Rings and Fields

Lecture 19: Rings

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Rings and ideals; prime/maximal follow from the quotient.

Quotient rings

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Prime ideals

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Prime ⇔ the quotient is an integral domain.

Maximal ideals, integral domains

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Maximal ⇔ the quotient is a field; maximal ⇒ prime, never the reverse in general.

Ideals, quotients and the Chinese remainder theorem free

In plain words

An ideal is the part you are allowed to declare zero. Prime means the quotient has no zero divisors, and maximal means the quotient is a field, which is the stronger condition.

Prime vs maximal

For a commutative ring R with 1:

I prime ⇔ R/I is an integral domain.
I maximal ⇔ R/I is a field.
maximal ⇒ prime; the converse fails: $(X) \subset \mathbb{Z}[X]$ is prime $(\mathbb{Z}[X]/(X) \cong \mathbb{Z}, a$ domain) but not maximal.
In a PID, non-zero prime ⇔ maximal.

Standard quotients

Quotient	Is
$\mathbb{Z}[X]/(X)$	$\mathbb{Z}$
$\mathbb{Z}[X]/(X, p)$	𝔽_p, so (X, p) is maximal
$\mathbb{R}[X]/(X^{2} + 1)$	$\mathbb{C}$
$\mathbb{Z}[X]/(X^{2} + 1)$	$\mathbb{Z}[i]$
𝔽_p[X]/(f), f irreducible of degree n	𝔽 $_{p^{n}}$
$\mathbb{Z}/n\mathbb{Z}$	field ⇔ n prime

Chinese remainder theorem

$\gcd(m, n) = 1 \Rightarrow \mathbb{Z}/$ mn $\cong \mathbb{Z}/m \times \mathbb{Z}/n$ as rings. Consequences used constantly:

Number of solutions of $x^{2} = 1$ in $\mathbb{Z}/105: 105 = 3\cdot5\cdot7$ , two roots each $\Rightarrow 2^{3} = 8$ . $- \varphi$ is multiplicative; the number of units is $\prod\varphi(p^{k})$ . $- \mathbb{Z}/n$ is a product of local rings; idempotents correspond to factorisations.

Nilpotents, units, idempotents

x nilpotent ⇒ 1 + x is a unit.
The nilradical is the intersection of all prime ideals. $- \mathbb{Z}/n$ has non-trivial nilpotents ⇔ n is not squarefree.
Fermat/Euler: $a^{\varphi(n)}$ ≡ 1 (mod n) for gcd(a, n) = 1; Wilson: (p−1)! ≡ −1 (mod p).
Power sums: $\sum_{k=1}^{p-1}$ kᵐ ≡ −1 (mod p) if (p − 1) | m, and ≡ 0 otherwise.

Key takeaways

Identify the quotient ring; prime/maximal follows instantly.
Maximal ⇒ prime always; converse only in PIDs (and for non-zero ideals).
CRT is the workhorse for every counting question modulo a composite.

Prime versus maximal, decided by the quotientinteractive

This visual is part of Trap Analysis

One test settles both — worked in ℤ[x], where an ideal can be prime and still not maximal.

Exam focus: Prime ⇔ integral domain quotient, maximal ⇔ field quotient. CRT converts counting questions mod n into products over prime powers.

Practice

Part C2025 Decembershow ▾

Consider a commutative ring R with unity with at least five elements such that for any two elements a,b∈R, there exists c∈R such that a=bc or b=ac. Which of the following statements are necessarily true?

A.R has a unique maximal ideal.✓ Correct
B.Every finitely generated ideal of R is principal.✓ Correct
C.R is an integral domain.
D.R is a euclidean domain.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

Let A be a subring of the field of rationals

\mathbb{Q}

such that for any nonzero rational

r \in \mathbb{Q}, r \in A

or 1/r ∈ A. Which of the following statements is FALSE?

A.The set {a ∈ A : 1/a ∉ A} ∪ {0} is an additive subgroup of $\mathbb{Q}$ .
B.A has at most one maximal ideal.
C.If $A \ne \mathbb{Q}$ , then A has infinitely many prime ideals.✓ Correct
D.For any nonzero a, b ∈ A, a divides b or b divides a in A.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2025 Juneshow ▾

Which of the following statements is true?

A.The ideal $2\mathbb{Z}[i]$ is maximal in $\mathbb{Z}[i]$ .
B.The ideal $X\mathbb{C}[X, Y]$ is maximal in $\mathbb{C}[X, Y]$ .
C.The set of all polynomials in $\mathbb{C}[X]$ whose coefficients add up to 0 is a maximal ideal in $\mathbb{C}[X]$ .✓ Correct
D.The ideal $(\sqrt{2} - 1)\mathbb{Z}[\sqrt{2}]$ is maximal in $\mathbb{Z}[\sqrt{2}]$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Which of the following rings are integral domains?

A.The ring of polynomials in 3 variables over $\mathbb{R}$ .✓ Correct
B.The ring of complex analytic functions on the open unit disc in $\mathbb{C}$ .✓ Correct
C.The ring of 2 × 2 real matrices.
D.The ring of continuous functions from [0, 1] to $\mathbb{R}$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let R be a nonzero ring with unity such that

r^{2} = r

for all r ∈ R. Which of the following statements are true?

