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Paper 1 : ABSTRACT ALGEBRA

Chapter 7 : in Integral

Module 3 : Euclidean and principal Domains

Anjan Kumar Bhuniya Department of Mathematics Visva-Bharati; Santiniketan West Bengal

1 Euclidean and Domains

Learning Objectives: 1. Introduction to Euclidean domains and principal ideal domains. 2. Every Euclidean is a . 3. gcd in principal ideal domain. 4. Prime and irreducible elements coincide in a PID.

The existence of gcd, Euclid’s Lemma and Unique Theorem in Z and in F [x], where F is a field, all are consequences of the . In this , we consider integral domains having a .

Definition 0.1. Let E be an . A function v : E∗ −→ Z] is called a Euclidean on E if

(i) for all a, b ∈ E with a 6= 0, there exist q, r ∈ E such that b = aq + r, where either r = 0 or v(r) < v(a) and

(ii) v(a) ≤ v(ab), for all a, b ∈ E∗.

An integral domain E together with a valuation v on E is called a Euclidean domain. It is denoted by (E, v).

Example 0.2. The Z of all can be considered as a Euclidean domain with the valuation v : Z∗ −→ Z] defined by v(a) = |a|, for all a 6= 0. Example 0.3. Consider ring F [x], where F is a field. Then F [x] is an integral domain. Define v : F [x] \{0} −→ Z] by

v(f(x)) = deg f(x), for all f(x) ∈ F [x] \{0}.

Let f(x), g(x) ∈ F (x), g(x) 6= 0. Since F is a field, so every nonzero element in F and hence the leading coefficient of g(x) is a unit. It follows, by the Division Algorithm for , that there exist unique q(x), r(x) ∈ F [x] such that f(x) = q(x)g(x) + r(x), where either r(x) = 0 or deg r(x) < deg g(x). Hence, we have

f(x) = q(x)g(x) + r(x), where either r(x) = 0 or v(r(x)) < v(g(x)).

Since there is no zerodivisors in F , so for any two nonzero elements f(x) and g(x) in F [x], we have deg(f(x)g(x)) = deg f(x) + deg g(x) ≥ deg f(x), that is v(f(x)g(x)) ≥ v(f(x)). Hence, F [x] is a Euclidean domain.

2 Now we show that every field is a Euclidean domain. In search of a suitable Euclidean valuation on a field, we first prove the following result:

Theorem 0.4. Let E be a Euclidean domain with the valuation v. Then for every a ∈ E∗, v(a) = v(1) if and only if a is a unit in E.

Proof. For all a ∈ E∗, v(a) = v(1a) ≥ v(1). Suppose a is a unit. Then there exists an element b ∈ E such that ab = 1 which implies that v(1) = v(ab) ≥ v(a). Hence v(a) = v(1). Conversely, suppose that v(a) = v(1) for every a ∈ E∗. Now a 6= 0 implies that there exist q, r ∈ E such that 1 = qa + r, where r = 0 or v(r) < v(a) = v(1). Since v(r) < v(1) is impossible, we have r = 0. Thus 1 = qa and hence a is a unit.

Hence, if it is possible to define a Euclidean valuation v on a field F , then v(a) = v(1) for every a ∈ F ∗, since every nonzero element of F is a unit. Thus v is a constant mapping.

Example 0.5. Let F be a field. Then v : F ∗ −→ Z] given by:

v(a) = 2, for every a ∈ F ∗ is a Euclidean valuation. (Note that image of v may be any nonnegative .)

Thus C, R and Q all are Euclidean domain. Every ideal of Z is a principal ideal. Now we consider the integral domains such that every ideal is a principal ideal.

Definition 0.6. An integral domain R is called a principal ideal domain (PID) if every ideal of R is a principal ideal.

Thus Z is a PID. Also every field is a PID.

Theorem 0.7. Every Euclidean domain is a principal ideal domain.

