Ideals of Z[x] with xZ[x]⊆J⊆Z[x]

Find all ideals J of Z[x] with xZ[x]⊆J⊆Z[x].

Assignment 1
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.

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+2 i)(1−2 i)(2+i)( 2−i)(3+2i)(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

Let x , y ∊ R . We show that γ ( x+ y ) =γ ( x ) + γ ( y ) and γ ( xy ) =γ ( x ) γ ( y ).
Now γ ( x+ y )=¿
¿ ¿ since α and β are ring homomorphisms
¿ ( α ( x ) , β ( x ) ) + ( α ( y ) , β ( y ) )
¿ γ ( x ) + γ( y)
Thus, the first property is satisfied.
Again, γ ( xy )=¿
¿ ¿ since α and β are ring homomorphisms
¿ ( α ( x ) , β ( x ) ) ( α ( y ) , β ( y ) )
¿ γ ( x ) γ ( y )
Thus, the second property is satisfied and the map γ :R → S ×T is a ring
homomorphism.
(d) Show that if n=pa qb with p , q>1 distinct primes and a , b ≥ 1 integers then
Z /nZ ≌( Z ¿ pa Z)×(Z /qb Z) .
Solution
Let s+nt , c+ nd ∊ Z /nZ.
We show that f ((s+ n1 t )(c+ n2 d))=f (s+nt) f (c +nd) where f is the map such that
f : Z /nZ →(Z ¿ pa Z)×(Z /qb Z). Then f ¿
2. a) Let a ( x), b(x )∊ Q[ x ] such that

a ( x)=x6 −2 x5−x4 +5 x3−2 x2−2 x +2
b ( x)=x5−3 x4 +3 x3−2 x+2
Find a generator for the principal ideal a ( x) Q[x ]+ b(x) Q[ x]
Solution
The generator for this principal ideal is a linear combination of its prime factors.
So, we first find the prime factors for a ( x)=x6 −2 x5−x4 +5 x3−2 x2−2 x +2
and b (x)=x5−3 x4 +3 x3−2 x+ 2 .
a ( x)=x6 −2 x5−x4 +5 x3−2 x2−2 x +2=( x3−x +1)(x3 −2 x2 +2)
b ( x)=x5−3 x4 +3 x3−2 x+ 2=( x2 −x+1)(x3−2 x2+2)
Thus ¿( x3 −2 x2 +2) is a common factor for a ( x) , b(x )∊ Q[ x ] .
Therefore, the generator is ((x3 −x+1)( x3−2 x2+2)(x2−x +1)).
That is ( x3−x +1)Q[ X ]+(x3−2 x2 +2)Q [ X ]+(x2−x +1)Q[ X ].
b) Prove or Disapprove:
i) If F is a field and R is a non-trivial ring, and φ : F → R is a non-trivial ring
homomorphism, then ϕ is injective.
Proof:
We need to show that ∀ x , y ∊ F , φ( x )=φ ( y)→ x = y .
So, suppose that φ (x)=φ( y ). Then φ−1 (φ(x ))=φ−1( φ( y ))→ x = y .
Thus, ϕ is injective.

ii) R with addition and multiplication defined by
a+b=min{a , b }∧a∗b=a+b is a ring.
Proof
For multiplication axioms, since + is the usual addition in R, clearly closure and
associativity holds. So, we only show the distributive laws.
a (b+c )=a (min {b , c }). Suppose min {b ,c }=c. Then a (min {b , c })=a(c )=a+ c
a (b+c )=a∗b+ a∗c=min {a+b , a+ c }¿=a+ c. Thus, the distributive law holds.
We now show addition axioms.
 Closure: clearly, ∀ a , b ∊ R, min {a ,b } is either a or b and thus in R.
 Associativity: It is clear that min {a ,(b , c) }=min {(a , b) , c }.
 Commutativity: It is clear that min {a ,b }=min{b , a }.
 Identity: There is no b ∊ R such that min {b , a} is a, ∀ a ∊ R . Therefore R is
not a ring since it does not have a additive identity.
iii) There exists an ideal I of Z2 [ X ] such that Z2 [ X ]/ I ≅ Z2 × Z2.
c) You are given that up to isomorphism there are exactly 4 distinct unital rings with
precisely 4 elements. Find them all.
Proof
Each of the following rings has four elements:
 R1=Z/4
 R2=F2×F2=F2[x]/(x2+x)

