Polynomial Rings - Units, Nilpotents and Zero divisors

Polynomial Rings - Units, Nilpotents and Zero divisors

M.Velrajan

Let R be a commutative ring with identity and R[x] be the ring of polynomials in

an indeterminate x, with coefficients in R. We find necessary and sufficient

conditions for a polynomial in x over R to be a unit, a nilpotent element and 

a zero divisor of the ring of polynomials R[x]. And we prove that

the nilradical of R[x] = the Jacobson radical of R[x].

Throughout R denotes a commutative ring with identity.

First we prove the property of nilpotent elements of R. 

1.  Let a be a nilpotent element in R. Then 1 + a is a unit in R. 

And in general, the sum of a unit and  a nilpotent element is a unit.

Proof    Since a is a nilpotent element in R, an = 0 for some integer n > 0. 

To prove 1 + a is a unit, we have to find y such that (1 + a)y = 1 using an = 0.

Think!! What is the relation between 1 + a, an and 1.

Yes, the simple algebraic relation . . .

(1 + a) (1 - a + a2 -  ∙ ∙ ∙ + (-1)n - 1an - 1) = 1 + (-1)n - 1an = 1 + 0 = 1.

Hence 1 + a is a unit in R.

Suppose u is a unit in R.

Suppose u is a unit and a is a nilpotent in R.

To prove u + a is a unit, we can use 1 + a is a unit if a is nilpotent.

So express u + a in terms of 1 + b. 

Yes, u + a = u(1 + u - 1a). Is u - 1a nilpotent ?

Then, since the set of all nilpotents in R is an ideal of R,

u - 1a is a nilpotent element in R.

Hence 1 + u - 1a is a unit in R.

Since the product of two units is a unit, u + a = u(1 + u - 1a) is a unit in R.

2.  Units of R[x]   

Let f = a0 + a1x + a2x2 + ∙ ∙ ∙ + anxn ∈ R[x].

Suppose f is a unit in R[x].

Then there exists b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm ∈ R[x] such that 

(a0 + a1x + a2x2 + ∙ ∙ ∙ + anxn )(b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) = 1.

i.e. a0b0 + (a0b1 + a1b0)x + ∙ ∙ ∙ + (a0br + a1br-1 + ∙ ∙ ∙ + arb0)xr + ∙ ∙ ∙ + anbmxn + m = 1,

where ai = 0 for all i > n and bj = 0 for all j > m.

Hence a0b0 = 1 i.e. a0 is a unit in R.

And a0br + a1br-1 + ∙ ∙ ∙ + arb0 = 0 for all r = 1, 2, ∙ ∙ ∙ , n + m, 

where ai = 0 for all i > n and bj = 0 for all j > m. — (A)

Taking r = n + m in (A), anbm = 0.

We use this with the coefficient of xn+m-1. 

Taking r = n + m - 1in (A) and multiplying both sides with an ,

(an - 1bm + anbm - 1)an = 0an = 0

i.e. an - 1bman + an2 bm - 1 = 0

i.e. an2 bm - 1 = 0.

Power of an increased by 1 and the suffix of b decreased by 1.

Taking r = n + m - 2 in (A) and multiplying both sides with an2 ,

(an - 2bm + an - 1bm - 1 + anbm - 2)an2 = 0an2 = 0

i.e. an - 2bman2 + an - 1bm - 1an2 + an3 bm - 2 = 0

i.e. an3 bm - 2 = 0.

Proceeding thus, we get an4 bm - 3 = 0 = an5 bm - 4 = ∙ ∙ ∙ = anm + 1 b0.

Combine this with a0b0 = 1

anm + 1 b0 = 0 ⇒ anm + 1 b0a0 = 0a0  

                   ⇒ anm + 1 = 0 (since a0b0 = 1)

                   ⇒ an is a nilpotent element in R.

Hence an is a nilpotent element in R[x].

Since – xn is in R[x] and the set of all nilpotents in R[x] is an ideal,

– anxn is a nilpotent element in R[x].

Since sum of a unit and a nilpotent element is a unit,

f – anxn is a unit in R[x].

i.e. a0 + a1x + a2x2 + ∙ ∙ ∙ + an - 1xn - 1 is a unit in R[x].

