On Ideals of Polynomial Rings

 

On Ideals of Polynomial Rings

M.Velrajan

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

an indeterminate x, with coefficients in R. Then R[x] is also a commutative ring

with identity.

Necessary and sufficient conditions for a polynomial in x over R to be a unit, 

a nilpotent element and to be a zero divisor of the ring of polynomials R[x] are

discussed in the post : 

https://velrajanm.blogspot.com/2025/03/polynomial-rings-units-nilpotents-and.html. 

Now we discuss some relations between the ideals of R and the ideals of R[x].

We prove that for each non negative integer m ≥ 0, the map I → I{m}[x] is a 

one to one map that preserves the order ⊆ from the set of all ideals of R onto 

the set of ideals of R[x] that contain 

{a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0},

the ideal of all polynomials in R[x] with coefficient of xm = 0, where

I{m}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / am ∈ I, ai ∈ R if i ≠ m and n ≥ 0}.

Also we prove that the polynomial ring R[x] has infinitely many maximal ideals.

In fact, suppose m ≥ 0 is a non negative integer then 

I is a maximal ideal of R if and only if I{m}[x] is a maximal ideal of R[X].

Throughout R denotes a commutative ring with identity.

Suppose I is an ideal of R. Then I ⊆ R ⊆ R[x]. But I is not an ideal of R[x].

So, what is the smallest ideal of R[x] containing I ? 

If we denote it by I[x] then : Is R[x]/I[x] isomorphic to (R/I)[x] ?

Think !!

1.  Let I be an ideal of R. Then 

I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0, a1, a2, . . . , an ∈ I and n ≥ 0}

is an ideal of R[x]. In fact, I[x] is the smallest ideal of R[x] containing I. 

i.e. I[x] is the ideal generated by I in R[x].

And R[x]/I[x] is isomorphic to (R/I)[x].

Proof  It can be verified that I[x] is an ideal of R[x].

Clearly I ⊆ I[x].

Suppose K is an ideal of R[x] and I ⊆ K.

Then for any n ≥ 0 and a0, a1, a2, . . . , an ∈ I, a0, a1, a2, . . . , an ∈ K.

Since K is an ideal of R[x], a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ K.

Hence I[x] ⊆ K.

Therefore, I[x] is the smallest ideal of R[x] containing I.

i.e. I[x] is the ideal generated by I in R[x].

Think how to define naturally a map from R[x]/I[x] into (R/I)[x].

Define 𝛗((a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn) + I[x]) 

                                          = (a0 + I) + (a1 + I)x + (a2 + I)x2 + ∙ ∙ ∙ + (an + I) xn.

Let (a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn), (b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) ∈ R[x].

Without loss of generality assume that n ≤ m. Then

(a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn) + I[x] = (b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) + I[x]

⟺ (a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn)  – (b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) ∈ I[x]

⟺ (a0 - b0) + (a1 - b1)x + ∙ ∙ ∙ + (an - bn) xn  – bn+1xn+1 – bn+2xn+2 – ∙ ∙ ∙ – bmxm ∈ I[x]

⟺ (a0 - b0), (a1 - b1), ∙ ∙ ∙ , (an - bn), – bn+1, – bn+2, ∙ ∙ ∙ , – bm ∈ I

⟺ a0 + I = b0 + I, a1 +I = b1 + I, ∙ ∙ ∙ , an + I =  bn + I, 

               bn+1 + I = I = bn+2 + I =  ∙ ∙ ∙ = bm + I = I

⟺ (a0 + I) + (a1 + I)x + (a2 + I)x2 + ∙ ∙ ∙ + (an + I) xn = (b0 + I) + (b1 + I)x + 

           (b2 + I)x2 + ∙ ∙ ∙ + (bn + I) xn  + (bn+1 + I)xn+1 + (bn+2 + I)xn+2 + ∙ ∙ ∙ + (bm + I) xm     

⟺ 𝛗((a0 + a1x + ∙ ∙ ∙ + an xn) + I[x]) = 𝛗((b0 + b1x + ∙ ∙ ∙ + bmxm) + I[x])

Hence 𝛗 is well defined and one-one.

Do you know, Why we are proving the well definedness of 𝛗 ?

Yes, the elements of R[x]/I[x] are cosets or equivalence classes, so they can be 

expressed by more than one element.

