Ring Theory and Polynomials PDF

1.1 Definition Ring: A ring 𝜙 ≠ 𝑅 be a set equipped with two binary operations +  If we define a relation 𝑆 on 𝑅(𝐶𝑅𝑈) as: (𝑎, 𝑏) ∈ 𝑆 ⟺ and ∙ (Here ‘+’ & ′ ∙ ′ are not always addition and multiplication) 𝑎 and b are associates. Then 𝑅 is an equivalence relation 𝑠. 𝑡....

1.1 Definition Ring: A ring 𝜙 ≠ 𝑅 be a set equipped with two binary operations +  If we define a relation 𝑆 on 𝑅(𝐶𝑅𝑈) as: (𝑎, 𝑏) ∈ 𝑆 ⟺ and ∙ (Here ‘+’ & ′ ∙ ′ are not always addition and multiplication) 𝑎 and b are associates. Then 𝑅 is an equivalence relation 𝑠. 𝑡. on ring 𝑅 and hence it partitions the ring 𝑅 into  (𝑅, +) is an abelian group equivalence classes.  (𝑅,∙) is a semi group  ∀ 𝑎, 𝑏, 𝑐 ∈ 𝑅 (𝑎 + 𝑏) ∙ 𝑐 = (𝑎 ∙ 𝑐) + (𝑏 ∙ 𝑐) 1.8 Definition Zero Divisor: If (𝑅, +,∙) be a ring and 0 ≠ 𝑎 ∈ & 𝑎 ∙ (𝑏 + 𝑐) = (𝑎 ∙ 𝑏) + (𝑎 ∙ 𝑐) 𝑅 then 𝑎 is called zero divisor if ∃ 0 ≠ 𝑏 ∈ 𝑅 𝑠. 𝑡. 𝑎 ∙ 𝑏 = 0 = Then (𝑅, +,∙) is called a Ring 𝑏 ∙ 𝑎 ∀𝑎, 𝑏 ∈ 𝑅 Ex: (ℂ, +,. ), (ℝ, +,. ) ,(ℚ, +,. ) , (ℤ, +,. ) , (ℤ[𝑖], +,∙) 1.9 Definition Integral Domain: A 𝐶𝑅𝑈 without zero 1.1.1 Definition Trivial Ring: divisors is called an Integral domain. A ring (𝑅, +,∙) is said to be trivial if 𝑅 = {0} & ′0′ is the identity  Many authors don’t take ring to be commutative with element of (𝑅, +). unity. They consider only ring without zero divisors. 1.1.2 Definition Zero Ring: 1.10 Definition Field: An integral domain is a field if each If (𝑅, +) be an abelian group and if we define the second non-zero element possess multiplicative inverse. binary operation ′ ∙′ on 𝑅 𝑠. 𝑡. 𝑎 ∙ 𝑏 = 0 = 𝑏 ∙ 𝑎 ∀𝑎, 𝑏 ∈ 𝐺 where & ′0′ is the identity element of (𝑅, +), then (𝑅, +,∙) 1.11 Definition Idempotent Element: If 𝑅 be a ring and forms a ring and called zero ring. 𝑎 𝜖 𝑅, then ′𝑎’ is said to be an idempotent element if 𝑎2 = 𝑎.  If R is a trivial ring, then R is a zero ring. (Converse need 1.12 Definition Nilpotent Element: If 𝑅 be a ring and 𝑎 ∈ not be true) 𝑅, then a is said to be nilpotent element if ∃ 𝑛 ∈ ℕ 𝑠. 𝑡. 𝑎𝑛 = 0 𝑖. 𝑒. ⏟ 𝑎 ∙ 𝑎 ∙∙∙ 𝑎 = 0. 1.2 Definition Zero Element: 𝑛−𝑡𝑖𝑚𝑒𝑠 Let (𝑅, +,∙) be a ring as (𝑅, +) is a group then the identity of  The smallest such 𝑛 is defined as the index of nilpotence. this group is called zero of ring and usually denoted by 0 or 0𝑅 and 𝑎 + 0 = 0 + 𝑎 = 𝑎 ∀𝑎 ∈ 𝑅 1.13 Definition Characteristic of Ring: The smallest positive integer 𝑛 is said to be characteristic of a ring 𝑅 if 1.3 Definition Unity Element: If (𝑅,. ) is a monoid 𝑖. 𝑒. ∃ an 𝑖𝑓 ⏟ 𝑎 + 𝑎+.. 𝑎 = 0 𝑖. 𝑒. 𝑛𝑎 = 0 ∀ 𝑎 ∈ 𝑅 If no such positive identity element in 𝑅 𝑤. 𝑟. 𝑡. ′ ∙′ then this identity element is 𝑛 𝑡𝑖𝑚𝑒 called the unity of the ring and usually denoted by 1 𝑜𝑟 1𝑅 integer exists, then characteristic of ring is 0. It is denoted by where 1 𝑜𝑟 1𝑅 ≠ 0 and ∙ 1 = 𝑎 = 1 ∙ 𝑎 ∀𝑎 ∈ 𝑅. 𝑐ℎ𝑎𝑟𝑅. 1.4 Definition Commutative Ring (CR): Let (𝑅, +,∙) be a 1.14 Definition Factor: If 𝑅 be a 𝐶𝑅𝑈 and 𝑎, 𝑏 ∈ 𝑅, then 𝑏 is ring and ′ ∙ ′ is a commutative B.O. then 𝑅 is called a said to be factor of 𝑎 if ∃ 𝑐 ∈ 𝑅 𝑠. 𝑡. 𝑏 = 𝑎𝑐 commutative ring  If 𝑎 ∈ 𝑅 then the associates of ′𝑎′ and units in 𝑅 are always the factor of ′𝑎′ and called improper factor & the other 1.5 Definition Ring with Unity (RU): If (𝑅, +,∙) be a ring and factors are called proper factor. unity element then 𝑅 is called a ring with unity 1.15 Definition Irreducible Element: If 𝑅 be a 𝐶𝑅𝑈 then a 1.6 Definition Commutative Ring with Unity (CRU): non-zero, non-unit element ′𝑎′ is said to be irreducible (𝑅, +,∙) be a commutative ring and has unity element then 𝑅 element if it has no proper factor in 𝑅. is called a commutative ring with unity 𝑖. 𝑒., whenever 𝑎 = 𝑏𝑐, then either 𝑏 is unit or 𝑐 is unit. 1.7 Definition Unit Element:  An element which is not irreducible is called reducible If (𝑅, +,∙) be a ring with unity and 0 ≠ 𝑎 ∈ 𝑅. Then 𝑎 is said element. to be unit element of 𝑅 if ∃ 0 ≠ 𝑏 ∈R such that 𝑎 ∙ 𝑏 = 𝑏 ∙ 𝑎 = 1𝑅. 1.16 Definition Prime Element: If 𝑅 is a 𝐶𝑅𝑈 then a non- zero, non-unit element 𝑝 ∈ 𝑅 is called prime element if 𝑎, 𝑏 ∈ 1.7.1 Definition Associate: If 𝑅 be a 𝐶𝑅𝑈 and 𝑎, 𝑏 ∈ 𝑅. Then a 𝑅 and 𝑝|𝑎𝑏 implies either 𝑝|𝑎 or 𝑝|𝑏. is called associate of b if ∃ a unit 𝑢 in 𝑅 𝑠. 𝑡. 𝑎 = 𝑏𝑢 2.1 Division Algorithm: If 𝑅 be a field and 𝑓(𝑥), 𝑔(𝑥) ∈ 𝑅[𝑥] with 𝑔(𝑥) ≠ 0. Then ∃ Important test(s) for irreducibility: unique polynomials 𝑞(𝑥)𝑎𝑛𝑑 𝑟(𝑥)𝑖𝑛 𝑅[𝑥] such that 𝑓(𝑥) =  Polynomials of degree 1 are irreducible over field 𝔽 𝑔(𝑥)𝑞(𝑥) + 𝑟(𝑥) 𝑠. 𝑡 either 𝑟(𝑥) = 0 or deg 𝑟(𝑥) < deg 𝑔(𝑥)  If a polynomial 𝑓(𝑥) is of degree > 1 and 𝑓(𝑎) = 0 for some 𝑎 ∈ 𝐹, Then 𝑓(𝑥) is reducible over 𝔽, where 𝔽 is a  If 𝔽 be a field, 𝑎 ∈ 𝔽 and 𝑓(𝑥) ∈ 𝔽[𝑥], then 𝑓(𝑎) is the field. remainder in the division of 𝑓(𝑥) by 𝑥 − 𝑎.  Let 𝔽 be a field if 𝑓(𝑥) 𝔽[𝑥] and 𝑑𝑒𝑔 𝑓(𝑥) = 2 or 3 then  If 𝔽 be a field, 𝑎 ∈ 𝔽 and 𝑓(𝑥) ∈ 𝔽[𝑥]. Then ′𝑎′ is zero of 𝑓(𝑥) is reducible over 𝐹 if and only if ∃ 𝑎 ∈ 𝐹 𝑠. 