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![Prime Ideal • Definition ○ Suppose R is commutative ○ An ideal P⊊R is prime if ○ a,b∈R,ab∈P⇒a∈P or b∈P • Example ○ Take R=Z ○ The prime ideals are ideals of the form (n), where n is prime or n=0 ○ Proof (⟹) § Let (n)⊆Z be a prime ideal, and n≠0 § We want to show that n is prime § Choose a,b∈Z s.t. n=ab § Then ab∈(n), so either a∈(n) or b∈(n) § Without loss of generality, suppose a∈(n) § Then ├ n┤ |a┤ § Choose q∈Z s.t. nq=a § n=ab⇒n=nqb⇒1=qb⇒b∈{±1} § So n is a prime ○ Proof (⟸) § (0) is prime □ Let a,b∈Z, and ab∈(0) □ Then ab=0 □ ⇒a=0 or b=0 □ ⇒a∈(0) or b∈(0) □ Therefore (0) is prime § (p) is prime for p∈Z prime □ Let a,b∈Z, and say ab∈(p) □ Then p|ab □ Since p is prime, this means p|a or p|b □ ⇒a∈(p) or b∈(p) Proposition 73: Criterion for Prime Ideal • Statement ○ Let R be a commutative ring, P⊆R an ideal, then ○ P is prime ⇔ R\/P is a domain ○ In particular, R is a domain ⇔ 0 ideal is prime • Proof (⟹) ○ Let a+P,b+P∈(R/P)∖\{P} ○ Then (a+P)(b+P)=ab+P=0 ○ So, ab∈P ○ Since P is prime, a∈P or b∈P ○ Therefore a+P=0 or b+P=0 • Proof (⟸) ○ Let a,b∈R, and suppose ab∈P, then ○ 0=ab+P=(a+P)(b+P) ○ Since R\/P is a domain, a+P=0 or b+P=0 ○ So a∈P or b∈P • Example ○ (x^2−1)⊆R[x] is not prime ○ R[x]\/(x^2−1)≅R×R, which is not a domain ○ Also, x^2−1∈(x^2−1), but x−1,x+1∉(x^2−1) Corollary 74 • Statement ○ If R is a commutative ring, and M⊆R is maximal, then M is prime • Proof ○ M is maximal ⇒R\/M is a field ⇒R\/M is a domain ⇒ M is prime Euclidean Domain • Definition ○ Let R be a domain ○ A norm on R is a function N:R→Z(≥0) s.t. N(0)=0 ○ R is called a Euclidean domain if R is equipped with a norm N s.t. ○ ∀a,b∈R with b≠0, ∃q,r∈R s.t. § a=qb+r § Either r=0 or N(r) N(b) • Example 1 ○ Z is a Euclidean domain, N(a)=|a| • Example 2 ○ If F is a field, then F is trivially a Euclidean domain ○ Take N:F→Z(≥0) to be any function s.t. N(0)=0 ○ Then, if a,b∈F, where b≠0, take q=a/b,r=0 • Example 3 ○ If F is a field, then F[x] is a Euclidean domain, with N(p)=degp ○ The division algorithm have is just polynomial division ○ Note § deg0=−∞∉Z(≥0), so this definition isn t quite right § To handle this problem, define a norm to take vales not in Z(≥0) § But any total ordered set in order-preserving bijection with Z(≥0) § (For instance, Z(≥0)∪{−∞}) Proposition 75 • Statement ○ Every ideal in a Euclidean domain R is principal ○ More precisely, if I⊆R is an ideal, then I=(d), where ○ d is an element of I with minimum norm • Proof ○ Let I⊆R be an ideal ○ If I=(0), then I is principal, so assume I≠(0) ○ {N(a)│a∈I∖{0} } has a minimal element, by well-ordering principal ○ Choose d∈I∖{0} s.t. N(d) is minimal ○ Certainly, (d)⊆I ○ Let a∈I, write a=qd+r, where q,r∈R and either r=0 or N(r) N(d) ○ Since r=a−qd∈I, N(r) can t be smaller than N(d) ○ So r=0⇒a=qd⇒a∈(d) • Example 1 ○ We haven t yet proven that F[x] is a Euclidean domain, when F is a field ○ Once we show this, we ll know F[x] has the property that ○ All of its ideals are principal • Definition ○ A domain in which every ideal is principal is called a principal ideal domain • Example 2 ○ Z[x] cannot be a Euclidean domain, since (2,x)⊆Z[x] is not principal Theorem 76 • Statement ○ Let F be a field ○ Then F[x] is a Euclidean domain ○ More specifically, if a,b∈F[x] where b≠0, then ○ ∃!q,r∈F[x] s.t. a=bq+r and degr degb • Proof (Existence) ○ We argue by induction on dega ○ If a=0, take q,r=0, so assume a≠0 ○ Set n≔dega, m≔degb ○ If n m, then take q=0,r=a ○ Assume n≥m ○ Write a=a_n x^n+…+a_1 x+a_0,b=b_m x^m+…+b_1 x+b_0 ○ Set a^′=a−a_n/b_m x^(n−m) b ○ Since a and a_n/b_m x^(n−m) b have the same leading coefficient, dega′ dega ○ By induction, ∃q^′,r∈F[x] with a^′=q^′ b+r with degr degb ○ Set q=q^′+a_n/b_m x^(n−m) b ○ Then a=a^′+a_n/b_m x^(n−m) b=q^′ b+r+a_n/b_m x^(n−m) b=(q^′+a_n/a_m x^(n−m) )b+r=qb+r • Proof (Uniqueness) ○ Suppose bq^′+r^′=a=bq+r where degr degb, and degr′ degb ○ Then deg(a−bq) degb and deg〖(a−bq′) degb 〗 ○ ⇒deg((a−bq)−(a−bq^′ ))=deg(bq^′−bq)=degb+deg(q^′−q) degb ○ ⇒deg(q^′−q) 0⇒q^′=q ○ It follows immediately that r^′=r](https://shawnzhong.com/wp-content/uploads/2018/05/img_5aee501817d08.png)