Proceeding thus, we get an - 1 is a nilpotent element in R and 

a0 + a1x + a2x2 + ∙ ∙ ∙ + an - 2xn - 2 is a unit in R[x];

an - 2 is a nilpotent element in R and 

a0 + a1x + a2x2 + ∙ ∙ ∙ + an - 3xn - 3 is a unit in R[x];

 ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ;

a1 is a nilpotent element in R.

Hence 

f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a unit in R[x] ⇒ a0 is a unit and a1, a2, . . . , an are

nilpotent elements in R.

Conversely, suppose a0 is a unit, a1, a2, . . . , an are nilpotent elements in R and

f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn.

Then a0 is a unit and a1, a2, . . . , an are nilpotent elements in R[x].

Since the set of all nilpotents elements in R[x] is an ideal,

a1x, a2x2, ∙ ∙ ∙ , an xn are nilpotent elements in R[x] and hence 

a1x + a2x2 + ∙ ∙ ∙ + an xn is a nilpotent element in R[x].

Since a0 is a unit in R[x], f = a0 + (a1x + a2x2 + ∙ ∙ ∙ + an xn) is a unit in R[x].

Therefore, 

f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a unit in R[x] ⟺ a0 is a unit and a1, a2, . . . , an

are nilpotent elements in R.

3.  Nilpotents of R[x]

Suppose f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a nilpotent element in R[x].

We know the units of R[x], so use the property of nilpotents.

Then 1 + f = (1 + a0) + a1x + a2x2 + ∙ ∙ ∙ + an xn is a unit in R[x].

Hence a1, a2, . . . , an are nilpotent elements in R and hence a1x, a2x2, ∙ ∙ ∙ , an xn are

nilpotent elements in R[x] and a1x + a2x2 + ∙ ∙ ∙ + an xn is a nilpotent element in R[x].

So a0 = f - (a1x + a2x2 + ∙ ∙ ∙ + an xn) is a nilpotent element in R[x].

i.e. a0 is a nilpotent element in R.

Hence f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a nilpotent element in R[x] 

                                ⇒ a0, a1, a2, . . . , an are nilpotent elements in R.

Conversely, suppose a0, a1, a2, . . . , an are nilpotent elements in R and 

f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn.

Then a0, a1, a2, . . . , an are nilpotent elements in R[x].

Since the set of all nilpotents elements in R[x] is an ideal,

a0, a1x, a2x2, ∙ ∙ ∙ , an xn are nilpotent elements in R[x] and 

f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a nilpotent element in R[x].

Therefore, 

f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a nilpotent element in R[x] ⟺ a0, a1, a2, . . . , an

are nilpotent elements in R.

4.  Zero Divisors of R[x]

Suppose f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a zero divisor in R[x].

Then there exists g ≠ 0 in R[x] such that fg = 0.

Choose a polynomial g = b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm of least degree m such that

fg = (a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn)(b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) = 0. 

Then a0b0 + (a0b1 + a1b0)x + ∙ ∙ ∙ + (a0br + a1br-1 + ∙ ∙ ∙ + arb0)xr + ∙ ∙ ∙ + anbmxn + m = 0,

where ai = 0 for all i > n and bj = 0 for all j > m.

Hence a0br + a1br-1 + ∙ ∙ ∙ + arb0 = 0 for all r = 0, 1, 2, . . . , n + m. — (A)

Taking r = n + m in (A), anbm = 0.

Hence ang = anb0 + anb1x + anb2x2 + ∙ ∙ ∙ + anbm - 1xm - 1 is a polynomial of 

degree m - 1 and f(ang) = anfg = an0 = 0.

Therefore, by the choice of g, ang = 0.

Claim :  aig = 0, for all i = 0, 1, 2, . . . , n.

Suppose it is proved that ang = 0, an - 1g = 0, . . . , an - kg = 0.