𝑡. 𝑓(𝑥) 𝑓(𝑥) 𝑖𝑓𝑓 𝑥 − 𝑎 is a factor of 𝑓(𝑥) has a zero in 𝔽. Ex: 2.2 Definition Greatest Common Divisor (GCD):  If 𝑓(𝑥) ∈ ℤ[𝑥] 𝑡ℎ𝑒𝑛 𝑓(𝑥) is irreducible over, ℤ 𝑖𝑓𝑓 it is Let 𝑅 be a commutative ring and 𝑓(𝑥), 𝑔(𝑥) be two non-zero irreducible over ℚ. elements of 𝑅[𝑥] then a non-zero element 𝑑(𝑥) ∈ 𝑅[𝑥] is Ex: called GCD of 𝑓(𝑥) and 𝑔(𝑥)if  Mod 𝑝 irreducibility Test: Let 𝑝 be a prime and suppose  ℎ(𝑥)/𝑓(𝑥), ℎ(𝑥)/𝑔(𝑥)and that 𝑓(𝑥) ∈ ℤ[𝑥] with 𝑑𝑒𝑔 𝑓 ≥ 1. Let ̅̅̅̅̅̅̅ 𝑓 (𝑥) be the  Whenever we have 0 ≠ ℎ(𝑥) ∈ 𝑅[𝑥]such that ℎ(𝑥)/ polynomial in ℤ𝑝 [𝑥] obtained from 𝑓(𝑥) by reducing all 𝑓(𝑥) & ℎ(𝑥)/𝑔(𝑥) then ℎ(𝑥)/𝑑(𝑥). ̅̅̅̅̅̅ is the co-efficients of 𝑓(𝑥) under modulo 𝑝 then if 𝑓(𝑥)  GCD of 𝑓(𝑥) & 𝑔(𝑥) is denoted by It is denoted by ̅̅̅̅̅̅ irreducible over ℤ𝑝 , and deg 𝑓(𝑥) = 𝑑𝑒𝑔𝑓(𝑥), then 𝑓(𝑥) (𝑓(𝑥), 𝑔(𝑥)) = ℎ(𝑥) is irreducible over ℤ 𝑎𝑛𝑑 ℚ. (Converse need not be true) 2.3 Definition Least Common Multiple (LCM): Ex: Let 𝑅 be a commutative ring and 𝑓(𝑥), 𝑔(𝑥) be two non-zero  Let 𝔽 be a field and 𝑎 ∈ 𝔽 and 𝑎 ≠ 0 elements of 𝑅[𝑥]. A non-zero element 𝑙(𝑥) ∈ 𝑅[𝑥] is called - 𝑓(𝑥) is irreducible over 𝔽 then 𝑎𝑓(𝑥) is irreducible over 𝔽 LCM of 𝑓(𝑥) and 𝑔(𝑥) if - 𝑓(𝑥) is irreducible over 𝔽 then 𝑓(𝑎𝑥) is irreducible over 𝔽  𝑓(𝑥)/𝑙(𝑥), 𝑔(𝑥)/𝑙(𝑥) - 𝑓(𝑥) is irreducible over 𝔽 then 𝑓(𝑥 + 𝑎) is irreducible  Whenever we have 0 ≠ ℎ(𝑥) ∈ 𝑅[𝑥] such that 𝑓(𝑥)/ℎ(𝑥) over & 𝑔(𝑥)/ℎ(𝑥). Then 𝑙(𝑥)/ ℎ(𝑥),.  Eisenstein's Criterion: Let 𝑝(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑛−1  LCM of 𝑓(𝑥) & 𝑔(𝑥) is denoted by [𝑓(𝑥), 𝑔(𝑥)] = 𝑙(𝑥) 𝑎𝑛−1 𝑥 +... +𝑎0 ∈ ℤ[𝑥] if ∃ a prime 𝑝 such that 𝑝 ∤ 𝑎𝑛 , 𝑝\𝑎𝑛−1 , 𝑝\𝑎𝑛−1 … 𝑝\𝑎0 𝑎𝑛𝑑 𝑝2 ∤ 𝑎0 then 𝑝(𝑥) is 2.4 Definition Irreducible Polynomial: irreducible over ℚ. Let 𝑅 be an integral domain. A polynomial 𝑓(𝑥) ∈ 𝑅[𝑥] of Ex: positive degree (𝑖. 𝑒. 𝑑𝑒𝑔 ≥ 1) is said to be an irreducible  If 𝑝 is a prime number, then the polynomial 𝑥 𝑛 − 𝑝 is an polynomial over 𝑅 if it can not be expressed as the product of irreducible over the rational numbers ℚ two polynomials of positive degree. 𝒊. 𝒆. if 𝑓(𝑥) = 𝑔(𝑥)ℎ(𝑥) then 𝑒𝑖𝑡ℎ𝑒𝑟 𝑑𝑒𝑔(𝑔) = 0 𝑜𝑟 𝑑𝑒𝑔(𝑓) = 0  A polynomial of positive degree which is not irreducible is called reducible over 𝑅  If 𝑅 is an integral domain & 𝑝(𝑥) ∈ 𝑅[𝑥] then - 𝑝(𝑥) is an irreducible element ⟹ 𝑝(𝑥) is an irreducible polynomial (converse need not be true) Ex: 𝑝(𝑥) = 2𝑥² + 2 ∈ ℤ[𝑥]  If 𝔽 is a field and 𝑝(𝑥) ∈ 𝔽[𝑥] is an irreducible polynomial of 𝔽[𝑥] ⇔ 𝑝(𝑥) ∈ 𝔽[𝑥] is an irreducible element  If 𝑝 is prime number, then the number of reducible 𝑝(𝑝+1) polynomials of the form 𝑥² + 𝑎𝑥 + 𝑏 over ℤ𝑝 are. 2  If 𝑝 is prime number. Then number of irreducible 𝑝(𝑝−1) polynomials over ℤ𝑝 of the form 𝑥² + 𝑎𝑥 + 𝑏 are 2 3.1 Subring & Ideal: 3.1.8 Ideal Test: 3.1.1. Subring: A Non-empty subset A of a ring R is an ideal of R if Let (𝑅, +,∙) be a ring. Then non-empty subset S of R is called a (a) 𝑎 − 𝑏 𝜖 𝐴 whenerver 𝑎, 𝑏 ∈ 𝐴 subring of R, if (𝑆, +,. ) is a ring. (b) 𝑟𝑎 & 𝑎𝑟 ∈ 𝐴 whenever 𝑎 ∈ 𝐴 & 𝑟 ∈ 𝑅 3.1.2. Subring Test: Note: 𝑆 ≠ 𝜙 & 𝑆 ⊆ 𝑅 then S is subring of R if and only if ∀ 𝑎, 𝑏, 𝜖 𝑆  Let R be a ring then subring R and {0} are ideal of R and (i) 𝑎𝑏 𝜖 𝑆 called improper ideal of R (ii) 𝑎 − 𝑏 ∈ 𝑆  An ideal 𝐼 ≠ 𝑅 of a ring R is called proper ideal  Many authors consider {0} as proper ideal and many Ex: ℤ , ℚ, ℚ(√2) are subring of ℝ improper ideal. (i) (ℤ8 ,⊕6 , ⨀ 6) is not subring of (ℤ16 ,⊕12 , ⨀ 6) Ex: (ii) 𝑆 = {𝑎 + 𝑏𝑖 + 𝑐𝑗 + 𝑑𝑘; 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℤ } & 𝑅 = {𝑎 + 𝑏𝑖 + (i) {0} & R are always ideals of R & called Trivial ideals 𝑐𝑗 + 𝑑𝑘; 𝑎, 𝑏, 𝑐, 𝑑 ∈ ℚ} where S is set of integral (ii) 𝑀𝑛 (𝐾ℤ) is ideal of 𝑀𝑛 (ℤ)∀ 𝑘 ∈ ℤ quaternions & R is set of rational quaternions then S is (iii) Let 𝑅 be ring of all real valued functions of a real variable. subring of R. Then the subset S of all differentiable functions is a (iii) 𝑆 = {(𝑟, 𝑟)|𝑟 𝜖 𝑅} is subring of R×R where R is a ring. subring of R but not an ideal of R. (iv) Intersection of two subring of R is a subring of R (iv) 𝐴 = {(𝑝𝑥, 𝑦)| 𝑥, 𝑝 𝜖 ℤ} is ideal of ℤ × ℤ where 𝑝 is prime. Some Important Observations: Some Important properties or results related of ideals:  Subring of a commutative ring is commutative.  In a commutative ring R every left ideal or right ideal is  Subring of a ring without zero devisor is also without zero ideal of R divisors.  Every ideal of a ring R is a subring of R but converse need  There exist rings with unity 1 having a subring with unity not be true. not equal to 1.  Arbitrary intersection of ideals of R is an ideal of R.  There exist rings with unity having a subring without unity.  Union of two ideals of a ring R need not be an ideal of R.  There exist rings without unity having subring with unity.  