = 0 or d(r) < d(b), The way we have chosen d(b), d(r) < d(b) is not possible.

Therefore, r = 0, and hence, a = bq  Rb.
Thus, I  Rb. ...(2)
From (1) and (2) we get
I = Rb.
Thus, every ideal I of a Euclidean domain R with Euclidean valuation d is principal, and is
generated by a  I, where d(a) is a minimal element of the set {d(x) | x  I \ (0) }.

Tasks 1. Show that every ideal of F[x] is principal, where F is a field.
2. Using Z as an example, show that the set
3. S = (a  R\ (0) | d(a) > d(1) }  (0) is not an ideal of the Euclidean domain R
with Euclidean valuation d.

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