Then, taking r = m + n - k +1, in (A), 

a0bm + n - k + 1 + a1bm + n - k  + ∙ ∙ ∙ + am + n - k + 1b0 = 0

an - k-1bm + an - kbm -1 + ∙ ∙ ∙ + anbm-k-1 = 0

Since ang = 0, an - 1g = 0, . . .  and an - kg = 0, 

anbm-k-1 = 0, an - 1bm -k = 0,  ∙ ∙ ∙ , an - kbm -1 = 0. Hence an - k-1bm = 0.

Hence an - k - 1g = an - k - 1b0 + an - k - 1b1x + an - k - 1b2x2 + ∙ ∙ ∙ + an - k - 1bm - 1xm - 1 is a

polynomial of degree m - 1 and f(an - k - 1g) = an - k - 1fg = an - k - 10 = 0.

Therefore, by the choice of g, an - k - 1g = 0.

Hence, by proceeding thus, aig = 0, for all i = 0, 1, 2, . . . , n and hence the claim.

Therefore, aibj = 0, for all i = 0, 1, 2, . . . , n and j = 0, 1, 2, . . . , m.

Since g ≠ 0, there exists at least one j such that bj ≠ 0.

Let a = bj ≠ 0. Then aia = 0, for all i = 0, 1, 2, . . . , n and hence af = 0.

Conversely, suppose there exists a ≠ 0 ∈ R such that af = 0 then, since a ≠ 0 ∈ R[x],

f is a zero divisor in R[x].

Therefore,

f  = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a zero divisor in R[x] 

         ⟺ there exists a ≠ 0 ∈ R such that af = 0.

         ⟺ there exists a ≠ 0 ∈ R such that aa0 = aa1 = aa2 = ∙ ∙ ∙ = aan  = 0.

4.1. R[x] has non zero zero divisors ⟺ R has non zero zero divisors.

i.e. R[x] is an integral domain ⟺ R is an integral domain.

5.  Suppose the ring R has no nonzero nilpotents. Then, by 2.and 3., 

i) only the units of R are the units of R[x].

ii) R[x] has no nonzero nilpotent elements.

5.1.  Suppose the ring R has a nonzero nilpotent. Then, by 2.and 3., 

i) There are infinite numbers of units of R[x] (irrespective of whether R has a finite

number of units or infinite number of units). 

ii) R[x] has infinite numbers of nonzero nilpotent elements and hence infinite

numbers of nonzero zero divisors.

6.  Suppose the ring R is an integral domain. Then R has no nonzero zero divisors.

Since any nonzero nilpotent is a nonzero zero divisor, R has no nonzero nilpotents.

Hence, by 5., 

i) only the units of R are the units of R[x].

ii) R[x] has no nonzero nilpotent elements.

And by 4.1.,  

iii) R[x] has no nonzero zero divisors and hence R[x] is an integral domain.

7.  If F is a field then F is an integral domain. Hence, by 6., 

i) only the units of F(i.e. the nonzero elements of F) are the units of F[x].

 Hence F[x] is not a field.

ii) F[x] has no nonzero nilpotent elements.

iii) F[x] has no nonzero zero divisors and hence F[x] is an integral domain.

8.  Recall :

The ideal of all the nilpotent elements of R is called the nilradical of R.

And the nilradical of R is the intersection of all prime ideals of R.

The intersection of all maximal ideals of R is called the Jacobson radical of R.

We denote the nilradical of R by 𝕹 and the Jacobson radical of R by 𝕽.

Since any maximal ideal of R is a prime ideal of R, 𝕹 ⊆ 𝕽.

In general 𝕹 ≠ 𝕽. But,

9.  In the polynomial ring R[x], 𝕹 = 𝕽.

Proof  We have to prove 𝕽 ⊆ 𝕹.

Let f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ 𝕽, the Jacobson radical of R[x].

Then 1 - fg is a unit in R[x], for all g ∈ R[x].

(since in a ring R, a ∈ 𝕽 ⟺ 1- ab is a unit in R, for all b ∈ R)

Hence 1 + fx = 1 - f(-x) is a unit in R[x].

i.e. 1 + a0x + a1x2 + a2x3 + ∙ ∙ ∙ + an xn+1 is a unit in R[x].