The sum of two ideals of a ring R is an ideal of R i.e. if I and  If S and T are two subring of a ring R, then their sum is J are two ideals of a ring R, then 𝐼 + 𝐽 = {𝑎 + 𝑏: 𝑎 ∈ 𝐼, 𝑏 ∈ defined as 𝑆 + 𝑇 = {𝑎 + 𝑏: 𝑎 ∈ 𝑆, 𝑏 ∈ 𝑇} may not be a 𝐽} is an ideals of 𝑅. subring.  If I and J are two ideals of a ring R, then their product IJ  Subring of integral domain will be integral domain. defined  If R be a ring with char 𝑅 = 𝑛 then 𝑀2 (𝑅) has  As {𝐼𝐽 = 𝑎₁𝑏₁ + 𝑎₂𝑏₂ +... + 𝑎𝑛 𝑏𝑛: 𝑎𝑖 ∈ 𝐼, 𝑏¡ ∈ 𝐽, 1 ≤ characteristic n. 𝑖 ≤ 𝑛 is an ideal of R and n being a positive integer}  If S is a subring of a ring R then.  If 𝐼 and 𝐽 are two ideals of a ring 𝑅, then 𝐼𝐽 ⊆ 𝐼 + 𝐽.  If char S & char R are finite then char 𝑆 ≤ 𝑐ℎ𝑎𝑟 𝑅  Let R be a ring with unity 1 and I be an ideal of R. If 1∈ I  If S & R have same unity, then char 𝑆 = 𝑐ℎ𝑎𝑟 𝑅. then I = R  Let R be a ring with unity 1 and I be an ideal of R. If u ∈ I 3.1.5 Left Ideal: then 𝐼 = 𝑅, where u is any unit element of R. Let R be a ring and I be a subset of R then I is said to be left  Let F be a field then {0} and F are only ideals of F ideal of R if I is subring of R and for each 𝑟 ∈ 𝑅 and 𝑎 ∈ 𝐼, 𝑟𝑎 ∈  Let R be a ring then Centre of ring 𝑍(𝑅) is subring of 𝑅 but 𝐼 it need not be ideal of R. 3.1.6. Right Ideal: Let R be a ring and I be subset of R then I is said to be right ideal  If 𝑅 is a commutative ring then of R if I is subring of R and for each 𝑎 𝜖𝐼, 𝑟𝜖𝑅, 𝑎𝑟𝑒 𝐼 𝑁(𝑅) = {𝑎|𝑎 ∈ 𝑅 & 𝑎𝑚 = 0 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑚 ≥ 1} is an ideal in 𝑅 & called nil-radical of 𝑅 3.1.7. Ideal:  If 𝑅 is a commutative ring then √𝐼 = {𝑎 ∈ A subset I of a ring R is said to be ideal if I is both left ideal and 𝑛 𝑅|𝑎 𝜖 𝐼 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑛 ≥ 1} right ideal. Note: If 𝐼 = 0, then √𝐼 is the Nil-radical.  𝑁(ℤ𝑛 ) i.e nil-radical of ℤ𝑛 is non-zero iff 𝑛 is divisible by  Let 𝐼 be an ideal of ring 𝑅 and 𝑆 be subring of 𝑅. Then 𝐼 ∩ square of a prime. 𝑆 is ideal of 𝑆  Any ideal of ℤ × ℤ is of from 𝑚ℤ × 𝑛ℤ; 𝑚, 𝑛 𝜖 ℤ  Let 𝑍 be ring of integers and 〈𝑚〉, 〈𝑛〉 be ideals of ℤ. Then  If 𝑅 is commutative ring , 𝑎1 , 𝑎2 …. 𝑎𝑛 ∈ 𝑅 then (i) 〈𝑚〉 + 〈𝑛〉 = 〈𝑎〉, where a is 𝐺𝐶𝐷 of 𝑚, 𝑛 𝐼 = 𝑅𝑎1 + 𝑅𝑎2 + ⋯ + 𝑅𝑎𝑛 is an ideal in 𝑅. (ii) 〈𝑚〉 ∩ 〈𝑛〉 = 〈𝑏〉, where b is 𝐿𝐶𝑀 of 𝑚, 𝑛  𝑅 be commutative ring & 𝐴 ⊂ 𝑅. 𝐴𝑛𝑛(𝐴) = {𝑟 ∈ 𝑅|𝑟𝑎 =  (xi) Every ideal of 𝑀𝑛 (𝑅) ring of all (𝑛𝑥𝑛) matrix over ring 0 ∀ 𝑎 ∈ 𝐴} is an ideal. 𝑅, is of the form 𝑀𝑛(𝐽) where 𝐽 is an ideal of 𝑅. Note: In ℤ𝑛 , ∃ (non-zero) nilpotent element if and only if 𝑛 is 3.1.11. Maximal Ideal: divisible by square of a prime. An ideal 𝑀 in a ring 𝑅 is called maximal ideal of R if 𝑀 ≠ 𝑅 and the only ideals containing 𝑀 are 𝑀 and 𝑅 ie: if ∃ ideal 𝑈 of 𝑅 3.1.9 Ideal Genrated by a Set such that 𝑀 ⊆ 𝑈 ⊆ 𝑅. Then either 𝑀 = 𝑈 𝑜𝑟 𝑈 = 𝑅. Let 𝑆 be any subset of a ring 𝑅. An ideal 𝐼 𝑜𝑓 𝑅 is said to be Ex: generated by 𝑆 if (i) 𝐼 = {(𝑝𝑥, 𝑦)|𝑥, 𝑦𝑒 ℤ} is maximal ideal of ℤ × ℤ where p is (i) 𝑆 ⊆ 𝐼 prime. (ii) If 𝐽 is any ideal of 𝑅 such that 𝑆 ⊆ 𝐽, then 𝐼 ⊆ 𝐽. (ii) 𝑝ℤ is maximal ideal of ℤ, where p is prime. We write ideal 𝐼 𝑎𝑠 𝐼 = 〈𝑆〉. Indeed 〈𝑆〉 is the smallest ideal 1 1 (iii) 𝐼 = {𝑓 ∈ 𝐶[0, 1]: 𝑓 ( ) = 0 = 𝑓 ( )} is not maximal 3 5 containing 𝑆 ideal. 1 3.1.10 Co-maximal Ideals (iv) 𝐼 = {𝑓 ∈ 𝐶[0, 1]: 𝑓 ( ) = 0} is maximal ideal. 3 Two ideals 𝐼 𝑎𝑛𝑑 𝐽 of a ring 𝑅 satisfying 𝐼 + 𝐽 = 𝑅 are called co-maximal ideals. 3.1.12. Prime Ideal: Results: Let 𝑅 be a commutative ring. An ideal 𝑃 is called a prime ideal (i) Let 𝐼 𝑎𝑛𝑑 𝐽 be any two ideals of a ring 𝑅. The 𝐼 + 𝐽 is an if 𝑃 ≠ 𝑅 and whenever the product 𝑎𝑏 of two elements 𝑎, 𝑏 ∈ Ideal Of 𝑅 generated by 𝐼 ∪ 𝐽. 𝑅 is an element of 𝑃, then at least one of them is an element (ii) If 𝐼 𝑎𝑛𝑑 𝐽 are two ideals of a ring 𝑅 then 𝐼 ∪ 𝐽 is an ideal of of 𝑃 i.e. whenever 𝑎𝑏 ∈ 𝑃 then either 𝑎 ∈ 𝑃 or 𝑏 ∈ 𝑃. 𝑅 iff either 𝐼 ⊆ 𝐽 or 𝐽 ⊆ 𝐼 Examples: (iii) Let 𝐼 𝑎𝑛𝑑 𝐽 be two ideal of a commutative ring 𝑅 with (i) 𝑝ℤ is prime ideal of ℤ. Where 𝑝 is prime number. unity such that 𝐼 + 𝐽 = 𝑅. Then 𝐼𝐽 = 𝐼 ∩ 𝐽. 1 1 (ii) 𝐼 = {𝑓 ∈ 𝐶[0, 1]: 𝑓 ( ) = 0 = 𝑓 ( )} is not prime ideal. 3 5 1 Some Important Results: (iii) 𝐼 = {𝑓 ∈ 𝐶[0,1]: 𝑓 ( ) = 0}is prime ideal. 2  Let 𝐼 be left ideal and 𝐽 is right ideal of a ring then 𝐼𝐽 is always an ideal of 𝑅 but 𝐽𝐼 may not be even one sided ideal. Results:  The intersection of two left (right) ideals of a ring 𝑅 is a  Let 𝑅 be commutative ring. Then 𝑅 is field if and only if {0} left (right) ideals of 𝑅. is maximal ideal of 𝑅.  The intersection of a left ideal and right ideal of a ring R  An ideal 𝑀 = 𝑛ℤ = {𝑛𝑥: 𝑥 ∈ ℤ} is maximal ideal of ℤ if may not be even a one sided ideal of R. and only if n is prime number.  The sum of two left (right) ideals of a ring 𝑅 is left (right)  Let 𝑅 be a ring of all real-valued continuous function on ideal of 𝑅. the closed interval [a,b]. Then for 𝑐 ∈ (𝑎, 𝑏), 𝑀 = {ƒ ∈  The sum of a left ideal and a right ideal of a ring 𝑅 may not 𝑅: ƒ(𝑐) = 0} is maximal ideal of 𝑅. ideal of 𝑅.  