Whatever be they are represented by, as 𝛗 is a map the image should be

unique.

That is why we have proved

 (a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn) + I[x] = (b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) + I[x]

⇒ 𝛗((a0 + a1x + ∙ ∙ ∙ + an xn) + I[x]) = 𝛗((b0 + b1x + ∙ ∙ ∙ + bmxm) + I[x]),

for 𝛗 is well defined.

It can be verified that 𝛗 is a homomorphism and onto.

Now if I is a prime or maximal ideal of R, what about I[x] ?

2.  An ideal I of R is prime if and only if I[x] is a prime ideal in R[X]. 

Proof   By 1., I[x] is an ideal of R[x] and R[x]/I[x] is isomorphic to (R/I)[x].

Now I is prime ⟺ R/I is an integral domain

                         ⟺ (R/I)[x] is an integral domain (ref. 4.1. of

https://velrajanm.blogspot.com/2025/03/polynomial-rings-units-nilpotents-and.html) 

                         ⟺ R[x]/I[x] is an integral domain

                         ⟺ I[x] is a prime ideal of R[x].

Aliter : 

Suppose P is a prime ideal of R. 

Then P[x] is an ideal of R[x] and, since P ≠ R, P[x] ≠ R[x].

Suppose (a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn)(b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm) ∈ P[x] and

a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn ∉ P[x].  

Then a0b0, a0b1 + a1b0,  ∙ ∙ ∙  , a0br + a1br-1 + ∙ ∙ ∙ + arb0, ∙ ∙ ∙ , anbm ∈ P and

ai ∉ P for at least one i.

Suppose i is the least integer such that ai ∉ P. 

Then 0 ≤ i ≤ n, a0, a1, a2, . . . , ai-1 ∈ P and ai ∉ P.

Since a0bi + a1bi-1 + ∙ ∙ ∙ + aib0 ∈ P and a0, a1, a2, . . . , ai-1 ∈ P,

aib0 = (a0bi + a1bi-1 + ∙ ∙ ∙ + aib0) - (a0bi + a1br-1 + ∙ ∙ ∙ + ai-1b1) ∈ P.

Since P is prime and ai ∉ P, b0 ∈ P.

Since a0bi+1 + a1bi + ∙ ∙ ∙ + aib1 + ai+1b0 ∈ P and a0, a1, a2, . . . , ai-1, b0 ∈ P,

aib1 = (a0bi+1 + a1bi + ∙ ∙ ∙ + ai+1b0) - (a0bi+1 + a1br-1 + ∙ ∙ ∙ + ai-1b2 + ai+1b0) ∈ P.

Since P is prime and ai ∉ P, b1 ∈ P.

Suppose b0 ∈ P, b1 ∈ P, . . . , bj ∈ P for some 0 ≤ j < m.

Since a0bi+j+1 + a1bi+j + ∙ ∙ ∙ + ai-1bj+2 + aibj+1 + ai+1bj + ∙ ∙ ∙ + ai+j+1b0 ∈ P and 

a0, a1, a2, . . . , ai-1, b0, b1, b2, . . . , bj ∈ P,

aibj+1 = (a0bi+j+1 + a1bi+j + ∙ ∙ ∙ + ai-1bj+2 + aibj+1 + ai+1bj + ∙ ∙ ∙ + ai+j+1b0) 

            - (a0bi+j+1 + a1bi+j + ∙ ∙ ∙ + ai-1bj+2 + ai+1bj + ∙ ∙ ∙ + ai+j+1b0) ∈ P.

Since P is prime and ai ∉ P, bj+1 ∈ P.

Hence, proceeding thus b0, b1, b2, . . . , bm ∈ P, and hence 

b0 + b1x + b2x2 + ∙ ∙ ∙ + bmxm ∈ P[x].

Therefore P[x] is a prime ideal of R[x].

Conversely, suppose P[x] is a prime ideal of R[x].

Then P[x] ≠ R[x], hence P ≠ R.

And ab ∈ P ⊆ P[x]. Then either a ∈ P[x] or b ∈ P[x].

Since a and b are constant polynomials, either a ∈ P or b ∈ P.

Hence P is prime.