A.R is never an integral domain.
B.r = −r for all r ∈ R.✓ Correct
C.Every nonzero prime ideal of R is maximal.✓ Correct
D.R must be a commutative ring.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Let R and S be non-zero commutative rings with multiplicative identities 1_R, 1_S, respectively. Let f : R → S be a ring homomorphism with f(1_R) = 1_S. Which of the following statements are true?

A.If f(a) is a unit in S for every non-zero element a ∈ R, then S is a field
B.If f(a) is a unit in S for every non-zero element a ∈ R, then f(R) is a field
C.If R is a field, then f(a) is a unit in S for every non-zero element a ∈ R✓ Correct
D.If a is a unit in R, then f(a) is a unit in S✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Juneshow ▾

Consider the ring R = {

\sum_(n\in\mathbb{Z}) a_{n}X^{n}

a_{n} \in \mathbb{Z}

, and

a_{n} \ne 0

only for finitely many

n \in \mathbb{Z}

}, with the usual addition and multiplication of such sums. Which of the following statements is true?

A.R is not commutative
B.The ideal (X − 1) is a maximal ideal in R
C.The ideal (X − 1, 2) is a prime ideal in R✓ Correct
D.The ideal (X, 5) is a maximal ideal in R

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Let

\mathbb{C}[x, y]

be the polynomial ring in two variables over

\mathbb{C}

. For which of the following ideals I, the quotient ring

\mathbb{C}[x, y]/I

is NOT an integral domain?

A.I = (x, y)
B.I = (x + y)
C. $I = (x^{2} + y^{2})$ ✓ Correct
D.I = (xy − 1)

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Let

f \in \mathbb{R}[x]

be a product of distinct monic irreducible polynomials

P_{1}, P_{2}, \dots, P_{n}

, where n ≥ 2. Let (f) denote the ideal generated by f in the ring

\mathbb{R}[x]

. Which of the following statements are true?

A. $\mathbb{R}[x]/(f)$ is a field.
B. $\mathbb{R}[x]/(f)$ is a finite dimensional $\mathbb{R}-$ vector space.✓ Correct
C. $\mathbb{R}[x]/(f)$ is a direct sum of fields, each of which is isomorphic to $\mathbb{R}$ or $\mathbb{C}$ .✓ Correct
D.There are no non-zero elements $u \in \mathbb{R}[x]/(f)$ such that u^m = 0 for some m ≥ 1.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

Let I be an ideal of the ring 𝔽

_{2}[t]/(t^{2}(1 - t)^{2})

. Which of the following are the possible values for the cardinality of I?

A.1✓ Correct
B.8✓ Correct
C.16✓ Correct
D.24

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Let

R = \mathbb{Z}[X]/(X^{2} + 1)

and

\psi : \mathbb{Z}[X] \to R

be the natural quotient map. Which of the following statements are true?

A.R is isomorphic to a subring of $\mathbb{C}$ .✓ Correct
B.For any prime number $p \in \mathbb{Z}$ , the ideal generated by $\psi(p)$ is a proper ideal of R.✓ Correct
C.R has infinitely many prime ideals.✓ Correct
D.The ideal generated by $\psi(X)$ is a prime ideal in R.

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾

The number of solutions of the equation

x^{2} = 1

in the ring

\mathbb{Z}/105\mathbb{Z}

A.0
B.2
C.4
D.8✓ Correct

Solution

By CRT, $\mathbb{Z}/105\mathbb{Z} \cong \mathbb{Z}/3 \times \mathbb{Z}/5 \times \mathbb{Z}/7; x^{2} = 1$ has exactly 2 solutions in each odd prime field, giving $2^{3} = 8$ .

Part B2023 Decembershow ▾

Which one of the following is equal to

1^{37} + 2^{37} + 3^{37} + \cdots + 88^{37}

\mathbb{Z}/89\mathbb{Z}

A.88
B.−88
C.−2
D.0✓ Correct

Solution

89 is prime and $(\mathbb{Z}/89)$ * is cyclic of order 88.

For a generator $g, \sum_{k}$ $k^{37} = \sum_j$ $g^{37j}$ is a geometric sum equal to 0 because 88 ∤ 37 (so $g^{37} \ne 1)$ .

Part C2023 Decembershow ▾

Let

n \in \mathbb{Z}

be such that n ≡ 1 (mod 7) and n ≡ 4 (mod 15). Which of the following statements are true?

A.n ≡ 1 (mod 3).✓ Correct
B.n ≡ 1 (mod 35).
C.n ≡ 1 (mod 21).✓ Correct
D.n ≡ 1 (mod 5).

Solution

n ≡ 4 (mod 15) gives n ≡ 1 (mod 3) and n ≡ 4 (mod 5).

With n ≡ 1 (mod 7), CRT gives n ≡ 1 (mod 21); n ≢ 1 (mod 5) kills the mod-35 claim.

Part Cshow ▾

Consider the ring

\mathbb{Z}[x]

. Which of the following are true?

A.(x) is a prime ideal.✓ Correct
B.(x) is a maximal ideal.
C.(2, x) is a maximal ideal.✓ Correct
D. $\mathbb{Z}[x]$ is a principal ideal domain.

Solution

$(1) \mathbb{Z}[x]/(x) \cong \mathbb{Z}$ , an integral domain, so (x) is prime.

$(2) \mathbb{Z}$ is not a field, so (x) is not maximal.

$(3) \mathbb{Z}[x]/(2, x) \cong$ 𝔽 $_{2}, a$ field, so (2, x) is maximal — note this also shows a prime ideal can be properly contained in another.