Let 𝑅 be a commutative ring with unity. Then an ideal 𝑀 𝑅  There exist ideals 𝐼 and 𝐽 of a ring 𝑅 such that 𝐼 ⊆ 𝐽 ⊆ 𝑅 of 𝑅 is maximal if and only if is a field. 𝑀 where 𝐼 is ideal of 𝐽 and 𝐽 is ideal of 𝑅 but. 𝐼 is not ideal of  Let 𝑅 be a commutative ring. Then an ideal 𝑀 of 𝑅 is 𝑅. 𝑅 maximal if and only hif is a simple ring. 𝑀  Let 𝑅 be a ring with unity and 𝐼 be proper ideal of 𝑅 then ℤ  For each prime number 𝑝, 〈𝑝〉 is a field. no element of 𝐼 can have multiplicative inverse.  Let 𝐼 be an ideal of ring 𝑅 such that 𝐼 ≠ 𝑅 and 𝑅 has unity  Let 𝑅 be a ring with unity. Then an ideal 𝑀 of 𝑅 is maximal 1 then 1 ∉ 𝐼 if and only if 𝑀 + 〈𝑎〉 = 𝑅 ∀𝑎 ∉ 𝑀.  Let 𝑅 be a commutative ring. Then the ideal 𝑃 of 𝑅 is prime = {𝑟 ∈ 𝑅|𝑟 − 𝑎 = 𝑥; 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥 ∈ 𝐼} 𝑅 ideal IN 𝑅 if and only if is an integral domain. = {𝑟 ∈ 𝑅|𝑟 = 𝑎 + 𝑥; 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑥 ∈ 𝐼} 𝑝 = {𝑎 + 𝑥|𝑥 ∈ 𝐼}  Let 𝑅 be a commutative ring with unity then every =𝑎+𝐼 maximal ideal of 𝑅 is prime ideal of 𝑅. But the converse 𝑅 Thus, the quotient ring is nothing but the ring of all 𝐼 need not be true. equivalence classes as defined above. Ex: {0} is prime ideal of ℤ but not maximal ideal of ℤ. In fact, the binary operations defined earlier would translate  The prime ideals of ℤ are just the ideals generated by to prime numbers 𝑝 together with the ideal {0}. 𝑐𝑙(𝑎) + 𝑐𝑙(𝑏) = 𝑐𝑙((𝑎 + 𝑏); 𝑎, 𝑏, ∈ 𝑅  Let 𝑅 be a commutative ring and 𝐼, 𝐽 be ideals of 𝑅 and P 𝑐𝑙(𝑎). 𝑐𝑙(𝑏) = 𝑐𝑙(𝑎𝑏) 𝑅 is prime ideal of 𝑅 such that 𝐼𝐽 ⊆ 𝑃. Then either 𝐼 ⊆ 𝑃 It would be interesting exercise for the reader to verify that 𝐼 or 𝐽 ⊆ 𝑃. thus defined forms a ring. 𝑅  Let 𝑅 be finite commutative ring with unity. An ideal 𝐼 of In fact, if 𝑅 has unity 1 then 𝑐𝑙(1) will be unity of. 𝐼 𝑅 is maximal if and only if 𝐼 is prime ideal of 𝑅. 𝑅 is therefore also called Quotient ring of 𝑅 modulo 𝐼. 𝐼 Observation: Some Important Observations about Quotient Ring  There exists a finite commutative ring which has maximal If 𝑅 be a ring and I be an ideal of R. ideal which is not prime ideal. 𝑅  If 𝑅 is a commutative ring, then is also commutative ring. Ex: 𝑅 = ({0,2,4,6}, +8 ,×8 ) 𝐼 (Converse need not be true)  The intersection of two prime ideals of a ring R may not be 𝑅  If 𝑅 has unity 1, Then has unity 1 + 𝐼(Converse need not prime ideal of R. 𝐼 be true)  The intersection of two maximal ideals of a ring R may not 𝑅  If 𝑅 is Boolean ring then is also Boolean ring. be maximal ideal of R 𝐼 𝑅  The sum of two prime (maximal) ideals of a ring R may not  If 𝑆 = {𝑎𝑏 − 𝑏𝑎|𝑎. 𝑏 ∈ 𝑅} Then is commutative ring if 𝐼 be prime (maximal) ideal of R. and any if 𝑆 ⊆ 𝐼  The product of two prime (maximal) ideal of a ring 𝑅 may not be prime (maximal) ideal. 3.3 Ring Homomorphism:  Let 𝑅 be commutative ring with ideal 𝐼 then if 𝑃 is prime Let 𝑅 & 𝑅′ are two rings and then a map 𝜙: 𝑅 → 𝑅′ is defined as ring homomorphism if ideal of 𝐼 then 𝑃 is prime ideal of 𝑅. (i) 𝑓(𝑎 + 𝑏) = 𝑓(𝑎) ⊕ 𝑓(𝑏) ∀ 𝑎, 𝑏 ∈ 𝑅  Let 𝑅 be Boolean ring. Then each proper prime ideal 𝑃 of (ii) 𝑓(𝑎. 𝑏) = 𝑓(𝑎) ⊙ 𝑓(𝑏) ∀ 𝑎, 𝑏 ∈ 𝑅 𝑅 is maximal ideal. Where + &. are addition & multiplication of ring 𝑅 and ⊕ & 3.2. Quotient Rings: ⊙ are addition & multiplication of ring 𝑅′ respectively. Let 𝑅 be a ring and let 𝐼 be an ideal of 𝑅 & if we define Ex: 𝑅 = {𝑟 + 𝐼|𝑟 ∈ 𝑅}=set of all cosets of 𝐼 in 𝑅 (i) 𝐹 be the ring of all functions ℝ into ℝ with point wise 𝐼 𝑅 We already know forms a group under ‘addition’ defined by. addition and point wise multiplication. Then ∀ 𝑎 ∈ ℝ, we 𝐼 can define 𝜙𝑎 (𝑓) = 𝑓(𝑎) ∀ 𝑓 ∈ 𝐹 and called evaluation (𝑟 + 𝐼) + (𝑠 + 𝐼) = (𝑟 + 𝑠) + 𝐼 𝑅 homomorphism Now, define a binary composition (product) on by 𝐼 (ii) The map 𝜙: ℤ → ℤ𝑛 (𝑟 + 𝐼). (𝑠 + 𝐼) = 𝑟𝑠 + 𝐼 Where 𝜙(𝑎) is the remainder of a modulo 𝑛 then 𝜙 is a 𝑅 Then, forms a ring, called the Quotient Ring or Residue Class ring homomorphism for each positive integer 𝑛 and called 𝐼 ring of 𝑅 by 𝐼. natural homomorphism from ℤ to ℤ𝑛 Another way to see quotient ring Let 𝑅 be a ring and 𝐼 an ideal of 𝑅. Some fundamental results related to homomorphism Define 𝑎, 𝑏 ∈ 𝑅, 𝑎 ≡ 𝑏(𝑚𝑜𝑑 𝐼) if 𝑎 − 𝑏 ∈ 𝐼 If 𝜙 be a homomorphism from a ring 𝑅 to a ring 𝑅′. Note that this relation is an equivalence relation on 𝑅. Hence, Let 𝑆 be a subring of 𝑅 and I be an ideal of 𝑅 then it will partition 𝑅 into equivalence classes.  𝜙(0) = 0′ Then, 𝑐𝑙(𝑎) = {𝑟 + 𝑅|𝑟 = 𝑎(𝑚𝑜𝑑 𝐼)}  𝜙(−𝑎) = −𝜙(𝑎) ∀ 𝑎 ∈ 𝑅 = {𝑟 ∈ 𝑅|𝑟 − 𝑎 ∈ 𝐼}  For any 𝑟 ∈ 𝑅 and any positive integer 𝑛, 𝜙(𝑛𝑟) = 𝑛𝜙(𝑟)  Any Homomorphism from a field 𝐹 to a ring 𝑅 is either 𝑛 and 𝜙(𝑟 𝑛) = (𝜙(𝑟)) one-one or zero map.  