3.  Suppose M is a maximal ideal of R. Then M[x] is not a maximal ideal of R[x].

Proof     By 1., M[x] is an ideal of R[x] and R[x]/M[x] is isomorphic to (R/M)[x].

Since M is maximal, R/M is a field and hence 

(R/M)[x] is an integral domain but it is not a field (ref. 7. of 

https://velrajanm.blogspot.com/2025/03/polynomial-rings-units-nilpotents-and.html). 

Hence R[x]/M[x] is not a field.

Therefore, M[x] is not a maximal ideal of R[x].

The natural question to be arised within us is :

4.  Suppose K ≠ R[x] is an ideal of R[x]. 

Does there exist an ideal I of R such that K = I[x] ?

i.e. Is K the ideal generated by some ideal I of R in R[x] ?

Let K = {xf / f ∈ R[x]} = (x), the principal ideal generated by x in R[x].

Then K ≠ R[x] and K ≠ {0}. In fact,

(x) = {a1x + a2x2 + ∙ ∙ ∙ + an xn / a1, a2, . . . , an∈ R and n ≥ 1}

consists of all polynomials in R[x] with constant term 0.

For any ideal I ≠ {0}of R, I[x] contains the constant polynomial a, for all a ∈ I, 

and (x) contains no nonzero constant polynomial.

Hence K ≠ I[x] for all ideals I ≠ {0} of R. Also {0}[x] = {0}.

Thus, there exists no ideal I of R such that K = I[x].    

Hence, 

if K is an ideal of R[x] and K ≠ R[x] then K need not be I[x], for some ideal I

of R. i.e. K need not be the ideal generated by some ideal I of R in R[x].

4.1. We note that (x) = {a1x + a2x2 + ∙ ∙ ∙ + an xn / a1, a2, . . . , an ∈ R and n ≥ 1}

consists of only non units of R[x]. (ref. 2 of

https://velrajanm.blogspot.com/2025/03/polynomial-rings-units-nilpotents-and.html). 

For each ideal K of R[x] we can get associated ideals of R.

5.  Suppose K is an ideal of R[x] and n ≥ 0. 

Let In(K) be the set of coefficients of xn of all the polynomials in K.

Then In(K) is an ideal of R. In particular the set I0(K) of constant terms of all the

polynomials in K is an ideal.

Proof  Since K is non empty, In(K) is non empty.

If a, b ∈ In(K) and r ∈ R then there exist f, g ∈ K such that a and b are the

coefficients of xn of the polynomials f, g respectively.

Since K is an ideal of R[x] and r ∈ R ⊆ R[x], f + g ∈ K and rf ∈ K.

And a + b and ra are the coefficients of xn of the polynomials f + g and rf

respectively. Hence a + b, ra ∈ In(K).

5.1.  Suppose K is an ideal of R[x].

Let Ilead(K) be the set of leading coefficients of all the polynomials in K.

Then Ilead(K) is an ideal of R.

Proof  If a, b ∈ Ilead(K) and r ∈ R then there exist f, g ∈ K such that a and b are the

leading coefficients of the polynomials f, g respectively.

Suppose degree of f is m and degree of g is n. 

Without loss of generality we assume that m ≤ n.

Since K is an ideal of R[x] and xn - m ∈ R[x] and r ∈ R ⊆ R[x], 

xn - mf + g ∈ K and rf ∈ K. 

And a + b and ra are the leading coefficients of the polynomials xn - mf + g and rf

respectively. 

Hence a + b, ra ∈ Ilead(K).

6. We call I0(K) as the ideal of constant terms of all the polynomials in K 

or the ideal of constant terms of K.

And In(K) as the ideal of coefficients of xn of all the polynomials in K 

or the ideal of coefficients of xn of K.

6.1.  Suppose K is an ideal of R[x]. Then In(K) ⊆ In+1(K), for all n ≥ 0.

For,   If a ∈ In(K) then there exists f ∈ K such that a is the coefficient of xn of f.

Since K is an ideal of R[x], xf ∈ K and the coefficient of xn+1 of xf is a.

Hence a ∈ In+1(K) and hence In(K) ⊆ In+1(K).

6.2.  Suppose K = (x) = {a1x + a2x2 + ∙ ∙ ∙ + an xn / a1, a2, . . . , an ∈ R and n ≥ 1}

is the principal ideal (x) generated by x in R[x].