(4) Fails: (2, x) itself is not principal, so $\mathbb{Z}[x]$ is not a PID, even though $\mathbb{Z}$ and 𝔽[x] both are.

Part Bshow ▾

The number of units in the ring

\mathbb{Z}/20\mathbb{Z}

A.4
B.8✓ Correct
C.10
D.19

The worked solution and the per-option analysis are in Trap Analysis.

2. Euclidean, PID, UFD hierarchy

Irreducible, prime elements

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Irreducible vs prime — they coincide only in a UFD.

Principal Ideal Domains

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Unique Factorization Domains 1

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Then see the notes for one counterexample at each strict inclusion.

The ring hierarchy and where each inclusion is strict free

In plain words

A ladder of niceness: division with remainder, then every ideal from a single element, then unique factorisation. Irreducible and prime only start agreeing once factorisation is unique.

The chain

Field ⊂ Euclidean domain ⊂ PID ⊂ UFD ⊂ integral domain

Inclusion	Strict because
Euclidean ⊊ PID	$\mathbb{Z}[(1 + \surd-19)/2]$ is a PID, not Euclidean
PID ⊊ UFD	$\mathbb{Z}[X]$ and 𝔽[X, Y]: (2, X) and (X, Y) are not principal
UFD ⊊ domain	$\mathbb{Z}[\surd-5]: 6 = 2\cdot3 = (1+\surd-5)(1-\surd-5)$

Facts that follow

R UFD ⇒ R[X] UFD (Gauss). So $\mathbb{Z}[X_{1}, \dots, X_{n}]$ is a UFD.
R PID ⇏ R[X] PID: $\mathbb{Z}$ is a PID, $\mathbb{Z}[X]$ is not.
𝔽 field ⇒ 𝔽[X] Euclidean (degree), hence a PID; 𝔽[X, Y] is not. $- \mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$ are Euclidean (norm), hence PIDs and UFDs.
In a UFD: irreducible ⇔ prime. In a general domain only prime ⇒ irreducible; in $\mathbb{Z}[\surd-5], 2$ is irreducible but not prime.
Every PID is Noetherian; $\mathbb{Z}[X_{1}, X_{2}, \dots] ($ infinitely many variables) is a UFD that is not Noetherian.

Norms for quadratic rings

$N(a + b\sqrt{d}) =$ | $a^{2} -$ db $^{2}$ | is multiplicative; units are the elements of norm 1. In $\mathbb{Z}[\surd-5], N(2) = 4$ and no element has norm 2, so 2 is irreducible; yet 2 | $(1+\surd-5)(1-\surd-5)$ without dividing either factor.

Key takeaways

Learn one counterexample per strict inclusion: $\mathbb{Z}[(1+\surd-19)/2], \mathbb{Z}[X], \mathbb{Z}[\surd-5]$ .
UFD is preserved by adjoining variables; PID is not.
Norm arguments settle irreducibility in quadratic rings.

Euclidean ⊂ PID ⊂ UFD, with a separating example at each stepinteractive

This visual is part of Trap Analysis

The chain is strict — here is the ring that sits in each gap.

Exam focus: Memorise the chain and the counterexample at each strict inclusion — that single table answers most Part-C ring questions.

Practice

Part C2024 Juneshow ▾

Let R be a principal ideal domain with a unique maximal ideal. Which of the following statements are necessarily true?

A.Every quotient ring of R is a principal ideal domain
B.There exists a quotient ring S of R and an ideal I ⊆ S which is not principal
C.R has countably many ideals✓ Correct
D.Every quotient ring S(≠ {0}) of R has a unique maximal ideal which is principal✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Which of the following statements are true?

A.The value of the Euler $\varphi-$ function is even for all integers n ≥ 3.✓ Correct
B.Let G be a finite group and S a subset of G with |S| > |G|/2. Then {ab : a, b ∈ S} = G.✓ Correct
C.The polynomial ring $\mathbb{R}[x_{1}, \dots, x_{n}]$ is a Euclidean domain for all integers n ≥ 1.
D.The subset {f ∈ C([0, 1]) : f(1/2) = 0} of the ring C([0, 1]) of continuous functions from [0, 1] to $\mathbb{R}$ is a prime ideal.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

R = \mathbb{Z}[\surd-5] =

{

a + b\surd-5 : a, b \in \mathbb{Z}

}. Which of the following are true?

A. $6 = 2\cdot3 = (1+\surd-5)(1-\surd-5)$ gives two genuinely different factorizations of 6 into irreducibles.✓ Correct
B.R is not a UFD.✓ Correct
C.R is not a PID.✓ Correct
D.R is a Euclidean domain with respect to $N(a+b\surd-5) = a^{2} + 5b^{2}$ .

Solution

(1) is the classical example — $2, 3, 1+\surd-5, 1-\surd-5$ are all irreducible and pairwise non-associate. (2) follows directly. (3) PID ⇒ UFD, so failing to be a UFD means failing to be a PID too. (4) is false for the same reason one level further down the hierarchy: Euclidean ⇒ PID, so a ring that isn't even a PID cannot be Euclidean.

Part Cshow ▾

Let R be a commutative ring. Which of the following are true?