𝜙(𝑆) = {𝜙(𝑎)|𝑎 ∈ 𝐴} is a sub ring of 𝑅′  If 𝑚 & 𝑛 are distinct positive integer then 𝑚ℤ and 𝑛ℤ are  If 𝑅 is commutative then 𝜙(𝑅) is commutative. NOT Isomorphic rings.  ′ If 𝑅 has a unity 1, 𝑅 ≠ {0} and 𝜙 is onto, then 𝜙(1 ) is the  There exists a homomorphism from a non-commutative unity of 𝑅 ′ ring whose image is a commutative ring.  If 𝑎 ∈ 𝑅 is nilpotent element then 𝜙(𝑎) ∈ 𝑅′ is nilpotent.  There exists homomorphism from a ring without unity  If 𝑒 ∈ 𝑅 is idempotent element then 𝜙(𝑒) ∈ 𝑅′ is whose image is ring with unity. idempotent.  The ring homomorphism from ring of integers to itself is  Let R, S, T be three rings such that 𝑓: 𝑅 → 𝑆 and 𝑔: 𝑆 → 𝑇 either zero mapping or identity mapping. are ring homomorphism then their composition 𝑔𝑜𝑓: 𝑅 →  There exists a ring homomorphism 𝜙: 𝑅 → 𝑅′ such that 𝑇 is also a ring homomorphism (if defined) ring 𝑅 has unity but 𝑅′ does not have unity.  There exists a ring homomorphism 𝜙: 𝑅 → 𝑅′ such that 3.4 Kernel of Homomorphism ring 𝑅 has unity 1 but f(1) is not unity of ring 𝑅′. Let 𝜙 be a homomorphism from a ring 𝑅 to a ring 𝑅′  Let 𝑅 & 𝑅′ be rings & 𝜙: 𝑅 → 𝑅′ be a homomorphism then Then ker 𝜙 is defined as: - If 𝑈 is left ideal of 𝑅, then 𝜙(𝑈) is left ideal of 𝑅′ ker 𝜙 = {𝑟 ∈ 𝑅|𝜙(𝑟) = 0} - If (𝑈) is right ideal of 𝑅, then 𝜙(𝑈) is right ideal of 𝑅′ Some Important Properties of Kernel of Homomorphism:  Let 𝑅 & 𝑅′ be rings If 𝜙 is a homomorphism from a ring 𝑅 to a ring 𝑅′ , then  The mapping 𝜙: 𝑅 × 𝑅′ → 𝑅 given by 𝜙((𝑎, 𝑏)) = 𝑎 is an  ker 𝜙 = {0} if and only if 𝑓 is one-one onto ring homomorphism.  ker 𝜙 is an ideal in 𝑅  𝑅 × 𝑅′ & 𝑅′ × 𝑅 are Isomorphic rings.  Every ideal in ring 𝑅 is the kernel of a ring homomorphism  There exists a ring homomorphism from ℤ2 to a sub ring of 𝑅. In particular, an ideal 𝐼 is the kernel of the mapping. 𝑅 𝑅 of ℤ2𝑛 if and only if 𝑛 is odd positive integer. 𝜙: 𝑅 → ; Such that 𝜙(𝑟) = 𝑟 + 𝐼, Where is quotient 𝐼 𝐼  The homomorphic image of commutative ring is ring. commutative.  If ker 𝜙 ℎ𝑎𝑠 m elements then 𝜙 is 𝑚 to 1 map.  Let 𝑅 be commutative ring and suppose 𝑝𝑥 = 0 ∀ 𝑥 ∈ 𝑅, where 𝑝 is a prime number. Then the mapping 𝜙(𝑥) = 3.5 Isomorphism 𝑥 𝑝 ∀𝑥 ∈ 𝑅 is a homomorphism & this map is known as Two rings 𝑅 & 𝑅′ are said to be isomorphic if ∃ one-one onto Frobenius map. homomorphism between them and then Rings 𝑅 & 𝑅′ are said to be isomorphic i.e. they represent same structure (object) in  The relation of Isomorphism is an equivalence relation. two different notations (language).  let 𝑅 & 𝑅′ be commutative rings with unity if 𝜙: 𝑅 → 𝑅′ be ring homomorphism from 𝑅 onto 𝑅′. Then if char 𝑅 is non- Some Important Properties or Results related to Ring zero ⇒ 𝑐ℎ𝑎𝑟 𝑅′ divides 𝑐ℎ𝑎𝑟 𝑅. Isomorphism: If 𝜙 is a homomorphism from 𝑅 to 𝑅′ then: 3.6 List of Some Important Theorems  𝜙 is an isomorphism if and only if 𝜙 is onto and ker 𝜙 =  Let 𝜙 be a onto ring homomorphism from R to R’ then the {𝑟 ∈ 𝑅|𝜙(𝑟) = 0′ } = {0} 𝑅 mapping from to R’ Given by  If 𝜙 is an isomorphism from 𝑅 onto 𝑅′ Then 𝜙 −1 is an ker 𝜙 Ψ(𝑟 + ker 𝜙)= 𝜙(𝑟) an isomorphism & isomorphism from 𝑅′ onto 𝑅. 𝑅  Every isomorphic image of an integral domain is an ≅ 𝑅′ ker 𝜙 integral domain. 𝑅 𝑅  Every isomorphic image of a field is a field  Let 𝐽 ⊆ 𝐼 be two ideals of a ring 𝑅. Then ≅ 𝐽 𝐼 𝐼 𝐽  Let 𝑅 be a commutative ring of characteristic 2. Then the 𝐼+𝐽 𝐽  Let 𝐼, 𝐽 be two ideals of a ring 𝑅, then ≅ mappin 𝐽 𝐼∩𝐽 𝑅 𝜙: 𝑅 → 𝑅 such that  If 𝑃 is an ideal of ring 𝑅.Then is an integral domain if and 𝑃 𝜙(𝑎) = 𝑎2 is a ring homomorphism. only if 𝑃 is prime ideal.  𝑅 𝐽  The ideal of are always of the form , where 𝐽 is an ideal 𝐼 𝐼 of 𝑅 and 𝐼 ⊂ 𝐽. 3.8 ED, PID, UFD 𝑅 3.8.1 Square Free Number:  Let 𝑅 be a 𝐶𝑅𝑈 & 𝑀 ≠ 𝑅 be an ideal in 𝑅. Then is a field 𝑀 Let 𝑑 be a non-zero rational number, then it is said to be square if and any if 𝑀 is maximal ideal. free if it is not divisible by square of any prime number.  The set 𝑁 of all nilpotent elements in a commutative ring 𝑅 forms an ideal of 𝑅 and then 𝑅 has no non-zero 3.8.2. Quadratic Field: 𝑁 Let 𝑑 be a square free number. Let us define ℚ(√𝑑) = nilpotent elements. (𝑁 is called nil radical of 𝑅) {𝑎 + 𝑏√𝑑: 𝑎, 𝑏 ∈ ℚ} as a subset of ℂ, the set of complex 3.7. Chinese Remainder Theorem: number. Clearly, ℚ(√𝑑) is subfield of ℂ. Let 𝐴1 , 𝐴2 , …. , 𝐴𝑘 be ideals in 𝑅. Examples: 𝑅 𝑅 𝑅 (a) ℚ[√−1] = ℚ[𝑖] is a quadratic field. The map 𝜙: 𝑅 → × × …× 𝐴1 𝐴2 𝐴𝑘 (b) ℚ[√5] = {𝑎 + 𝑏√5: 𝑎, 𝑏 ∈ ℚ} is a quadratic field. Defined by 𝜙(𝑟) = (𝑟 + 𝐴1 , 𝑟 + 𝐴2. …. 𝑟 + 𝐴𝑘 ) 3.8.3 Principal Ideal: Is a ring homomorphism with ker 𝜙 = 𝐴1 ∩ 𝐴2 ∩ … ∩ 𝐴𝑘 An ideal generated by a single element is said to be principal If for each 𝑖, 𝑗 ∈ {1, , …. , 𝑘} with 𝑖 ≠ 𝑗 the ideal 𝐴𝑖 and 𝐴𝑗 are ideal. co-maximal, then this map is onto and 𝐴1 ∩ 𝐴2 ∩ … ∩ 𝐴𝑘 = Examples: 𝐴1 𝐴2 …. 𝐴𝑘 𝑅 𝑅 𝑅 𝑅 𝑅 (a) Every ideal of ℤ𝑚 is principal ideal. So, = ≅ × × …× (b) Every ideal of (ℤ, +, ) is principal. 