Then the ideal of constant terms of K = {0} and 

the ideal of coefficients of xn of K = R. But K ≠ {0}[x] = {0} and K ≠ R[x].

6.3.  If I is an ideal of R then 

the ideal of constant terms of I[x] = I = the ideal of coefficients of xn of I[x].

6.4.  Suppose I is an ideal of R. Then the ideal of constant terms of (x) + I[x] = I.

For,  I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0, a1, a2, . . . , an ∈ I and n ≥ 0} and the

principal ideal (x) generated by x in R[x] is given by

(x) = {a1x + a2x2 + ∙ ∙ ∙ + an xn / a1, a2, . . . , an ∈ R and n ≥ 1}. It can be verified that 

(x) + I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0 ∈ I, a1, a2, . . . , an ∈ R and n ≥ 0}.

Clearly, 

the ideal of constant terms of (x) + I[x] = I = the ideal of constant terms of I[x].

6.5.  Suppose K is an ideal of R[x] and I = I0(K) is the ideal of constant terms of K.

Then K ⊆ (x) + I[x].

For,  If f = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ K then a0 ∈ I and hence f ∈ (x) + I[x].

Therefore, K ⊆ (x) + I[x].

From 6.4. and 6.5. it follows that : 

6.6.  Suppose I is an ideal of R. Then (x) + I[x] is the largest ideal of R[x] having I as the

ideal of constant terms.

6.7.  Suppose K is an ideal of R[x] that contains the principal ideal (x) and I = I0(K) is the ideal of constant terms of K. Then K = (x) + I[x].

For,  By 6.5., K ⊆ (x) + I[x].

If a ∈ I then there exists a1, a2, . . . , an ∈ R such that 

a + a1x + a2x2 + ∙ ∙ ∙  + an xn ∈ K. 

Since a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ (x) ⊆ K,

a = (a + a1x + a2x2 + ∙ ∙ ∙ + an xn) - (a1x + a2x2 + ∙ ∙ ∙ + an xn) ∈ K.

Hence I ⊆ K.

Since I[x] is the smallest ideal of R[x] that contains I, I[x] ⊆ K.

Since (x) ⊆ K, (x) + I[x] ⊆ K.

Therefore, K = (x) + I[x].

7.  We know that in the ring of integers ℤ, the ideal 2ℤ, generated by the prime

number 2 is a maximal ideal. By 3., 2ℤ[x] is not a maximal ideal of ℤ[x].

So, What is a maximal ideal of ℤ[x] that contains 2ℤ[x] ? 

To get a maximal ideal that contains 2ℤ[x] we can add non units of ℤ[x] with

2ℤ[x]. So, we try with (x) an ideal of non units of ℤ[x].

Let (x) = {a1x + ∙ ∙ ∙ + an xn / ai ∈ ℤ for all i and n ≥ 0} be the principal ideal

generated by x in ℤ[x]. 

Then 2ℤ[x] + (x) is an ideal of ℤ[x] containing 2ℤ[x].

We note that 2ℤ[x] + (x) consists of all polynomials in ℤ[x] with even constant

term i.e. 2ℤ[x] + (x) = {a0 + a1x + ∙ ∙ ∙ + an xn / a0∈ 2ℤ, ai ∈ ℤ if i ≠ 0 and n ≥ 0}.

Since x ∈ (x), x ∈ 2ℤ[x] + (x). But x ∉ 2ℤ[x]. Hence 2ℤ[x] + (x) ≠ 2ℤ[x].

Since 1 ∉ 2ℤ[x] + (x), 2ℤ[x] + (x) ≠ ℤ[x].

Suppose K is an ideal of ℤ[x] containing 2ℤ[x] + (x) and K ≠ 2ℤ[x] + (x).

Then there exists a polynomial f(x) = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ K, with the odd

constant term a0. Suppose a0 = 2m + 1. 

Then f(x) = 1 + 2m + a1x + a2x2 + ∙ ∙ ∙ + an xn = 1 + g(x),

where g(x) = 2m + a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ 2ℤ[x] + (x) ⊆ K.

Hence 1 = f(x) - g(x) ∈ K and hence K = ℤ[x]. 

Therefore, 2ℤ[x] + (x) is a maximal ideal of ℤ[x]. 