Proof. Let (E, v) be a Euclidean domain. Consider an ideal I of E. If I = {0}, then I =< 0 >. Let I 6= {0}. Then N = {v(x) | x ∈ I, x 6= 0} is a nonempty set of nonnegative integers; and so, by the well-ordering principle, it has the least element. Let a ∈ I, a 6= 0 be such that v(a) is the least element of N, i.e. v(a) ≤ v(x) for all x ∈ I, x 6= 0. We show that I = Ea. Since a ∈ I, it follows that Ea ⊆ I. Let b ∈ I. Since E is a Euclidean domain, there exist q, r ∈ E such that b = aq +r, where r = 0 or v(r) < v(a). If r 6= 0, then r = b−aq ∈ I shows that v(r) ∈ N; and since v(r) < v(a) this contradicts the minimality of v(a) in N. Therefore, r = 0 and so b = aq ∈ Ea. Thus I ⊆ Ea, and hence I = Ea.

Now we characterize the polynomial rings which are Euclidean domains.

3 Theorem 0.8. Let F be a with 1. Then the following conditions are equivalent:

1. F is a field;

2. F [x] is a Euclidean domain;

3. F [x] is a principal ideal domain.

Proof. (1) ⇒ (2) : Follows from Example 0.3. (2) ⇒ (3) : Follows from Theorem 0.7. (3) ⇒ (1) : First note that F [x] is an integral domain, since F is so. Let a ∈ F be a nonzero element of F . Consider I =< a, x >, the ideal of F [x] generated by a and x. Since F [x] is a principal ideal domain, there exists u(x) ∈ F [x] such that I =< u(x) >= {u(x)f(x) | f(x) ∈ F [x]}. Now a, x ∈< u(x) > implies that there exist g(x), h(x) ∈ F [x] such that u(x)g(x) = a and u(x)h(x) = x. Since F is an integral domain, u(x)g(x) = a shows that u(x) is nonzero and deg u(x) = 0 which implies that u(x) = b ∈ F . Again, bh(x) = x implies that bc = 1 for some c ∈ F . Thus b is a unit and so I =< b >= F [x]. Then 1 ∈ I which implies that 1 = af1(x) + xf2(x) for some f1(x), f2(x) ∈ F [x]. This implies that 1 = da for some d ∈ F . Thus a is a unit in F and hence F is a field.

Since Z is not a field, It follows that Z[x] is not a principal ideal domain. In the following example we give an example of an ideal of Z[x], which is not principal.

Example 0.9. In Z[x], we show that the ideal < x, 2 > is not a principal ideal. On the contrary, if possible, assume that < x, 2 > is a principal ideal and < x, 2 >=< u(x) >, u(x) ∈ Z[x]. Now 2 ∈< u(x) > implies that 2 = u(x)f(x) for some f(x) ∈ Z[x]. Since Z is an integral domain, so deg u(x) = 0 and u(x) = a ∈ Z. Since x ∈< a >, there is g(x) ∈ Z[x] such that x = ag(x). It follows that ab = 1 for some b ∈ Z, and so 1 ∈< a >=< x, 2 >. Hence there are h(x), k(x) ∈ Z[x] such that 1 = xh(x) + 2k(x) which implies 1 = 2c for some c ∈ Z, a contradiction. Therefore < x, 2 > is not a principal ideal.

In Module 2 of Chapter 7, we proved that gcd of any two elements not both zero exists in a principal ideal ring. Since every Euclidean domain is a principal ideal ring, we have

Theorem 0.10. Let R be a Euclidean domain (principal ideal domain) and a, b ∈ R not both zero. Then a and b have a gcd d. For every gcd d of a and b, there exist s, t ∈ R such that d = sa + tb. a, b are relatively prime if and only if there exist s, t ∈ R such that 1 = sa + tb.

Now we show that the irreducible elements and the prime elements coincide in a PID.

Theorem 0.11. Let R be a principal ideal domain and p ∈ R. Then p is irreducible if and only if it is prime.