𝐴1 𝐴2 …𝐴𝑘 𝐴1 ∩𝐴2 ∩….∩𝐴𝑘 𝐴1 𝐴2 𝐴𝑘 Where 𝑅 is commutative ring with unity. 𝜆 𝜆 (𝜆𝑘 ) Observation: If (𝑅, +, ) is a ring and (𝑅, +) is cyclic group then Note: Let 𝑛 be a positive integer and let 𝑝1 1 𝑝2 2 … 𝑝𝑘 be its every ideal is generated by single element hence principal factorization into powers of distinct primes. ℤ ℤ ℤ ℤ ideal. Then ≅ 𝜆 × 𝜆 × …× 𝜆 as rings. 𝑛ℤ 𝑝1 1 𝑝2 2 ℤ 𝑝𝑘 𝑘 ℤ 3.8.4 Principal Ideal Domain: 3.7.1. Embedding of Ring: An integral domain 𝑅 is said to be principal ideal domain if each A ring 𝑅 is said to be embedded in a ring 𝑅′. If there is an ideal 𝐼 of 𝑅 is a principal ideal. isomorphism from 𝑅 into 𝑅′. i.e., 𝐼 = 〈𝑎〉 = {𝑎𝑟: 𝑟 ∈ 𝑅} for some 𝑎 ∈ 𝐼 i.e., 𝑅 is isomorphic to a subring. We also say that 𝑅′ is an Ex: extension ring or over ring of 𝑅. (i) ℤ𝑝 [𝑥] is PID. (ii) ℤ[𝑥] is not PID. 3.7.2 Prime Field: 3.8.5. Norm on an Integral Domain: A field that does not have proper subfield is known as prime Let 𝑅 be an integral domain then function 𝑁: 𝑅 → ℤ+ ∪ {0} field. with 𝑁(0) = 0 is called norm on domain if 𝑁(𝑎) > 0∀0 ≠ 𝑎 ∈ Ex: 𝑅, 𝑁 is called positive norm also. (i) ℤ𝑝 is prime field if 𝑝 is prime. Ex: (ii) ℚ is prime field. (i) In ℤ[𝑖], the ring of Gaussian integers, the function Note: Every ring can be embedded in a ring with unity. 𝑁: 𝕫[𝑖] → ℤ+ ⊂ {0} defined as 𝑁(𝑎 + 𝑖𝑏) = 𝑎2 + 𝑏 2 ∀𝑎 + 𝑖𝑏 ∈ ℤ[𝑖] is a norm. List of some important observations or results: (ii) In any integral domain 𝑅, the function 𝑁: 𝑅 → ℤ+ ⊂  Every ring with or without unity can be embedded in a ring {0} defined as 𝑁(𝑎) = 0 ∀𝑎 ∈ 𝑅 is a norm on 𝑅. of endomorphism of some additive abelian group. 3.8.6. Euclidean Domain:  Every ring with unity has a subring either Isomorphic to Let 𝑅 be an integral domain if ∃ a norm 𝑁 on 𝑅 such that ℤ𝑚 or ℤ. (i) 𝑁(𝑎. 𝑏) ≥ 𝑁(𝑎)∀𝑎, 𝑏 ≠ 0 ∈ 𝑅  Every field has a subfield Isomorphic to either ℤ𝑝 or ℚ. (ii) For any 𝑎 ∈ 𝑅, 𝑏 ∈ 𝑅 − {0} ∃𝑞, 𝑟 ∈ 𝑅 such that 𝑎 = 𝑏𝑞 +  Every integral domain can be embedded in a field. 𝑟, 𝑟 = 0 or 𝑁(𝑟) < 𝑁(𝑏) then 𝑅 is said to be Euclidean  ℤ𝑝 can be embedded into a field 𝐹 if char 𝐹 = 𝑝 domain.  ℚ can be embedded into a field 𝐹 if char 𝐹 = 0 Examples: (i) ℤ is ED with norm 𝑁(𝑎) = |𝑎|, the usual absolute value of  Let 𝑅 be a PID, which is not a field, then an ideal 𝐼 = 〈𝑎〉 is 𝑎. maximal ideal if and only if 𝑎 is an irreducible element of (ii) Every field 𝐹 is ED with norm 𝑁 as 𝑁(𝑎) = 1∀0 ≠ 𝑎 ∈ 𝐹 R. (iii) ℤ[𝑖] is ED with norm 𝑁(𝑎 + 𝑖𝑏) = 𝑎2 + 𝑏 2 ∀𝑎, 𝑏 ∈ ℤ  Let 𝑅 be a PID and 𝐼 be non-zero ideal such that 𝐼 ≠ 𝑅, (iv) The ring ℤ[√2] = {𝑎 + 𝑏√2: 𝑎, 𝑏 ∈ ℤ} is an ED under the then 𝐼 is prime ideal if and only if 𝐼 is maximal ideal. norm 𝑁 defined as 𝑁(𝑎 + 𝑏√2) = |𝑎2 − 2𝑏 2 |∀𝑎, 𝑏 ∈ ℤ  If 𝑅 is a P.I.D. then every ascending chains of ideals 〈𝑎1 〉 ⊆ (v) The field of rational numbers ℚ is not an Euuclidean 〈𝑎2 〉 ⊆ 〈𝑎3 〉 ⊆ ⋯ ⊆ 〈𝑎𝑛 〉 ⊆ ⋯ is finite domain under the norm 𝑁 defined as 𝑁(𝑎) = |𝑎|, the  The quotient ring of a PID by a prime ideal is again a PID. usual absolute value.  If 𝑅 is an ED with norm 𝑁 and 𝑎 ∈ 𝑅 be an arbitrary 3.8.7 Unique Factorization Domain (U.F.D.): element then 𝑎 is unit if and only if 𝑁(𝑎) = 𝑁(1) An Integral domain 𝑅 with unity is called a unique factorisation  (xvi) As a consequence of above result it follows that the domain (U.F.D.), if satisfies the following conditions: only units of ℤ[𝑖] are ± 1, ±𝑖. (i) Each non-zero element of 𝑅 is either a unit or can be  In ℤ[𝑖], 𝑎 + 𝑖𝑏 is not a unit then 𝑎2 + 𝑏 2 > 1 expressed as a product of finite number of irreducible  If 𝑅 be an ED with norm 𝑁 and 𝑎, 𝑏 be two non-zero elements of 𝑅. elements of 𝑅. Then 𝑏 is not unit of 𝑅 if and only if 𝑁(𝑎) < (ii) The above Decomposition is unique up to the order and 𝑁(𝑎𝑏) associates of the irreducible elements it means if a non-  If 𝑅 be an ED with norm 𝑁, then zero, non-unit element 𝑎 ∈ 𝑅 is expressible as 𝑎 = - 𝑁(𝑎) = 𝑁(−𝑎) ∀ 0 ≠ 𝑎 ∈ 𝑅 𝑝1 𝑝2 … 𝑝𝑟 and 𝑎 = 𝑞1 𝑞2 … 𝑞𝑠 where 𝑝𝑖′ 𝑠 and 𝑞𝑖′ 𝑠 are - If 𝑁(𝑎) = 0&0 ≠ 𝑎 ∈ 𝑅, then 𝑎 is unit in 𝑅. irreducible elements of 𝑅, then ∃ a one-to-one - 𝑁(𝑎) = 𝑁(𝑎𝑏) if and only if 𝑏 is unit in 𝑅 where 0 ≠ correspondence between 𝑝𝑖′ 𝑠 and 𝑞𝑖′ 𝑠 such that the corresponding elements are associates. In particular 𝑟 = 𝑠. 𝑎, 0 ≠ 𝑏, 𝑎, 𝑏, ∈ 𝑅  (xx) 𝑅 is a U.F.D. ⇔ 𝑅[𝑥] is a U.F.D. Examples: (i) 𝔽[𝑥], ring of polynomials over field 𝔽 (ii) The ring ℤ[𝑖] of Gaussian integers. SOME IMPORTANT DEFINITIONS (iii) ℤ[𝑥] is U.F.D. 4.1 Field Extension : A field K is said to a field extension of a 1+√𝐷 field 𝐹 if 𝐹 is a subfield of 𝐾 or 𝐹 is (iv) For 𝐷 < 0, ℤ[√𝐷] = {𝑎 + ( ) 𝑏; 𝑎, 𝑏 ∈ ℤ} is U.F.D. if 2 isomorphic to a subfield of 𝐾. and only if 𝐷= e.g., ℂ / ℝ , ℂ/ℚ, ℝ/ℚ are field extension. −1, −2, −3, −7, −11, −19, −43, −67, −163 NOTE : K / F is a field extension but not a quotient ring. (v) For 𝑛 > 3, ℤ[√−𝑛] = {𝑎 + 𝑏√−𝑛: 𝑎, 𝑏 ∈ ℤ} is never U.F.D. Degree of a field extension : Let 𝐾 / 𝐹 be a field extension, 1+√−19 (vi) The quadratic ring ℤ [ ] is a P.I.D. but not E.D. then 𝐾 is a vector space over 𝐹 2 and therefore it must have a dimension. The dimension of 𝐾 Some important results related to E.D., P.I.D., U.F.D. over 𝐹 is called degree of K over F and it is denoted by [𝐾 ∶ 𝐹]  ED ⇒ P.I.D. ⇒ U.F.D. ⇒ 𝐼. 