So, for the maximal ideal 2ℤ of the ring ℤ,

2ℤ[x] is not a maximal ideal of ℤ[x], but 2ℤ[x] + (x) is a maximal ideal of ℤ[x]. 

We note that 2ℤ[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / ai∈ 2ℤ for all i and n ≥ 0} and

2ℤ[x] + (x) = {a0 + a1x + ∙ ∙ ∙ + an xn / a0∈ 2ℤ, ai ∈ ℤ if i ≠ 0 and n ≥ 0}.

The ideal of constant term of both 2ℤ[x] and 2ℤ[x] + (x) is 2ℤ.

Now the question arise is : 

Is I[x] + (x) a maximal ideal of R[x], for all maximal ideals I of a ring R ?

8.  Suppose I is a maximal ideal of R. Then 

(x) + I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0 ∈ I, a1, a2, . . . , an ∈ R and n ≥ 0} and

the ideal of constant terms of (x) + I[x] is I. — (i) 

Since I is maximal, I ≠ R hence there exists a ∈ R - I.  

By (i), a ∉ (x) + I[x]. But a ∈ R[x]. 

Hence (x) + I[x] ≠ R[x].

Suppose K is an ideal of R[x] such that (x) + I[x] ⊆ K ⊆ R[x] and K ≠ (x) + I[x].

Then there exists f(x) = b0 + b1x + b2x2 + ∙ ∙ ∙ + bn xn ∈ K - ((x) + I[x]).

Clearly, b0 ∉ I. Hence I + Rb0 ≠ I. Since I is maximal, I + Rb0 = R.

Hence 1 = a + rb0, for some a ∈ I and r ∈ R.

Since b1x + b2x2 + ∙ ∙ ∙ + bn xn ∈ (x) ⊆ (x) + I[x] ⊆ K,

b0 = (b0 + b1x + b2x2 + ∙ ∙ ∙ + bn xn) - (b0 + b1x + b2x2 + ∙ ∙ ∙ + bn xn) ∈ K.

Since a ∈ I ⊆ I[x] ⊆ (x) + I[x] ⊆ K, 1 = a + rb0 ∈ K.

Hence K = R]x].

Therefore, (x) + I[x] is a maximal ideal of R[x].

So, if I is a maximal ideal of R then (x) + I[x] is a maximal ideal of R[x].

Is the converse true?

Suppose (x) + I[x] is a maximal ideal of R[x]. 

Then (x) + I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0 ∈ I, a1, a2, . . . , an ∈ R and n ≥ 0}

                          ≠ R[x]. 

If I = R then (x) + I[x] = R[x]. Hence I ≠ R. 

Suppose J is an ideal of R such that I ⊆ J ⊆ R and J ≠ I.

Then I[x] ⊆ J[x] and I[x] ≠ J[x]. Hence

(x) + I[x] ⊆ (x) + J[x] and (x) + I[x] ≠ (x) + J[x].

Since (x) + I[x] is maximal, (x) + J[x] = R[x].

Hence any a ∈ R is a constant polynomial in (x) + J[x].

Since the ideal of constant terms of (x) + J[x] is J, a ∈ J. 

So, R ⊆ J and hence J = R.

Therefore I is a maximal ideal of R.

So, we have proved that : 

An ideal I of R is a maximal ideal of R if and only if (x) + I[x] is a maximal

ideal of R[x].

8.1.  The principal ideal (x) generated by x in R[x] is a maximal ideal if and only if

R is a field.

Proof   Since (x) = (x) + {0}[x], 

(x) is a maximal ideal ⟺ {0} is a maximal ideal in R ⟺ R is a field.

We observe in the proof of 8., that the map I → (x) + I[x]

is a one to one map that preserves the order ⊆ from the set of all ideals of R

into the set of ideals of R[x] that contain the principal ideal (x) generated by x

in R[x]. 

In fact, we have : 

9.  The map I → (x) + I[x] is a one to one map that preserves the order ⊆ from the

set of all ideals of R onto the set of ideals of R[x] that contain the principal ideal

(x) generated by x in R[x]. 

And {0}, R and I ≠ R are mapped respectively to (x), R[x] and (x) + I[x] ≠ R[x].