A.Every Euclidean domain is a PID.✓ Correct
B.Every PID is a UFD.✓ Correct
C.Every UFD is a PID.
D. $\mathbb{Z}[i]$ , the Gaussian integers, is a Euclidean domain with respect to N(a+bi $) = a^{2} + b^{2}$ .✓ Correct

Solution

(1)–(2) are the standard hierarchy. (4) is the standard Euclidean algorithm on $\mathbb{Z}[i]. (3)$ is false — the converse fails: $\mathbb{Z}[x]$ is a UFD, but the ideal (2, x) is not principal, so $\mathbb{Z}[x]$ is not a PID.

Part Cshow ▾

Which of the following rings are unique factorisation domains?

A. $\mathbb{Z}[x]$ ✓ Correct
B. $\mathbb{Z}[\surd-5]$
C. $\mathbb{Q}[x, y]$ ✓ Correct
D. $\mathbb{Z}[i]$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following are true about the hierarchy Euclidean ⊂ PID ⊂ UFD?

A. $\mathbb{Z}[x]$ is a UFD but not a PID.✓ Correct
B. $\mathbb{Z}[\surd-5]$ is not a UFD.✓ Correct
C.Every Euclidean domain is a PID.✓ Correct
D.𝔽[x, y] over a field 𝔽 is a PID.

The worked solution and the per-option analysis are in Trap Analysis.

3. Polynomial rings and irreducibility tests

Polynomial rings 1

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Gauss Lemma

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Transfers irreducibility between ℤ[X] and ℚ[X].

Eisenstein criterion and Problems 7

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Eisenstein, including the shift trick for cyclotomic polynomials.

Irreducibility tests free

In plain words

Polynomials over a field behave much like the integers. Every irreducibility test runs in one direction only: passing it proves the point, and failing it proves nothing at all.

The toolkit

Test	Use when
Rational root theorem	degree 2 or 3 — irreducible ⇔ no root
Eisenstein at p	some prime divides all but the leading coefficient, $p^{2}$ ∤ constant
Eisenstein after x ↦ x + 1	cyclotomic-looking, $e.g. \Phi_p(x) =$ $x^{p-1}$ + ⋯ + 1
Reduction mod p	leading coefficient not divisible by p; irreducible mod p ⇒ irreducible over $\mathbb{Q}$
Gauss's lemma	transfers between $\mathbb{Z}[X]$ and $\mathbb{Q}[X]$

Degree 2 or 3: reducible ⇔ has a root. Degree ≥ 4: a polynomial may factor with no roots $(x^{4} + 4 = (x^{2} - 2x + 2)(x^{2} + 2x + 2))$ .

The one-way trap

Irreducible over $\mathbb{Q}$ does not imply irreducible mod p for any p: $- x^{4} + 1$ is irreducible over $\mathbb{Q}$ but factors modulo every prime. $- x^{2} + x + 1$ is irreducible over $\mathbb{Q}$ but $\equiv (x - 1)^{2} mod 3$ .

The converse direction is the useful one: irreducible mod some p (with p not dividing the leading coefficient) ⇒ irreducible over $\mathbb{Q}$ .

Finite fields

𝔽 $_p^{n}$ exists and is unique; 𝔽 $_p^{n} \subseteq$ 𝔽_pᵐ ⇔ n | m.
xᵖ $^{n} - x$ is the product of all monic irreducibles over 𝔽_p whose degree divides n.
The number of monic irreducibles of degree n over 𝔽_p is $(1/n)\sum_{d|n}$ $\mu(d)$ $p^{n/d}$ .
𝔽 $_p^{n}$ * is cyclic of order $p^{n} - 1$ .

Key takeaways

Degree ≤ 3: roots decide. Degree ≥ 4: they do not.
Reduction mod p proves irreducibility, never reducibility over $\mathbb{Q}$ .
Eisenstein after a substitution handles the cyclotomic family.

Irreducibility tests, in the order worth trying theminteractive

This visual is part of Trap Analysis

Rational roots, Eisenstein, reduction mod p — and the quartic that factors despite having no real roots.

Exam focus: Pick the right test: rational root, Eisenstein (possibly after a shift), or reduction mod p. Irreducible over ℚ does not mean irreducible mod every p.

Practice

Part C2025 Juneshow ▾

Let

f(X) = X^{5} + X + 1 \in \mathbb{Q}[X]

and

g(X) = X^{5} - X + 1 \in \mathbb{Q}[X]

. Which of the following statements are true?

A.f(X) is irreducible in $\mathbb{Q}[X]$ , but g(X) is not.
B.g(X) is irreducible in $\mathbb{Q}[X]$ , but f(X) is not.✓ Correct
C.Both f(X) and g(X) are irreducible in $\mathbb{Q}[X]$ .
D.Neither f(X) nor g(X) is irreducible in $\mathbb{Q}[X]$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Juneshow ▾

Let

f : \mathbb{Q}[X] \to \mathbb{Q}[X]

be a ring homomorphism with f(1) = 1. For n ≥ 1, let

f^{n} = f

∘ ⋯ ∘ f (n times). Which of the following statements are true?

A.If f is onto, then so is $f^{n}$ for all n ≥ 1.✓ Correct
B.ker $f^{n+1}$ $= \ker f^{n}$ for some n ≥ 1.✓ Correct
C.If f is onto, then f is one-to-one.✓ Correct
D.If f is one-to-one, then f is onto.

The worked solution and the per-option analysis are in Trap Analysis.

Part C2025 Decembershow ▾

Which of the following statements are true?