Hence, by 2., a0, a1, a2, . . . , an are nilpotent elements in R and hence, by 3., 

f is a nilpotent element in R[x]. i.e. f ∈ 𝕹, the nilradical of R[x].

Hence 𝕽 ⊆ 𝕹.

But in general 𝕹 ⊆ 𝕽.

Therefore, 𝕹 = 𝕽 in R[x]. 

Without examples, learning is incomplete.

10.   i)  In the ring ℤ4 of integers modulo 4 only 1 and 3 are the units.

Since 22 = 0 only 2 is the nonzero nilpotent and nonzero zero divisor of ℤ4.

Hence 

{a0 + a1x + ∙ ∙ ∙ + an xn / n ≥ 0, a0 ∈ {1, 3} and ai ∈ {0, 2}, i = 1, . . . , n}

is the group GL1ℤ4[x] (or ℤ4[x]× ) of all units of ℤ4[x].

And {a0 + a1x + ∙ ∙ ∙ + an xn / n ≥ 0, ai ∈ {0, 2}, i = 0, 1, . . . , n}

is both the nilradical and the set of all zero divisors of ℤ4[x].

ii)  In the ring ℤ6 of integers modulo 6 only 1 and 5 are the units.

ℤ6 has no nonzero nilpotents. And 2, 3 and 4 are nonzero zero divisors of ℤ6.

In fact 2.3 = 0 = 3.4 and 2.4 = 2.

Hence only 1 and 5 are the units of ℤ6[x]. 

And ℤ6[x] has no nonzero nilpotent elements.

If n ≥ 0 and ai ∈ {0, 2, 4}, i = 0, 1, . . . , n and at least one ai ≠ 0 then 

a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a non zero zero divisor of ℤ6 (since 3.2 = 0 = 3.4).

And if n ≥ 0 and ai ∈ {0, 3}, i = 0, 1, . . . , n and at least one ai ≠ 0 then 

a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is a non zero zero divisor of ℤ6.

But if n ≥ 0 and ai ∈ {0, 2, 3}, i = 0, 1, . . . , n and at least one ai ≠ 0 and if ai = 2

and aj = 3 for at least one i ≠ j then a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is not a zero divisor

of ℤ6(since there is no a ≠ 0 ∈ ℤ6 such that a.2 = 0 and a.3 = 0).

Also if n ≥ 0 and ai ∈ {0, 3, 4}, i = 0, 1, . . . , n and at least one ai ≠ 0 and if ai = 3

and aj = 4 for at least one i ≠ j then a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn is not a zero divisor

of ℤ6(since there is no a ≠ 0 ∈ ℤ6 such that a.3 = 0 and a.4 = 0).

In particular, 2 + 3x is not a zero divisor of ℤ6 and 

3 + 4x is not a zero divisor of ℤ6.

See how examples enlighten the understanding of the theory.

iii)  In the ring ℤ8 of integers modulo 8 only 1, 3, 5 and 7 are the units.

Since 23 = 0, 42 = 0, 63 = 0, 2, 4 and 6 are the nonzero nilpotents of ℤ8.

And hence 2, 4 and 6 are nonzero zero divisors of ℤ8.

Also 2.4 = 0 = 4.4 = 6.4.

Hence 

{a0 + a1x + ∙ ∙ ∙ + an xn / n ≥ 0, a0 ∈ {1, 3, 5, 7} and ai ∈ {0, 2, 4, 6}, i = 1, . . . , n}

is the group GL1ℤ8[x] (or ℤ8[x]× ) of all units of ℤ8[x].

And {a0 + a1x + ∙ ∙ ∙ + an xn / n ≥ 0, ai ∈ {0, 2, 4, 6}, i = 0, 1, . . . , n}

is both the nilradical and the set of all zero divisors of ℤ8[x].

iv)  ℤ is an integral domain and only 1 and -1 are the units of ℤ. 

Hence only 1 and -1 are the units of ℤ[x]. And ℤ[x] has no nonzero nilpotent

elements. Also ℤ[x] is an integral domain.

So Learners :  

Understand, Think Mathematically, Raise questions, Try to get answers. 

This will make your learning Enjoyable.