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**3.2 Principal Ideal Domains**

**Definition 3.2.1** A principal ideal domain (PID) is an integral domain in which every ideal is principal.

**Lemma 3.2.2** Z is a PID.

**NOTE:** Showing that Z is a PID means showing that if I is an ideal of Z, then there is some integer n for which I consists of all the integer multiples of n.

**Proof:** Suppose that I __âŠ‚__ Z is an ideal. If I = {0} then I is the principal ideal generated by 0 and I is principal. If I â‰ {0} then I contains both positive and negative elements. Let m be the least positive element of I. We will show that I = (m).

Certainly (m) __âŠ‚__ I as I must contain all integer mulitples of m. On the other hand suppose a âˆˆ I. Then we can write

a = mq + r

where q âˆˆ Z and Then r = a - qm. Since a âˆˆ I and -qm âˆˆ I, this means r âˆˆ I. It follows that r = 0, otherwise we have a contradiction to the choice of m. Thus a = qm and a âˆˆ (m). We conclude I = (m).

**Note:** In fact every subring of Z is an ideal - think about this.

**Lemma 3.2.3** Let F be afield. Then the polynomial ring F[x] is a PID.

**NOTE: **Recall that F[x] has one important property in common with Z, namely a division algorithm. This is the key to showing that F[x] is a PID.

**Proof:** Let I __âŠ‚__ F[x] be an ideal. If I = {0} then I = (0) and I is principal. If I â‰ {0}, let f(x) be a polynomial of minimal degree m in I. Then (f(x)) __âŠ‚__ I since every polynomial multiple of f(x) is in I.

We will show that I = (f (x)). To see this suppose g(x) âˆˆ I. Then

g(x) = f(x)q(x) + r(x)

where q(x),r(x) âˆˆ F[x] and r(x) = 0 or deg(r(x)) < m. Now

r(x) = g(x) â€” f(x)q(x)

and so r(x) âˆˆ I. It follows that r(x) = 0 otherwise r(x) is a polynomial in I of degree strictly less than m, contrary to the choice of f(x).

Thus g(x) = f(x)q(x), g(x) âˆˆ (f(x)) and I = (f(x)).

Question for the Seminar: If R is a ring (not a field) it is not always true that R[x] is a PID.

Find an example of a non-principal ideal in Z[x].

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