A. $x^{2} +$ xy $^{2} - x - y + 1$ is irreducible in $\mathbb{Q}[x,y]$ .✓ Correct
B. $x^{3} + 100x^{2} + 25x + 10$ is irreducible in $\mathbb{Z}[x]$ .✓ Correct
C. $2x^{2} + 3$ xy $+ y^{2} + 3x + 2y + 1$ is irreducible in $\mathbb{Q}[x,y]$ .
D. $x^{4} + 4x^{2} + 3$ is irreducible in $\mathbb{Z}[x]$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Which of the following quotient rings are fields?

A. $\mathbb{Q}[x]/(x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1)$ ✓ Correct
B. $\mathbb{Q}[x]/(x^{5} + x^{4} + x^{3} + x^{2} + x + 1)$
C. $\mathbb{Z}[x]/(x - 101)$
D. $\mathbb{Q}[x]/(x^{43} - 41x + 41)$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

For two indeterminates x, y, let R = 𝔽

_{3}[x]

and S = R[y]. Which of the following statements are true?

A.S is a principal ideal domain
B. $S/(y^{2} + x^{2})$ is a unique factorization domain
C.S is a unique factorization domain✓ Correct
D.S/(x) is a principal ideal domain✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2024 Decembershow ▾

Consider the ring homomorphism

\psi : \mathbb{C}[x, y] \to \mathbb{C}[t]

defined by

\psi(f(x, y)) = f(t^{2}, t^{3})

. Which of the following statements is true?

A. $\psi$ is surjective.
B.If $p_{1}(x)y + p_{2}(x) \in \ker \psi$ for polynomials $p_{1}(x), p_{2}(x) \in \mathbb{C}[x]$ , then both $p_{1}$ and $p_{2}$ are the zero polynomial.✓ Correct
C.There exist non-zero polynomials $p_{1}(x), p_{2}(x) \in \mathbb{C}[x]$ such that $\psi(p_{1}(x)y + p_{2}(x)) = 0$ .
D.There exists $f \in \mathbb{C}[x, y]$ such that $\psi(f(x, y)) = 0$ and $\psi(f^{2}(x, y)) \ne 0$ .

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Juneshow ▾

How many real roots does the polynomial

x^{3} + 3x - 2023

have?

A.0
B.1✓ Correct
C.2
D.3

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Let

f(X) = X^{2} + X + 1

and

g(X) = X^{2} + X - 2

be polynomials in

\mathbb{Z}[X]

. Which of the following statements are true?

A.For all prime numbers p, f(X) mod p is irreducible in $(\mathbb{Z}/p\mathbb{Z})[X]$ .
B.There exists a prime number p such that g(X) mod p is irreducible in $(\mathbb{Z}/p\mathbb{Z})[X]$ .
C.g(X) is irreducible in $\mathbb{Q}[X]$ .
D.f(X) is irreducible in $\mathbb{Q}[X]$ .✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Which of the following polynomials are irreducible over

\mathbb{Q}

A. $x^{4} + 4$
B. $x^{3} - 2$ ✓ Correct
C. $x^{2} + x + 1$ ✓ Correct
D. $x^{4} + x^{3} + x^{2} + x + 1$ ✓ Correct

Solution

(1) is the Sophie Germain trap: $x^{4} + 4 = (x^{2} + 2x + 2)(x^{2} - 2x + 2)$ .

(2) Eisenstein at p = 2; alternatively, a cubic with no rational root is irreducible.

(3) Degree 2 with no rational roots.

(4) is the 5th cyclotomic polynomial $\Phi_{5}$ : substitute x → x + 1 and apply Eisenstein at p = 5.

4. Field extensions, splitting fields, finite fields

Lecture 20: Field

NPTEL · Introduction to Abstract and Linear Algebra (IIT Kharagpur)

Fields as a starting point; degrees and towers are in the notes.

Field extensions 1

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Degree of a field extension 1

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

The tower law — the first move in every degree computation.

Splitting fields

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

Finite fields 1

NPTEL · Introduction to Rings and Fields (K. Hanumanthu, CMI)

𝔽_pⁿ exists, is unique, and its subfields match divisors of n.

Field extensions, degrees and splitting fields free

In plain words

Adjoining a number gives a bigger field, and degrees multiply as you build up in stages. A splitting field holds all the roots, not merely the one you started by adding.

Degrees

[K : F] is the dimension of K as an F-vector space. Tower law: [L : F] = [L : K][K : F].

$- [F(\alpha) : F] = \deg$ of the minimal polynomial of $\alpha$ over F. $- \alpha$ algebraic of degree $n \Rightarrow F(\alpha) = F[\alpha]$ has basis $1, \alpha, \dots, \alpha^{n-1}$ . $- [\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4; [\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3; [\mathbb{Q}(\sqrt[3]{2}, \omega) : \mathbb{Q}] = 6; [\mathbb{Q}(\zeta_n) : \mathbb{Q}] = \varphi(n)$ .

Splitting fields and normality

A splitting field of f over F exists and is unique up to isomorphism, of degree dividing (deg f)!.
K/F is normal iff it is the splitting field of some family; separable iff minimal polynomials have distinct roots (automatic in characteristic 0 and for finite fields); Galois = normal + separable, and then |Gal(K/F)| = [K : F].

Finite fields

𝔽 $_p^{n}/$ 𝔽_p is Galois with cyclic group generated by the Frobenius x ↦ xᵖ. Subfields ↔ divisors of n.