Proof   Let I be an ideal of R. Then

(x) + I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0 ∈ I, a1, a2, . . . , an ∈ R and n ≥ 0}.

If I ≠ R then (x) + I[x] ≠ R[x].

If I = R then (x) + I[x] = R[x] and if I = {0} then (x) + I[x] = (x).

Suppose I, J are ideals of R. 

If I ⊆ J then I[x] ⊆ J[x] and hence (x) + I[x] ⊆ (x) + J[x].

If I ≠ J then I[x] ≠ J[x] and hence (x) + I[x] ≠ (x) + J[x]. 

Thus I → (x) + I[x] is a one to one map that preserves the order ⊆ from the set of

all ideals of R into the set of ideals of R[x] that contain the principal ideal (x)

generated by x in R[x].

Suppose K is an ideal of R[x] that contains the principal ideal (x).

Let I = I0(K), the ideal of constant terms of K.

Then, by 6.7., K = (x) + I[x].

Thus, I → (x) + I[x] is a one to one map that preserves the order ⊆ from the set of

all ideals of R onto the set of ideals of R[x] that contain the principal ideal (x)

generated by x in R[x]. 

And {0}, R and I ≠ R are mapped respectively to (x), R[x] and (x) + I[x] ≠ R[x].

 

9.1.  Using the one- one and onto map I → (x) + I[x] we can rewrite the proof

of if I is maximal then (x) + I[x] is maximal as follows : 

Suppose I is a maximal ideal of R. Then 

(x) + I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0 ∈ I, a1, a2, . . . , an ∈ R and n ≥ 0} 

Since I is maximal, I ≠ R hence there exists a ∈ R - I.  

Clearly, a ∉ (x) + I[x]. But a ∈ R[x]. 

Hence (x) + I[x] ≠ R[x].

Suppose K is an ideal of R[x] such that (x) + I[x] ⊆ K ⊆ R[x] and K ≠ (x) + I[x].

Then there exists an ideal J of R such that K = (x) + J[x].

Since I ⊆ I[x] ⊆ (x) + I[x] ⊆ (x) + J[x], any a ∈ I is a constant polynomial 

in (x) + J[x], hence a ∈ J. So, I ⊆ J.

Since (x) + J[x] = K ≠ (x) + I[x], J[x] ≠ I[x]. Hence J ≠ I.

Since I is maximal, J = R. Hence K = (x) + J[x] = (x) + R[x] = R[x].

Therefore (x) + I[x] is a maximal ideal of R[x].

We note that for any ideal I of R, the ideal

(x) + I[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / a0 ∈ I, a1, a2, . . . , an ∈ R and n ≥ 0}

is the largest ideal of R[x] having I as the ideal of constant terms.

Compare (x) + I[x] and R[x]. 

Yes, only the constant term of the polynomials in (x) + I[x] are restricted to I.

Of course constant term = coefficient of x0.

So, why constant terms ? Why not coefficients of xm, for any m ≥ 0 ?

First we generalise (x) + I[x] as follows :

10.  Let I be an ideal of R. Let L be any subset of {0, 1, 2, . . . , n, . . .}.

Let IL[x] = {a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn / ai ∈ I if i ∈ L, ai ∈ R if i ∉ L and n ≥ 0}.

Then IL[x] is an ideal of R[x] containing I[x].

If L is empty then IL[x] = R[x].

If L is any proper subset of {0, 1, 2, . . . , n, . . .} then IL[x] ≠ R[x]

If L = {0, 1, 2, . . . , n, . . .} then IL[x] = I[x].

If L ⊆ M ⊆ {0, 1, 2, . . . , n, . . .} then IM[x] ⊆ IL[x].

For any m ∈ {0, 1, 2, . . . , n, . . .},

I{m}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / am ∈ I, ai ∈ R if i ≠ m and n ≥ 0}

           = {a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0} + I[x]

           = {0}{m}[x] + I[x]. 

And I{0}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / a0 ∈ I, ai ∈ R if i ≠ 0 and n ≥ 0} = (x) + I[x].

So, in 9., can we replace (x) + I[x] = I{0}[x] by

I{m}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / am ∈ I, ai ∈ R if i ≠ m and n ≥ 0} 

           = {0}{m}[x] + I[x] ?

We start to do it from 6.6.  

