Ideals of Z[x] with xZ[x]⊆J⊆Z[x]
Added on 2023-04-21
9 Pages2414 Words126 Views
Assignment 1
Student Name
The University of Sydney
School of Mathematics and Statistics
Student Name
The University of Sydney
School of Mathematics and Statistics
1. a) Provide an example and proof for each of the following:
i) An irreducible element in the ring Q[x].
Example: x2+ 1.
Proof:
Suppose that x2+ 1 is reducible in Q[X].
Then the equation x2+ 1=0 has a root in Q. That is, there is a b in Q such that
b2=−1. This implies that b= √−1 is in Q. Clearly this is not true and thus
x2+ 1 is not reducible in Q[X].
ii) A maximal ideal in Z[x].
Example: (3 , x).
Proof:
We need to show that (3 , x) is a maximal ideal. Let us suppose that (3 , x) is
not maximal and thus there is another generator polynomial, q ∉(3 , x ).
Let Q be some polynomial. We can write p¿ k + x∗Q where k is an integer not
divisible by 3. Note that subtracting a multiple of one generator from another
generator does not change the ideal. Thus (3 , x , P)=(3 , x , n). Since n is not a
multiple of 3 , gcd (3 , n)=1 , so 1 ∈(3 , x , P).
Therefore(3 , x , P) is all of Z [x ]→ (3 , x) is maximal.
iii) A unit u ≠ 1 in the ring Z[x]/( x2 + 3x + 1)Z[x].
Since x2 + 3x + 1 irreducible in Z then x2 + 3x + 1 is a unit in Z[x]/(x2 + 3x +
1)Z[x].
iv) Example: 1− √−3.
i) An irreducible element in the ring Q[x].
Example: x2+ 1.
Proof:
Suppose that x2+ 1 is reducible in Q[X].
Then the equation x2+ 1=0 has a root in Q. That is, there is a b in Q such that
b2=−1. This implies that b= √−1 is in Q. Clearly this is not true and thus
x2+ 1 is not reducible in Q[X].
ii) A maximal ideal in Z[x].
Example: (3 , x).
Proof:
We need to show that (3 , x) is a maximal ideal. Let us suppose that (3 , x) is
not maximal and thus there is another generator polynomial, q ∉(3 , x ).
Let Q be some polynomial. We can write p¿ k + x∗Q where k is an integer not
divisible by 3. Note that subtracting a multiple of one generator from another
generator does not change the ideal. Thus (3 , x , P)=(3 , x , n). Since n is not a
multiple of 3 , gcd (3 , n)=1 , so 1 ∈(3 , x , P).
Therefore(3 , x , P) is all of Z [x ]→ (3 , x) is maximal.
iii) A unit u ≠ 1 in the ring Z[x]/( x2 + 3x + 1)Z[x].
Since x2 + 3x + 1 irreducible in Z then x2 + 3x + 1 is a unit in Z[x]/(x2 + 3x +
1)Z[x].
iv) Example: 1− √−3.
Proof:
We show that 1− √−3 is irreducible in Z[√−3].
Assume 1− √−3 is reducible, then there must exist a , b ∈ Z [ √ −3] so that N¿
b) Write 45+ 420 i as a product of irreducible Gaussian integers, showing all working.
Solution
In Z [i], N (a+ bi)=a2 +b2 .
N (45 420 i)=452 + 4202=178425=3∗3∗5∗5∗13∗61
Thus 45+ 420 i has factors a+ bi, c+di , s +ti∧m+ ¿ such that
N (a+bi) =32, N (c+ di)=5 , N ( s +ti )=13∧N ( m+¿ )=61.
If these factors exist, then they are clearly irreducible over Z.
We have a+ bi=32=9. 9 cannot be expressed as a sum of two elements of Z and thus
a+ bi=3.
We have c +di=5 which can be expressed as 12 +22 .
We have s+ti=13 which can be expressed as 32 + 42 .
We have m+¿=61 which can be expressed as 52 +62.
Thus 45+ 420 i=(3)(3)(1+ 2i)(1−2 i)(2+i)(2−i)(3+2 i)(3−2i)(5+6 i)(5−6 i)
c) Let R, S, T be rings, and suppose that α : R → S and β : R → T are ring
homomorphisms.
Show that the map γ :R → S ×T with γ (x) = (α(x), β(x)) is a ring homomorphism.
Solution
We show that 1− √−3 is irreducible in Z[√−3].
Assume 1− √−3 is reducible, then there must exist a , b ∈ Z [ √ −3] so that N¿
b) Write 45+ 420 i as a product of irreducible Gaussian integers, showing all working.
Solution
In Z [i], N (a+ bi)=a2 +b2 .
N (45 420 i)=452 + 4202=178425=3∗3∗5∗5∗13∗61
Thus 45+ 420 i has factors a+ bi, c+di , s +ti∧m+ ¿ such that
N (a+bi) =32, N (c+ di)=5 , N ( s +ti )=13∧N ( m+¿ )=61.
If these factors exist, then they are clearly irreducible over Z.
We have a+ bi=32=9. 9 cannot be expressed as a sum of two elements of Z and thus
a+ bi=3.
We have c +di=5 which can be expressed as 12 +22 .
We have s+ti=13 which can be expressed as 32 + 42 .
We have m+¿=61 which can be expressed as 52 +62.
Thus 45+ 420 i=(3)(3)(1+ 2i)(1−2 i)(2+i)(2−i)(3+2 i)(3−2i)(5+6 i)(5−6 i)
c) Let R, S, T be rings, and suppose that α : R → S and β : R → T are ring
homomorphisms.
Show that the map γ :R → S ×T with γ (x) = (α(x), β(x)) is a ring homomorphism.
Solution
End of preview
Want to access all the pages? Upload your documents or become a member.