Constructibility

A length is constructible ⇔ it lies in a tower of quadratic extensions ⇒ its degree over $\mathbb{Q}$ is a power of 2. Hence: doubling the cube (degree 3) and trisecting a general angle (degree 3) are impossible; squaring the circle fails because $\pi$ is transcendental.

Algebraic vs finite

Finite ⇒ algebraic; the converse fails: the field of all algebraic numbers is an infinite algebraic extension of $\mathbb{Q}$ .

Key takeaways

Degree of a simple extension = degree of the minimal polynomial; build up with the tower law.
Galois ⇔ normal + separable, and then the group order equals the degree.
Algebraic ⇏ finite.

Degrees multiply — computing [K:ℚ] and spotting the primitive elementinteractive

This visual is part of Trap Analysis

The tower law in practice, plus the subfield rule for finite fields that decides many options.

Exam focus: Tower law plus 'degree = degree of the minimal polynomial' answers most questions. Know the standard degrees: [ℚ(∛2, ω) : ℚ] = 6, [ℚ(√2, √3) : ℚ] = 4.

Practice

Part C2025 Juneshow ▾

Let p > 2 be a prime number. Let 𝔽_p denote the field with p elements and 𝔽̄_p an algebraic closure of 𝔽_p. Which of the following statements are true?

A.Let f(X) ∈ 𝔽_p[X] and $\alpha$ be a root of f in 𝔽̄_p. Then 𝔽 $_p(\alpha)$ is the splitting field of f in 𝔽̄_p.
B.Let f, g ∈ 𝔽_p[X] be irreducible polynomials of same degree and $\alpha$ be a root of f in 𝔽̄_p. Then 𝔽 $_p(\alpha)$ is the splitting field of g in 𝔽̄_p.✓ Correct
C.𝔽_p[X] has infinitely many irreducible polynomials.✓ Correct
D.The set {a + b | a, b ∈ 𝔽_p} is contained in { $a^{2} + b^{2}$ | a, b ∈ 𝔽_p}.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Consider the polynomial f(x) = x^2025 − 1 over 𝔽

_{5}

, where 𝔽

_{5}

is the field with five elements. Let S be the set of all roots of f in an algebraic closure of the field 𝔽

_{5}

. Which of the following statements are true?

A.S is a cyclic group.✓ Correct
B.S has $\varphi(2025)$ elements, where $\varphi$ denotes the Euler $\varphi-$ function.
C.S has $\varphi(2025)$ generators, where $\varphi$ denotes the Euler $\varphi-$ function.
D.S has 81 elements.✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Juneshow ▾

For which of the following values of q, does a finite field of order q have exactly 6 subfields?

A. $q = 2^{18}$ ✓ Correct
B. $q = 2^{32}$ ✓ Correct
C. $q = 2^{12}$ ✓ Correct
D. $q = 2^{243}$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part B2023 Decembershow ▾

Consider the field

\mathbb{C}

with the Euclidean topology. Let K be a proper subfield of

\mathbb{C}

that is not contained in

\mathbb{R}

. Which one of the following statements is necessarily true?

A.K is dense in $\mathbb{C}$ .✓ Correct
B.K is an algebraic extension of $\mathbb{Q}$ .
C. $\mathbb{C}$ is an algebraic extension of K.
D.The smallest closed subset of $\mathbb{C}$ containing K is NOT a field.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let 𝔽

_{8}

denote the field with 8 elements. Which of the following are true?

A.𝔽 $_{8}$ contains a subfield isomorphic to 𝔽 $_{4}$ .
B.The multiplicative group 𝔽 $_{8}$ ˣ is cyclic of order 7.✓ Correct
C.[𝔽 $_{8}$ : 𝔽 $_{2}] = 3$ .✓ Correct
D.𝔽 $_{8}$ is isomorphic to $\mathbb{Z}/8\mathbb{Z}$ as a ring.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let

K = \mathbb{Q}(\sqrt{2}, \sqrt{3})

. Which of the following are true?

A. $[K : \mathbb{Q}] = 4$ .✓ Correct
B. $K = \mathbb{Q}(\sqrt{2} + \sqrt{3})$ .✓ Correct
C. $\sqrt{6} \in K$ .✓ Correct
D. $K/\mathbb{Q}$ is not a normal extension.

The worked solution and the per-option analysis are in Trap Analysis.

Part Cshow ▾

Let K = 𝔽

_{2}(\alpha)

, where

\alpha

is a root of the irreducible polynomial

x^{3} + x + 1

over 𝔽

_{2}

. Which of the following are true?

A.[K : 𝔽 $_{2}] = 3$ .✓ Correct
B.|K| = 8.✓ Correct
C.K*, the multiplicative group of K, is cyclic of order 7.✓ Correct
D. $\alpha^{2} = \alpha + 1$ in K.

Solution

$x^{3}+x+1$ has no root in 𝔽 $_{2} ($ checking both elements) and has degree 3, so it is irreducible, giving (1). (2) |K| $= 2^{3} = 8. (3)$ the multiplicative group of any finite field is cyclic, here of order 7. (4) is false: since $\alpha$ has minimal polynomial of degree exactly 3, { $1, \alpha, \alpha^{2}$ } is linearly independent over 𝔽 $_{2}$ — a relation like $\alpha^{2} = \alpha + 1$ would be a nonzero degree-2 polynomial vanishing at $\alpha$ , contradicting that its minimal polynomial has degree 3.