11.   Suppose I is an ideal of R. Then I{m}[x] is the largest ideal of R[x]

having I as the ideal of coefficients of xm.

For,  Since I{m}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / am ∈ I, ai ∈ R if i ≠ m and n ≥ 0},

the ideal of coefficients of xm of  I{m}[x] = I.

Suppose K is an ideal of R[x] and I is the ideal of coefficients of xm of K.

If f(x) = a0 + a1x + a2x2 + ∙ ∙ ∙ + an xn ∈ K then am ∈ I and hence f(x) ∈ I{m}[x].

Therefore, K ⊆ I{m}[x].

Hence I{m}[x] is the largest ideal of R[x] having I as the ideal of coefficients of xm.

11.1.  Suppose K is an ideal of R[x] that contains {0}{m}[x] and I is the ideal of coefficients of xm of K. Then K = I{m}[x].

For,  By 11., K ⊆ I{m}[x].

If a ∈ I then there exists a0, a1, a2, . . . , am-1, am+1, . . . , an ∈ R such that 

a0 + a1x + a2x2 +∙ ∙ ∙  +  axm + ∙ ∙ ∙  + an xn ∈ K. 

Since a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn ∈ {0}{m}[x] ⊆ K,

axm = (a0 + a1x + ∙ ∙ ∙ + an xn) - (a0 + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn) ∈ K.

Hence if a ∈ I then axm ∈ K.

Now, suppose a0 + a1x + ∙ ∙ ∙ + an xn ∈ I{m}[x] then am ∈ I and ai ∈ R for i ≠ m.

Since am ∈ I, amxm ∈ K. Since ai ∈ R for i ≠ m, 

a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn ∈ {0}{m}[x] ⊆ K. Hence 

a0 + a1x + ∙ ∙ ∙ + an xn = (a0 + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn) + amxm ∈ K. 

Thus, I{m}[x] ⊆ K.

Therefore K = I{m}[x].

12.  Let m ≥ 0 be any non negative integer.

The map I → I{m}[x] is a one to one map that preserves the order ⊆ from the set of

all ideals of R onto the set of ideals of R[x] that contain 

{0}{m}[x] = {a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0},

the ideal of all polynomials in R[x] with coefficient of xm = 0.

And {0} and R are mapped respectively to {0}{m}[x] and R[x].

Proof  For each ideal I of R, 

I{m}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / am ∈ I, ai ∈ R if i ≠ m and n ≥ 0} is an ideal of R[x]

and I[x] ⊆ I{m}[x].

If I = R then I{m}[x] = R[x]. And  

{0}{m}[x] = {a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0}.

It is clear that {0}{m}[x] ⊆ I{m}[x] for all ideal I of R.

Suppose I, J are ideals of R.

If I ≠ J then clearly, I{m}[x] ≠ J{m}[x] and if I ⊆ J then I{m}[x] ⊆ J{m}[x].  

Thus I → I{m}[x] is a one to one map that preserves the order ⊆ from the set of all

ideals of R into the set of ideals of R[x] that contain 

{0}{m}[x] = {a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0},

the ideal of all polynomials in R[x] with coefficient of xm = 0.

Suppose K is an ideal of R[x] that contains {0}{m}[x].

Let I be the ideal of the coefficients of xm of  K.

Then, by 11.1., K = I{m}[x]. Thus 

The map I → I{m}[x] is a one to one map that preserves the order ⊆ from the set of

all ideals of R onto the set of ideals of R[x] that contain 

{0}{m}[x] = {a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0},

the ideal of all polynomials in R[x] with coefficient of xm = 0.

And {0} and R are mapped respectively to {0}{m}[x] and R[x]. 

13.  Let m ≥ 0 be any non negative integer. Then 

I is a maximal ideal of R if and only if I{m}[x] is a maximal ideal of R[X].

Proof  Suppose I is a maximal ideal of R. 

Then I ≠ R. Hence

I{m}[x] = {a0 + a1x + ∙ ∙ ∙ + an xn / am ∈ I, ai ∈ R if i ≠ m and n ≥ 0} ≠ R[x].

Suppose K is an ideal of R[x] such that I{m}[x] ⊆ K and K ≠ I{m}[x].

Then {0}{m}[x] ⊆ I{m}[x] ⊆ K. 