 R3=F4=F2[x]/(x2+x+1)
 R4=F2[x]/(x2)
They are non-isomorphic because only R3 is a field, only R1 has characteristic ≠2, and
only R2, R3 are reduced.
Conversely, let R be a ring with four elements. If a ∈ R ∖ {0,1 }, then the centralizer of
a is a subgroup of ¿ with at least three elements 0,1 , a , so by Lagrange also the
fourth element must commute with a. Thus, R is commutative. If R is reduced, then it
is a finite product of local artinian reduced rings, i.e. fields, so that R ≅ R 2 or R ≅ R 3.
If R is not reduced, there is some a ∈ R ∖ {0 } such that a2=0. Since 0,1 , a , a+1 are
pairwise distinct, these are the elements of R. If 2=0, then we get an injective
homomorphism F2 [x ]/( x2 )→ R , x ↦a. Since both sides have four elements, it is an
isomorphism. If 2≠0, the characteristic must be 4, i.e. we get an embedding
Z /4 → R , which again has to be an isomorphism.
3. (a) Find all ideals J of Z [x ] with xZ [ x]⊆ J ⊆ Z [ x ].
We need to find the maximal ideals of Z [x ].
The maximal ideals of Z [x ] are of the form ( p , f (x )) where p is a prime number and
f ( x) is a polynomial in Z [x ] which is irreducible modulo p.
Using this criterion, below are the ideals J of Z [x ] with xZ [ x]⊆ J ⊆ Z [ x ]:
(b) i) Let N : R→ ¿be given by N ( z )=¿ z∨¿2 ¿. We show that N ( z )∊ N for all z ∊ R .

Solution
Let z=a+b √ −11 where a and b are integers.
N ( z )=((a+ b √−11)(a−b √−11))2=(a¿ ¿2+ b2 (−11))2=(a¿¿ 2−11b2)2 ¿ ¿
Since a , b ∊ Z , then (a ¿¿ 2−11 b2)2 ∊ N .¿
ii) The units of R are 1 , √−11 ,−1 ,− √ −11.
1 and -1 are the units inherited from the set of integers whereas
√−11 ,− √−11 are the units brought about by the irrational extension of the
set of integers.
iii)Decompose 25
2 −1
2 √−11 into irreducible factors over R.
The irreducible factors of 25
2 −1
2 √−11 are simply its prime factors .
To get the prime factors of 25
2 −1
2 √−11 .We completely factorize or
decompose 25
2 −1
2 √−1 1. Below is the complete factorization of 25
2 −1
2 √−11into its prime
factors. This is the decomposition of 25
2 −1
2 √−11 .
25
2 −1
2 √−11= 1
2 (25−1)√−11= 1
2 (5−1)(5+1)√−11.
iv) Show that R is a principal ideal domain (PID).
R is a principal ideal domain if every ideal of R can be generated by a single
element. Therefore, we need to show that every ideal of R can be generated by a
single element.

Let I be an Ideal of R. If I ={0 }, then I is generated by 0 and so I is a principal
ideal.
I f I ≠ {0 } then let a be the smallest integer such that a\ >0anda∈I. We will show
that I =¿〈a〉
Since a ∈ I we have 〈 a 〉⊆ I. Conversely, if b ∈ I then we have b=qa+r for
some q ,r ∈ Z ,0 ≤ r ≤ a−1. This gives r =b−qa, so r ∈ I. Since a is the smallest
positive element of I, this implies that r =0. Therefore b=qa, and so b ∈〈 a 〉.