Part Cshow ▾

Let F be a finite field. Which of the following are true?

A.|F| $= p^{n}$ for some prime p and positive integer n.✓ Correct
B.The multiplicative group F* is cyclic of order |F| − 1.✓ Correct
C.For every prime power $q = p^{n}$ , there is a unique field of order q (up to isomorphism), 𝔽_q.✓ Correct
D.𝔽 $_{8}$ contains a subfield isomorphic to 𝔽 $_{4}$ .

Solution

(1)–(3) are the standard classification of finite fields. (4) is false: 𝔽 $_{p^{n}}$ contains a subfield 𝔽 $_{p^{m}}$ exactly when m divides n — the divisors of 3 are only 1 and 3, so 𝔽 $_{8} =$ 𝔽 $_{2^{3}}$ contains 𝔽 $_{2}$ and itself, but not 𝔽 $_{4} =$ 𝔽 $_{2^{2}}$ , since 2 does not divide 3.

Part Bshow ▾

The degree of the field extension

\mathbb{Q}(2^{1/3})

over

\mathbb{Q}

A.1
B.2
C.3✓ Correct
D.6

Solution

$x^{3} - 2$ is irreducible over $\mathbb{Q} ($ Eisenstein at 2) and has $2^{1/3}$ as a root, so it is the minimal polynomial and the degree is its degree, 3.

Note $\mathbb{Q}(2^{1/3})$ $\subset \mathbb{R}$ does not contain the complex cube roots — it is not the splitting field, which has degree 6.

5. Galois theory essentials

mod03lec14 - Galois extensions, Galois groups

NPTEL · Galois Theory (K. Hanumanthu, CMI)

mod04lec24 - Main theorem of Galois theory - Part 1

NPTEL · Galois Theory (K. Hanumanthu, CMI)

The correspondence: subgroups ↔ intermediate fields, normal ↔ normal.

mod06lec37 - Solvability by radicals

NPTEL · Galois Theory (K. Hanumanthu, CMI)

Why the general quintic is unsolvable.

mod07lec40 - Discriminants, Galois groups of polynomials

NPTEL · Galois Theory (K. Hanumanthu, CMI)

The cubic rule: discriminant a square ⇒ A₃, otherwise S₃.

Galois correspondence and the standard groups free

In plain words

A Galois extension carries a group whose subgroups match the fields sitting in between, exactly and in reverse order. That turns infinitely many possible fields into a finite list to check.

Fundamental theorem

For a Galois extension K/F with G = Gal(K/F):

Subgroups H ≤ G ↔ intermediate fields K^H, inclusion-reversing.
[K : K^H] = |H| and [K^H : F] = [G : H].
K^H/F is normal ⇔ H ⊴ G, and then Gal(K^H/F) ≅ G/H.

The standard computations

Polynomial over $\mathbb{Q}$	Splitting field	Galois group	Degree
$x^{2} - 2$	$\mathbb{Q}(\sqrt{2})$	$\mathbb{Z}_{2}$	2
$(x^{2}-2)(x^{2}-3)$	$\mathbb{Q}(\sqrt{2}, \sqrt{3})$	$\mathbb{Z}_{2} \times \mathbb{Z}_{2}$	4
$x^{3} - 2$	$\mathbb{Q}(\sqrt[3]{2}, \omega)$	$S_{3}$	6
$x^{3} - 2$ over $\mathbb{Q}(\omega)$	$\mathbb{Q}(\sqrt[3]{2}, \omega)$	$\mathbb{Z}_{3}$	3
$x^{4} - 2$	$\mathbb{Q}(2^{1/4}$ , i)	$D_{4}$	8
$x^{n} - 1$	$\mathbb{Q}(\zeta_n)$	$(\mathbb{Z}/n)$ *, abelian	$\varphi(n)$
x^p − x − a over 𝔽_p	—	$\mathbb{Z}_p ($ Artin–Schreier)	p

Cubic rule: the Galois group of an irreducible cubic is $A_{3} \cong \mathbb{Z}_{3}$ if the discriminant is a square, $S_{3}$ otherwise.

Solvability

f is solvable by radicals ⇔ its Galois group is solvable. $S_{3}$ and $S_{4}$ are solvable; $S_{5}$ is not, so the general quintic is unsolvable. A specific quintic with group $S_{5}: x^{5} - 6x + 3$ .

Key takeaways

Subgroups ↔ intermediate fields, and normal ↔ normal.
Adjoining $\omega$ first turns the $S_{3}$ of $x^{3} - 2$ into $a \mathbb{Z}_{3}$ .
Cyclotomic extensions are abelian with group $(\mathbb{Z}/n)$ *.

The Galois correspondence, used on one exampleinteractive

This visual is part of Trap Analysis

A degree-4 extension, its group, and the subfield lattice read straight off the subgroup lattice.

Exam focus: For an exam you need the Galois groups of x³ − 2, x⁴ − 2, xⁿ − 1 and the fundamental correspondence — subgroups ↔ intermediate fields, with normal subgroups ↔ normal extensions.

