Questions: Subrings and Ideals

Question 1

The integers ℤ form a subring of the rationals ℚ. Is ℤ also an ideal of ℚ?

• A. Yes — ℤ is closed under addition and subtraction, so it satisfies all ideal requirements
• B. No — because multiplying a rational number like 1/2 by an integer like 1 gives 1/2, which is not in ℤ
• C. Yes — every subring of a commutative ring is automatically an ideal
• D. No — because ℤ does not contain the multiplicative identity of ℚ

Answer: (B)

An ideal I of R must satisfy: for every r ∈ R and a ∈ I, both ra ∈ I and ar ∈ I. Testing ℤ in ℚ: take r = 1/2 ∈ ℚ and a = 1 ∈ ℤ. Then ra = 1/2 ∉ ℤ. So ℤ fails the absorbing property and is not an ideal of ℚ. Option A captures the subring requirement but ignores the additional absorbing property that ideals need. The key distinction: subrings must only be closed under multiplication of elements *within* the subring; ideals must absorb multiplication by *any* element of the ambient ring.

Question 2

Which property distinguishes an ideal from a mere subring?

• A. An ideal must contain the multiplicative identity 1, while a subring need not
• B. An ideal must be closed under multiplication of its own elements, while a subring need not
• C. An ideal must absorb multiplication by any element of the ambient ring, while a subring only needs closure under its own elements
• D. An ideal must be finite, while a subring can be infinite

Answer: (C)

A subring S ⊆ R is closed under ring operations on its own elements: for s₁, s₂ ∈ S, we need s₁s₂ ∈ S. An ideal I ⊆ R is strictly stronger: for any r ∈ R (not just r ∈ I) and any a ∈ I, we need ra ∈ I and ar ∈ I. This 'absorbing' property — the ideal swallows multiplication from the entire ambient ring — is exactly what allows quotient rings to be well-defined. Option A is backwards: ideals often do NOT contain 1 (if I is a proper ideal of a ring with 1, then 1 ∉ I).

Question 3

Nearly every subring of a commutative ring is also an ideal of that ring.

• T. True
• F. False

Answer: False

False. The integers ℤ are a subring of ℚ but not an ideal (as shown above). More generally, any ring R is a subring of itself; but the only ideal of a field (like ℚ) is {0} and the whole field. So ℤ ⊂ ℚ is a proper subring that is not an ideal. The confusion arises because closure under the subring's own multiplication is a weaker condition than absorbing all of R's multiplication. Every ideal is a subring, but not every subring is an ideal.

Question 4

In the ring of integers ℤ, every ideal is a principal ideal — generated by a single element.

• T. True
• F. False

Answer: True

True. ℤ is a principal ideal domain (PID). Any ideal I ⊆ ℤ that is not {0} contains a smallest positive integer n, and every element of I must be a multiple of n (by the division algorithm), so I = ⟨n⟩ = nℤ. This is a special property of ℤ — not all rings are PIDs. For example, in ℤ[x], the ideal generated by {2, x} cannot be expressed as a principal ideal ⟨f(x)⟩ for any single polynomial.

Question 5

Why are ideals — rather than subrings — the right substructure for forming quotient rings, and what goes wrong if you try to form a quotient by a subring that isn't an ideal?

Model answer: To form a quotient ring R/I, we define cosets r + I and need multiplication of cosets to be well-defined: (r₁ + I)(r₂ + I) = r₁r₂ + I. This requires that if we replace r₁ by r₁ + a (a different representative of the same coset), the product coset doesn't change — i.e., (r₁ + a)r₂ = r₁r₂ + ar₂ must land in the same coset, which requires ar₂ ∈ I. This is exactly the absorbing property of ideals. If I is only a subring, ar₂ might not be in I for arbitrary r₂ ∈ R, and coset multiplication becomes undefined.

The parallel to group theory is illuminating: normal subgroups (not arbitrary subgroups) are kernels of group homomorphisms, exactly because left and right cosets coincide. Ideals play the same role for rings: they are precisely the kernels of ring homomorphisms, and forming a quotient is constructing the codomain of the 'collapsing' homomorphism that sends all elements of I to zero.