Hence by 12., there exists an ideal J of R such that K = J{m}[x].

For any a ∈ I, axm ∈ I{m}[x] ⊆ K = J{m}[x]. Hence a ∈ J. So, I ⊆ J.

Since J{m}[x] = K ≠ I{m}[x], J ≠ I. 

Since I is a maximal ideal of R, J = R.

Hence K = J{m}[x] = R{m}[x] = R[x].

Therefore I{m}[x] is a maximal ideal of R[x].

Conversely, suppose I{m}[x] is a maximal ideal of R[x]. 

Then I{m}[x] ≠ R[x] = R{m}[x]. 

Since I → I{m}[x] is a map, I ≠ R. 

Suppose J is an ideal of R such that I ⊆ J ⊆ R and J ≠ I.

Then I{m}[x] ⊆ J{m}[x] and I{m}[x] ≠ J{m}[x].

Since I{m}[x] is maximal, J{m}[x] = R[x] = R{m}[x].

Since I → I{m}[x] is a one to one map, J = R.

Therefore I is a maximal ideal of R.

13. 1.  We note that if I is a maximal ideal of R then I{m}[x] is a maximal ideal of

R[x], for all non negative integers m ≥ 0.

And if I{l}[x] is a maximal ideal of R[x], for some non negative integer l ≥ 0 

then I is a maximal ideal of R and hence I{m}[x] is a maximal ideal of R[x], for all 

non negative integers m ≥ 0.

13. 2.  Since R is a commutative ring with identity, 

R has at least one maximal ideal. Hence R[x] has infinitely many maximal ideals.

Infact, for each maximal ideal I of R there are countable maximal ideals

I{0}[x], I{1}[x], I{2}[x], . . ., I{m}[x], . . .  of R[X]. 

13.3.  Let m ≥ 0 be any non negative integer. Then

an ideal of R[x] that contains {0}{m}[x] is a maximal ideal of R[x]

if and only if the ideal of coefficients of xm of K is a maximal ideal of R.

For,  Suppose K is an ideal of R[x] that contains {0}{m}[x] and I is the ideal of the coefficients of xm of  K. Then, by 11.1., K = I{m}[x]. 

Hence, by 13., 

K is a maximal ideal of R[X] if and only if I is a maximal ideal of R.

 

14.  Let K = R[x](1+x) = {f(x)(1+x) / f(x) ∈ R[x]} be the principal ideal generated

by 1+x in R[x]. 

For each non negative integer m ≥ 0, since xm ∈ R[x], xm(1+x) = xm + xm+1 ∈ K.

Hence for each non negative integer m ≥ 0, 1 ∈ Im(K), the ideal of coefficients of

xm of K and hence Im(K) = R.

Since 1+x is a non unit in R[x], 1 ∉ K and hence K ≠ R[x].

Therefore, 

even if Im(K) = R for all non negative integers m ≥ 0, K may not be R[x].

Is there any maximal ideal K of R[x] other than I{m}[x], where m ≥ 0 is any 

non negative integer and I is any maximal ideal of R ?

15.  There may be maximal ideals K of R[x] such that K ≠ I{m}[x], for all non

negative integers m ≥ 0 and for all maximal ideals I of R.

For example, If F is a field then 1+x is irreducible over F, hence K = F[x](1+x) is

a maximal ideal of F[x]. Since F is a field, (0) is the only maximal ideal of F.

For any negative integer m ≥ 0, 

{0}{m}[x] = {a0 + a1x + ∙ ∙ ∙ + am-1xm-1 + am+1xm+1 + ∙ ∙ ∙ + an xn / ai ∈ R and n ≥ 0} and 

xm + xm+1 ∉ {0}{m}[x]. But xm + xm+1 ∈ K. 

Hence K ≠ {0}{m}[x], for all negative integer m ≥ 0.  

Hence there is a maximal ideal K of F[x] such that K ≠ I{m}[x], for all 

non negative integers m ≥ 0 and for all maximal ideals I of F. 

Dear Young Learners :  

We can learn directly the generalised result 13.  

But it is not interesting and useful.

Understanding how the generalised result is developed from the base results

by raising questions, by observation of proofs and by Mathematical thinking

leads to enjoyable learning.

So,

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

Have Enjoyable Learning.