Practice

Part C2025 Decembershow ▾

Let p≥3 be a prime number and

f(x)\in\mathbb{Q}[x]

an irreducible polynomial of degree p. Suppose that

a_{1},\dots,a_{p}\in\mathbb{C}

are the roots of f and that

a_{1}\notin\mathbb{R}, a_{2}\notin\mathbb{R}

, and

a_{i}\in\mathbb{R}

for all 3≤i≤p. Let

K=\mathbb{Q}(a_{1},\dots,a_{p})

be the subfield of

\mathbb{C}

generated by the roots of f. Consider the Galois group G of K over

\mathbb{Q}

as a subgroup of

S_{p}

, the group of permutations of {

a_{1},\dots,a_{p}

}. Which of the following statements are true?

A.The transposition (1 2) belongs to G.✓ Correct
B.|G| is divisible by p.✓ Correct
C.A p-cycle belongs to G.✓ Correct
D. $G = S_{p}$ ✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2024 Decembershow ▾

Which of the following statements are true?

A.If K is the splitting field of a non-constant polynomial over $\mathbb{Q}$ , then K is Galois over $\mathbb{Q}$ .✓ Correct
B.If K is a normal extension of $\mathbb{Q}$ , then K is Galois over $\mathbb{Q}$ .✓ Correct
C.If K is the set of all the roots of the polynomial $x^{121} - x$ in an algebraic closure of 𝔽 $_{11}$ , then K is Galois over 𝔽 $_{11}$ , where 𝔽 $_{11}$ is the field with 11 elements.✓ Correct
D.If $K = \mathbb{Q}(2^(1/13), \zeta_{13})$ , where $\zeta_{13}$ is a primitive 13th root of unity in $\mathbb{C}$ , then K is Galois over $\mathbb{Q}$ .✓ Correct

The worked solution and the per-option analysis are in Trap Analysis.

Part C2023 Decembershow ▾

Let

f(X) = X^{3} - 2 \in \mathbb{Q}[X]

and let

K \subset \mathbb{C}

be the splitting field of f(X) over

\mathbb{Q}

. Let

\omega =

e^{2\pi{}i/3}

. Which of the following statements are true?

A.The Galois group of K over $\mathbb{Q}$ is the symmetric group $S_{3}$ .✓ Correct
B.The Galois group of K over $\mathbb{Q}(\omega)$ is the symmetric group $S_{3}$ .
C.The Galois group of K over $\mathbb{Q}$ is $\mathbb{Z}/3\mathbb{Z}$ .
D.The Galois group of K over $\mathbb{Q}(\omega)$ is $\mathbb{Z}/3\mathbb{Z}$ .✓ Correct

Solution

$K = \mathbb{Q}(\sqrt[3]{2}, \omega)$ has degree 6 over $\mathbb{Q}$ with non-abelian Galois group $S_{3}$ ; over $\mathbb{Q}(\omega)$ the degree is 3, so the group is $\mathbb{Z}/3$ .

Part Cshow ▾

Let K =

\mathbb{Q}(2^{1/4})

(the real fourth root of 2) over

\mathbb{Q}

. Which of the following are true?

A. $[K:\mathbb{Q}] = 4$ .✓ Correct
B. $K/\mathbb{Q}$ is a Galois extension.
C.The splitting field of $x^{4} - 2$ over $\mathbb{Q}$ has degree 8 over $\mathbb{Q}$ .✓ Correct
D.K contains a primitive 4th root of unity.

Solution

$(1) x^{4} - 2$ is irreducible over $\mathbb{Q}$ by Eisenstein at p = 2.

(3) the splitting field is $\mathbb{Q}(2^{1/4}$ , i): degree 4 from $2^{1/4}$ , and since $\mathbb{Q}(2^{1/4})$ $\subset \mathbb{R}$ does not contain i, adjoining i costs another factor of 2 — total 8 (the Galois group is the dihedral group of order 8).

(2) is false: K is real, so it does not contain the non-real roots ±i· $2^{1/4}$ of $x^{4} - 2$ — K is not the splitting field, so the extension is not normal, hence not Galois.

(4) is false for the same reason: a primitive 4th root of unity is ±i, and $K \subset \mathbb{R}$ contains no non-real numbers.

Part Cshow ▾

Let

f(x) = x^{3} - 2

over

\mathbb{Q}

, with splitting field L =

\mathbb{Q}(2^{1/3}

\omega)

where

\omega =

e^{2\pi{}i/3}

. Which of the following are true?

A. $[L:\mathbb{Q}] = 6$ .✓ Correct
B.Gal $(L/\mathbb{Q})$ is isomorphic to $S_{3}$ .✓ Correct
C. $\mathbb{Q}(2^{1/3})/\mathbb{Q}$ is a normal extension.
D.Gal $(L/\mathbb{Q})$ is abelian.

Solution

(1) [ $\mathbb{Q}(2^{1/3})$ : $\mathbb{Q}] = 3$ , and $\omega \notin$ $\mathbb{Q}(2^{1/3})$ (that field is real) with minimal polynomial $x^{2} + x + 1$ of degree 2 over it — since gcd(2,3) = 1, that degree cannot drop, giving $[L:\mathbb{Q}] = 6$ .

(2) this is the standard example of a degree-6 splitting field with Galois group $S_{3}$ .

(3) is false: $\mathbb{Q}(2^{1/3})$ contains only one of the three roots of $x^{3} - 2 ($ the other two, $\omega\cdot2^{1/3}$ and $\omega^{2}\cdot2^{1/3}$ , are non-real), so it is not the splitting field and the extension is not normal.

(4) is false directly from $(2): S_{3}$ is the smallest non-abelian group.