𝐷. ⇒ 𝐶𝑅𝑈 EX: [ ℂ: ℂ] = 1 ; [ℂ ∶ ℝ] = 2; [ℂ ∶ ℚ] = ∞ ; [ℝ ∶ ℚ] = ∞  If 𝔽 is a field, then 𝔽[𝑥] is ED ⇒ P.I.D. ⇒ U.F.D. Finite Extension : The extension 𝐾 / 𝐹 is said to be finite if its  If 𝐼, 𝐽 be a non-zero ideals of a PID 𝑅, generated by 𝑎 and degree is finite. 𝑏 respectively then 𝐼 ∙ 𝐽 is ideal of 𝑅 generated by 𝑎𝑏 i.e. if 𝐼 = 〈𝑎〉& 𝐽 = 〈𝑏〉; then 𝐼𝐽 = 〈𝑎𝑏〉 Infinite Extension : The extension 𝐾 / 𝐹 is said to be infinite if  Let 𝑅 be a integral domain and 𝑎, 𝑏 be non-zero elements its degree is infinite. of 𝑅 then Algebraic Element : Let 𝐾 / 𝐹 be any field extension. An - 𝑎/𝑏 and 𝑏/𝑎 ⇒ 〈𝑎〉 = 〈𝑏〉 element 𝑎 ∈ 𝐾 is said to be algebraic over 𝐹 if ‘𝑎’ satisfies - 𝑎 and 𝑏 are associate ⇒ 〈𝑎〉 = 〈𝑏〉 some polynomial over 𝐹.  Let 𝑅 be PID & 𝑎 ∈ 𝑅 then 𝑎 is prime element ⇔ 𝑎 is Algebraic Extension : An extension 𝐾 / 𝐹 is said to be irreducible element. algebraic extension if every element of K is algebraic over F. Non-Algebraic Extension : An extension 𝐾 / 𝐹 is said to be Results : non-algebraic if there is atleast one element in K which is not 1. If 𝛼 is a primitive 𝑝𝑡ℎ root of unity where p is a prime, then algebraic over F. minimal polynomial of 𝛼 over ℚ is 𝑚(𝑥) = 𝑥 𝑝−1 + 𝑥 𝑝−2 +...... + 𝑥 2 + 𝑥 + 1. Minimal Polynomial of an element : Let 𝐾 / 𝐹 be any 2. Degree of minimal polynomial of any real number over ℝ is extension and 𝑎 ∈ 𝐾 be any algebraic element. A polynomial 1 and any non-real complex number is 2. 𝑚(𝑥) ∈ 𝐹[𝑥] is said to be ‘minimal polynomial’ of a over F if 3. A minimal polynomial of 𝑛𝑡ℎ primitive root of unity over ℚ (i) m(x) is monic is 𝑔𝑛 (𝑥). (ii) 𝑚(𝑎) = 0 (iii) 𝑚(𝑥) is irreducible over F. i.e, 𝑚(𝑥)is lowest degree monic polynomial which is satisfied by ‘a’. Sr. No Element Minimal Polynomial Over ℚ Over ℝ Over ℂ 1. 1 1 1 1 𝑥− 𝑥− 𝑥− 2 2 2 2 2. √2 𝑥2 − 2 𝑥 − √2 𝑥 − √2 3. 𝑖 𝑥2 + 1 𝑥2 + 1 𝑥−𝑖 2 2 4. 𝜔 𝑥 + 𝑥 +1 𝑥 +𝑥+1 𝑥−𝜔 5. 3 √2 𝑥3 − 2 3 𝑥 − √2 3 𝑥 − √2 6. √2 + √3 𝑥 4 − 10𝑥 2 + 1 𝑥 − √2 − √3 𝑥 − √2 − √3 7. 𝜋 does not exist 𝑥−𝜋 𝑥−𝜋 8. 𝑒 does not exist 𝑥−𝑒 𝑥−𝑒 9. 2𝜋𝑖 𝑔3 (𝑥) = 𝑥 2 + 𝑥 + 1 2 𝑥 +𝑥+1 𝑥−𝜔 𝑒 3 =𝜔 10. 2𝜋𝑖 𝑔4 (𝑥) = 𝑥 2 + 1 𝑥2 + 1 𝑥−𝑖 𝑒 4 =𝑖 Results : 1. Every finite extension is algebraic but converse may not be true. 2. Finite extension of finite extension is finite. 3. Let 𝐾 / 𝐸 and 𝐸 / 𝐹 are two finite extension then 𝐾 / 𝐹 is also a finite extension and [𝐾 ∶ 𝐹] = [𝐾 ∶ 𝐸][𝐸 ∶ 𝐹] 4. Algebraic extension of algebraic extension is algebraic. 5. Let 𝐾 / 𝐹 be any extension. Suppose a and b are two algebraic element of 𝐾 over 𝐹, then 𝑎 + 𝑏, 𝑎 − 𝑏, 𝑎𝑏, 𝑎𝑏 −1 (𝑏 ≠ 0) are also algebraic over F. Def. 𝐹(𝑎) = Smallest field containing 𝐹 and 𝐹(𝑎, 𝑏) = Smallest field containing 𝐹, 𝑎 and 𝑏. 𝐹(𝑎1 , 𝑎2 , ….. 𝑎𝑛 ) = Smallest field containing 𝐹 and. 𝑎1 , 𝑎2 , … … 𝑎𝑛 6. ℚ(√𝑝) = {𝑎 + 𝑏 √𝑝 ∶ 𝑎, 𝑏 ∈ ℚ} 7. ℚ[𝑖] = {𝑎 + 𝑖𝑏 ∶ 𝑎, 𝑏 ∈ ℚ} 8. Let K / F be any extension and a K be any algebraic element then (i) [𝐹 (𝑎) ∶ 𝐹] = degree of minimal polynomial of a over F. (ii) If [𝐹 (𝑎) ∶ 𝐹] = n, then {𝑎1 , 𝑎, 𝑎2 , … …. , 𝑎𝑛−1 } is a basis of 𝐹 (𝑎) over F. 9. Here i 𝑝𝑖 ’𝑠 and 𝑞𝑖 ’𝑠 all are distinct primes. (i) [ ℚ(√𝑝1 , √𝑝2 , ….. √𝑝𝑛 ): ℚ] = 2𝑛 (ii) [ ℚ(√𝑝1 , √𝑝2 , ….. √𝑝𝑛 ): ℚ](√𝑝1 , √𝑝2 , ….. √𝑝𝑚 ) = 2𝑛−𝑚 , 0 ≤ 𝑚 ≤ 𝑛 (iii) [ ℚ( 3√𝑝1 , 3√𝑝2 , ….. 3√𝑝𝑛 ): ℚ] = 3𝑛 (iv)[ ℚ( 3√𝑝1 , 3√𝑝2 , ….. 3√𝑝𝑛 ): ℚ( 3√𝑝1 , 3√𝑝2 , ….. 3√𝑝𝑚 )] = 3𝑛−𝑚 , 0 ≤ 𝑚 ≤ 𝑛 (v) [ℚ(√𝑝1 , √𝑝2 , ….. √𝑝𝑛 , 3√𝑞1 , 3√𝑞2 , ….. 3√𝑞𝑚 )]: ℚ = 2𝑛 3𝑚 ℚ(√𝑝1 , √𝑝2 , ….. √𝑝𝑛 , 3√𝑞1 , 3√𝑞2 , ….. 3√𝑞𝑚 ) (vi) [ ] = 2𝑛−𝑟. 3𝑚−𝑠 , 0 ≤ 𝑟 ≤ 𝑛, 0 ≤ 𝑠 ≤ 𝑚 : ℚ√𝑝1 , √𝑝2 , ….. √𝑝𝑟 , 3√𝑞1 , 3√𝑞2 , ….. 3√𝑞𝑠 NOTE: (i) The above six results are also true if some or all commas are replaced by plus (+) or minus (-). (ii) 𝑖 and 𝜔 behave as √𝑝 because they are of degree 2. 𝑝𝜋𝑖 𝑝𝜋 𝑝𝜋 10. If 𝑝 and 𝑞(≠ 0) are any integers then 𝑒 𝑞 , sin ( ),sin and cos ( )are algebraic over ℚ. 𝑞 𝑞 11. sin 𝑚°and cos 𝑚° are algebraic over ℚ where 𝑚 ∈ ℚ. 3 EXAMPLE: The degree of the extension ℚ(√2 + √2) over the field ℚ(√2) is (a) 1 (b) 2 (c) 3 (d) 6 Explanation (c) We have field extension 3 ℚ(√2 + √2) 3 Take 𝑥 − (√2 + √2) = 0 ⇒ 𝑥 − √2 = 21/3 3 ⇒ (𝑥 − √2) = 2 ⇒ 𝑥 3 − 2√2 + 6𝑥 − 3√2𝑥 2 = 2 ⇒ 𝑥 3 + 6𝑥 − 2 = 3√2𝑥 2 + 2√2 = √2(3𝑥 2 + 2) ⇒ (𝑥 3 + 6𝑥 − 2)2 = 2(3𝑥 2 + 2)2 We get monic polynomial of degree 6 whose root is 3 (√2 + √2).  3 ⇒ Degree of extension (ℚ(√2 + √2): ℚ) = 6 And degree of extension (ℚ(√2): (ℚ)) = 2 3 6 ⇒ Degree of extension (ℚ(√2 + √2): (ℚ√2)) = = 3 2 Hence option (c) is correct. 