4 Proof. Already we have proved that every prime element is irreducible in an integral domain. (Module 2, Chapter 7) Suppose that p is an irreducible element in R. Consider a, b ∈ R and assume that p | ab. Then ab = pc for some c ∈ R. Since R is a principal ideal ring, there is d ∈ R such that < p, b >=< d >. Then there exists q ∈ R such that p = dq. Since p is irreducible, either d or q must be a unit. If d is a unit, then < p, b >=< d >= R. Hence 1 ∈< p, b >, and so 1 = sp + tb for some s, t ∈ R. This implies that a = asp+atb = asp+tcp = (as+tc)p. Thus p | a. If q is a unit, then d = q−1p ∈< p >. This implies that < d >⊆< p >⊆< p, b >=< d >. Hence < p >=< p, b > and so b ∈< p >. Thus p | b.

Recall that if F is a field then a polynomial p(x) is irreducible if and only if F [x]/ < p(x) > is a field. Hence p(x) is irreducible if and only if < p(x) > is a in F [x]. We show that this result holds in every PID.

Theorem 0.12. Let R be a principal ideal domain. Then M is a maximal ideal of R if and only if M =< q > for some irreducible element q ∈ R.

Proof. Already we have proved that if M =< q > is a maximal ideal then q is an irreducible element. (Module 2, Chapter 7) Conversely suppose that q is an irreducible element and M =< q >. Consider an ideal I of R such that M ⊆ I ⊆ R. Since R is a principal ideal domain, there exists a ∈ R such that I =< a >. Now q ∈ M ⊆< a > shows that q = ab for some b ∈ R. Irreducibility of q implies that either a or b is a unit. If a is a unit then I =< a >= R. If b is a unit then a = qb−1 ∈< q >= M. This implies that < a >⊆ M and hence M = I. Thus M is a maximal ideal of R.

If a ring R becomes a PID then primality and irreducibility are no longer different, and as a consequence we have the following important corollary.

Corollary 0.13. Let R be a principal ideal domain. Then a nonzero proper ideal P of R is prime if and only if it is maximal.

1 Summary

• Let E be an integral domain. A function v : E∗ −→ Z] is called a Euclidean valuation on E if (i) for all a, b ∈ E with a 6= 0, there exist q, r ∈ E such that b = aq + r, where either r = 0 or v(r) < v(a) and (ii) v(a) ≤ v(ab), for all a, b ∈ E∗.

An integral domain E together with a valuation v on E is called a Euclidean domain. It is denoted by (E, v).

5 • The ring Z of all integers can be considered as a Euclidean domain with the valuation v : Z∗ −→ Z] defined by v(a) = |a|, for all a 6= 0.

• Every field is a Euclidean domain.

Thus C, R and Q all are Euclidean domain.

• If F is a field, then F [x] is a Euclidean domain with the valuation v : F [x]\{0} −→ Z] defined by: v(f(x)) = deg f(x), for all f(x) ∈ F [x] \{0}.

• Let E be a Euclidean domain with the valuation v. Then for every a ∈ E∗, v(a) = v(1) if and only if a is a unit in E.

• An integral domain R is called a principal ideal domain (PID) if every ideal of R is a principal ideal.

• Every Euclidean domain is a principal ideal domain.

• Let F be a commutative ring with 1. Then the following conditions are equivalent:

(i) F is a field; (ii) F [x] is a Euclidean domain; (iii) F [x] is a principal ideal domain.

• Z[x] is not a principal ideal domain, since Z is not a field.

• Let R be a Euclidean domain (principal ideal domain) and a, b ∈ R not both zero. Then a and b have a gcd d. For every gcd d of a and b, there exist s, t ∈ R such that d = sa + tb. a, b are relatively prime if and only if there exist s, t ∈ R such that 1 = sa + tb.

• Let R be a principal ideal domain and p ∈ R. Then p is irreducible if and only if it is prime.

• Let R be a principal ideal domain. Then M is a maximal ideal of R if and only if M =< q > for some irreducible element q ∈ R.

• Let R be a principal ideal domain. Then a nonzero proper ideal P of R is prime if and only if it is maximal.

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