The ring ℤ[x] (polynomials with integer coefficients) and the ring ℚ[x] (polynomials with rational coefficients) differ in an important structural way. Which statement correctly identifies this difference?
Aℤ[x] is a field but ℚ[x] is not, because integers are more fundamental than rationals
Bℚ[x] is a principal ideal domain with a division algorithm, while ℤ[x] is only a UFD — division with remainder is not generally possible in ℤ[x]
Cℤ[x] is a PID because ℤ is a UFD, while ℚ[x] is only a UFD because ℚ is a field
DBoth rings are fields, but ℚ[x] has more elements since ℚ is larger than ℤ
This question targets the key structural hierarchy. ℤ is a UFD but not a field, so ℤ[x] is also a UFD — but not a PID, because not every ideal in ℤ[x] is principal. For example, the ideal generated by (2, x) in ℤ[x] cannot be generated by a single polynomial. ℚ is a field, so ℚ[x] is a PID — every ideal is generated by a single polynomial, and the division algorithm (dividing any polynomial by any nonzero polynomial with a remainder of smaller degree) works. This is why F[x] over a field F mirrors the structure of ℤ so closely: both are PIDs with Euclidean algorithms.
Question 2 Multiple Choice
What does Gauss's lemma say, and why is it the key tool for proving that ℤ[x] is a UFD given that ℤ is a UFD?
AGauss's lemma says that if a polynomial f ∈ ℤ[x] factors in ℚ[x], then it factors in ℤ[x] — this lifts factorization from ℚ to ℤ[x]
BGauss's lemma says that ℤ[x] has a division algorithm, making it a Euclidean domain
CGauss's lemma says that every prime ideal in ℤ[x] is generated by a prime integer
DGauss's lemma says that the product of two primitive polynomials is primitive, which ensures unique factorization is preserved under multiplication
Gauss's lemma states that if a polynomial with integer coefficients factors as a product of two polynomials with rational coefficients, it also factors as a product of two polynomials with integer coefficients (up to rational scalar factors). This is the bridge: it shows that factoring in the larger ring ℚ[x] (which is a PID and easy to work in) implies factoring in ℤ[x] itself. Combined with the unique factorization in ℤ and ℚ[x], this establishes that irreducible elements in ℤ[x] are genuinely irreducible and factor uniquely. The primitive polynomial version (option D) is related but is more commonly the statement of Gauss's lemma in textbooks; either version makes option A essentially correct.
Question 3 True / False
If R is a unique factorization domain, then the polynomial ring R[x] is also a unique factorization domain.
TTrue
FFalse
Answer: True
This is one of the central theorems of polynomial ring theory. The proof uses Gauss's lemma to show that irreducible elements in R[x] are either irreducible elements of R (viewed as constant polynomials) or primitive polynomials that are irreducible in F[x], where F is the fraction field of R. Together these generate all elements of R[x], and the factorization can be shown to be unique up to units and order. The result extends iteratively: if R is a UFD, so are R[x], R[x,y], R[x,y,z], and so on — polynomial rings in any number of variables over a UFD are UFDs.
Question 4 True / False
The polynomial ring F[x] over a field F is itself a field, because most nonzero polynomial has a multiplicative inverse.
TTrue
FFalse
Answer: False
F[x] is a principal ideal domain but not a field. A field requires that every nonzero element has a multiplicative inverse within the ring. But no polynomial of degree ≥ 1 has a multiplicative inverse in F[x]: the product of two polynomials of degrees m and n has degree m+n ≥ 1, so it can never equal the constant polynomial 1. F[x] does inherit very rich structure from F being a field — it has a division algorithm, unique factorization, and a Euclidean algorithm — but these properties make it an analogue of ℤ, not an analogue of ℚ. To get a field from F[x], you must quotient by an irreducible polynomial: F[x]/(p(x)) is a field when p(x) is irreducible.
Question 5 Short Answer
What does it mean for x to be a 'formal variable' in R[x], and how does this differ from treating polynomials as functions?
Think about your answer, then reveal below.
Model answer: In R[x], x is not a number or element of R — it is a formal symbol used to index coefficients. A polynomial a₀ + a₁x + a₂x² is really just the finite sequence (a₀, a₁, a₂) with a specific addition and multiplication rule, where x tracks the positions. You do not evaluate it; x has no value. As a ring, R[x] is about the algebraic structure of expressions. This differs from polynomial functions: a polynomial function takes an element c ∈ R and returns the value a₀ + a₁c + a₂c². Over infinite rings or fields, the two notions coincide (equal functions ↔ equal polynomials), but over finite rings they diverge: two different polynomials in R[x] can define the same function.
The formal/function distinction matters algebraically because it clarifies what R[x] actually is. The evaluation homomorphism — 'plug in c' — is a specific ring homomorphism from R[x] to R, not an identity. Its kernel (polynomials that evaluate to 0 at c) is an ideal generated by (x − c) when R is a field, and studying this kernel is how you connect polynomial algebra to root-finding and field extensions. Conflating the ring of expressions with the set of functions obscures this structure.