4 8 EXAMPLE: Find the degree of the field extension ℚ(√2, √2, √2) over ℚ (a) 4 (b) 8 (c) 14 (d) 32 Explanation (b) Note that 4 8 4 3⁄ 1⁄ 3⁄ 5⁄ 7⁄ ℚ(√2, √2, √2) = ℚ + ℚ(√2) + ℚ( √2 + ℚ (2 4) + ℚ (2 8) + ℚ (2 8) + ℚ (2 8) + ℚ (2 8) 4 4 ℚ(√2, √2) = ℚ + ℚ(√2) + ℚ( √2) + ℚ(23/4 ) ℚ(√2) = ℚ + ℚ(√2) Then, 4 8 4 8 4 [ℚ(√2, √2, √2): ℚ] = [ℚ(√2, √2, √2): (ℚ√2, √2)] 4 × [ℚ(√2, √2): ℚ(√2)] × [ℚ(√2): ℚ] =2×2×2=8 EXAMPLE. Which of the following statements are true for 𝛼 ∈ ℝ? (a) If 𝛼 3 is algebraic over ℚ then 𝛼 is algebraic over ℚ (b) 𝛼 could be algebraic over ℚ[√2] but may not be algebraic over ℚ (c) 𝛼 need not be algebraic over any subfield of ℝ (d) There is an 𝛼 which is not algebraic over ℚ[√−1] Explanation: (a), (b), (d) For option (a), If 𝛼 3 is algebraic over ℚ then there exists a non-zero polynomial 𝑝(𝑥) = 𝑎0 + 𝑎1 𝑥 + ⋯ + 𝑎𝑛 𝑥 𝑛 ∈ ℚ[𝑥] such that 𝑝(𝛼 3 ) = 0. ⇒ 𝑎0 + 𝑎1 𝛼 3 + 𝑎2 𝛼 6 … + 𝑎𝑛 𝛼 3𝑛 = 0 Multiply 𝛼 both side ⇒ 𝑎0 𝛼 + 𝑎1 𝛼 4 + 𝑎2 𝛼 7 … + 𝑎𝑛 𝛼 3𝑛+1 = 0 …….(i) Take 𝑞(𝑥) = 𝑎0 𝑥 + 𝑎1 𝑥 4 + 𝑎2 𝑥 7 … + 𝑎𝑛 𝑥 3𝑛+1. Observe that 0 ≠ 𝑞(𝑥) ∈ ℚ[𝑥] and from (i), 𝑞(𝛼) = 0. That is 𝛼 is algebraic over ℚ Hence option (a) is correct. For option (c), Note that ℝ is also subfield of ℝ. For any 𝛼 ∈ ℝ, we have polynomial 0 ≠ 𝑝(𝑥) = 𝑥 − 𝛼 ∈ ℝ[𝑥] such that 𝑝(𝛼) = 0. For option (d), take 𝛼 = 𝜋.Observe that 𝛼 is not algebraic over ℚ[√−1]. EXAMPLE: Let 𝑧 = 𝑒 2𝜋𝑖/7 and 𝜃 = 𝑧 + 𝑧 2 + 𝑧 4. Then (a) 𝜃 ∈ ℚ (b) 𝜃 ∈ ℚ(√𝐷) for some 𝐷 > 0 (c) 𝜃 ∈ ℚ (√𝐷) for some 𝐷 < 0 (d) 𝜃 ∈ 𝑖ℝ Explanation (c) We have 𝑧 = 𝑒 2𝜋𝑖/7 and 𝜃 = 𝑧 + 𝑧 2 + 𝑧 4. We know that 1 + 𝑧 + 𝑧2 + 𝑧3 + 𝑧4 + 𝑧5 + 𝑧6 = 0. 𝜃 2 = 𝑧 2 + 𝑧 4 + 𝑧 8 + 2𝑧 3 + 2𝑧 6 + 2𝑧 5 = 𝑧 2 + 𝑧 4 + 𝑧 + 2𝑧 3 + 2𝑧 6 + 2𝑧 5 = 𝑧2 + 𝑧4 + 𝑧 + 𝑧3 + 𝑧6 + 𝑧5 + 1 + 𝑧3 + 𝑧6 + 𝑧5 − 1 = 𝑧3 + 𝑧6 + 𝑧5 − 1 = 𝑧 + 𝑧2 + 𝑧4 − 2 = θ-2 1±√−7 ⇒ 𝜃 2 − 𝜃 + 2 = 0, ⇒ 𝜃= 2 Clearly, option (c) is only correct. 4.2 Splitting Field and Multiple Roots Def. Splitting Field : Let 𝑓 (𝑥) ∈ 𝐹 [𝑥] be any polynomial where F is a field. The splitting field of 𝑓 (𝑥)over 𝐹 is the smallest extension of 𝐹 which contains all the roots of (𝑥). Results : 1. Kronecker’s Theorem : Let 𝐹 be a field and 𝑓 (𝑥)a non constant polynomial 𝐹 (𝑥). Then there is an extension field E of F in which 𝑓 (𝑥)has a zero (roots). 2. Let F be a field and let 𝑝 (𝑥) ∈ 𝑓 (𝑥)be irreducible over 𝐹. If ‘𝑎’ is a zero of 𝑝(𝑥) in some extension 𝐸 of F 𝐹[𝑥] 𝐹[𝑥] then 𝐹 (𝑎) is isomorphic to 〈𝑝(𝑥)〉 i.e, 𝐹(𝑎) ≅ 〈𝑝(𝑥)〉.Furthermore, if deg 𝑝(𝑥) = 𝑛, then every member of 𝐹 (𝑎) can be uniquely expressed in the form 𝑐𝑛−1 𝑎𝑛−1 + 𝑐𝑛−2 𝑎𝑛−2 + ⋯ + 𝑐1 𝑎 + 𝑐0 , where 𝑐0 , 𝑐1 ….. 𝑐𝑛−1 ∈ 𝐹 3. Splitting Fields are unique i.e, isomorphic. 4. Criterion for multiple zeros: A polynomial 𝑓 (𝑥) over a field 𝐹 has a multiple zero in some extension 𝐸 iff 𝑓 (𝑥) and 𝑓 [𝑥] have a common factor of positive degree in 𝑓 [𝑥] 5. Working Rule to find the splitting field over ℚ : (i) Find out the roots of the given polynomial. (ii) Adjoin all the roots with ℚ. (iii) Simplify to obtain splitting field. EXAMPLE: Which of the following statements are correct? (a) The fields ℚ(√5) and ℚ(√7) are isomorphic as vector spaces over ℚ (b) The fields ℚ(√5) and ℚ(√5) are isomorphic as fields (c) The Galois group of ℚ(√5)/ℚ is isomorphic to the Galois group of ℚ(√7)ℚ (d) The fields ℚ(√5, √7) and ℚ(√5 + √7) are isomorphic as fields Explanation: (a), (c), (d) ℚ[√5] over ℚ is vector space and ℚ[√7] is also vector space over ℚ. ⇒ [ℚ[√5]: ℚ] = 2 and [ℚ[√5]: ℚ] = 2 Vector space of save dimensions and isomorphic. ⇒ ℚ[√5] ≅ ℚ [√5] But ℚ[√5] and ℚ[√5] ore not isomorphic as field. Hence (a) is correct but (b) is incorrect. For option (d), ℚ[√5, √7] ≅ ℚ [√5 + √7] as fields. Hence option (d) is correct. 𝑄√5 For option (c), |cot ( ) | = [ 𝑄[√5]: 𝑄] = 2 𝑄 𝑄√7 |cot ( ) | = [ 𝑄[√7]: 𝑄] = 2 𝑄 𝑄√5 𝑄√7 ⇒ 𝐺𝑎𝑙 ( ) ≅ 𝐺𝑎𝑙 ( ) 𝑄 𝑄 Hence option (c) is correct. 4.3 Finite Fields Results: 1. 𝐹 is finite iff 𝐹 ∗ is cyclic where 𝐹 ∗ = 𝐹 − {0}. OR 𝐹 is infinite iff 𝐹 ∗ is non cyclic. 2. ℚ∗ , ℝ∗ , ℂ∗ are always group w.r.t. multiplication but not cyclic as ℚ, ℝ and ℂ are field of infinite order. 3. ℤ∗𝑝 is a cyclic group with respect to ×𝑝 , where 𝑝 is a prime number. 4. Characteristic of an integral domain is either zero or a prime number. Def. Prime Field: A field having no proper subfield is called a prime field 5. Up to isomorphism there are only two prime fields namely ℚ and ℤ𝑝. 6. [ℝ ∶ ℝ] = 1 and [ℂ ∶ ℝ] = 2, so there are no field between ℝ and ℂ 7. [ℝ ∶ ℚ] = ∞ and [ℚ ∶ ℚ] = 1, so there are infinite many field between ℚ and ℝ. 8. The number of subfields of a field of order 𝑝𝑛 is 𝑑(𝑛) i.e. number of divisors of 𝑛. 9. Let 𝔽 be a field with order 𝑝𝑛 then it has a unique subfield of order 𝑝𝑚 iff m divides n 10. Let 𝐾 be a field of order 𝑝𝑛 and 𝐹 be its subfield of order 𝑝𝑚 , where 𝑚 divides 𝑛, then 𝐾/𝐹 is a finite (and 𝑛 hence algebraic) extension and [𝐾 ∶ 𝐹] =. 𝑚