Questions: First Isomorphism Theorem for Rings

Question 1

The evaluation map φ: R[x] → R defined by φ(p(x)) = p(1) is a surjective ring homomorphism. Which ring is R[x]/(x−1) isomorphic to, and why?

• A. R[x], because the quotient just removes the root at x=1
• B. R, because φ is surjective with kernel exactly (x−1), so R[x]/ker(φ) ≅ im(φ) = R
• C. Z, because evaluation at 1 always produces an integer
• D. R[x]/(x−1) cannot be simplified further without more information

Answer: (B)

The First Isomorphism Theorem says R/ker(φ) ≅ im(φ). Here ker(φ) = {p(x) : p(1) = 0} = (x−1) (all multiples of x−1), and im(φ) = R since every real number is p(1) for some polynomial. So R[x]/(x−1) ≅ R. The theorem's power is exactly this: rather than reasoning directly about cosets, you build the right surjection and let the theorem do the work.

Question 2

A student wants to prove Z[x]/(x²+1) ≅ Z[i] by defining φ: Z[x] → Z[i] via φ(p(x)) = p(i). Which statement correctly explains why the First Isomorphism Theorem applies?

• A. Because Z[x] and Z[i] have the same cardinality, so any bijection between them gives an isomorphism
• B. Because φ is a surjective ring homomorphism with kernel (x²+1), so Z[x]/ker(φ) ≅ im(φ) = Z[i]
• C. Because (x²+1) is a prime ideal, which automatically forces the quotient to be a known ring
• D. Because φ maps the generator x to i, and isomorphisms need only be defined on generators

Answer: (B)

The theorem requires φ to be a ring homomorphism, φ to be surjective onto Z[i], and ker(φ) = (x²+1). Checking: φ(p(x)+q(x)) = p(i)+q(i) ✓, φ(p·q) = p(i)q(i) ✓ (ring hom); every a+bi ∈ Z[i] equals (a+bx)(i) = φ(a+bx) ✓ (surjective); and p(i) = 0 iff (x²+1) | p(x) ✓ (kernel). Cardinality arguments (option A) don't establish ring isomorphism — you need the algebraic structure to match.

Question 3

If φ: R → S is a surjective ring homomorphism, then R/ker(φ) is isomorphic to S.

• T. True
• F. False

Answer: True

When φ is surjective, im(φ) = S, so the First Isomorphism Theorem gives R/ker(φ) ≅ im(φ) = S. Surjectivity is what collapses the general statement R/ker(φ) ≅ im(φ) to the cleaner R/ker(φ) ≅ S. If φ were not surjective, you would only get R/ker(φ) ≅ im(φ), a proper subring of S.

Question 4

For any ring homomorphism φ: R → S, the First Isomorphism Theorem gives an isomorphism R/ker(φ) ≅ S.

• T. True
• F. False

Answer: False

The theorem gives R/ker(φ) ≅ im(φ), not R/ker(φ) ≅ S. These coincide only when φ is surjective. If φ: Z → Z is multiplication by 2, then im(φ) = 2Z ≠ Z, and Z/ker(φ) = Z/{0} = Z ≅ 2Z, not all of Z. Conflating im(φ) with S is the most common misstatement of the theorem.

Question 5

What practical strategy does the First Isomorphism Theorem enable for proving that a quotient ring R/I is isomorphic to a known ring S?

Model answer: Find a surjective ring homomorphism φ: R → S whose kernel is exactly I. Then the theorem immediately gives R/I = R/ker(φ) ≅ im(φ) = S, without needing to construct the isomorphism directly.

This strategy — build the homomorphism, check surjectivity, identify the kernel — is far more tractable than directly defining a map on cosets and verifying all isomorphism properties. It separates the work into recognizable pieces: what ring do I want? What map sends R onto it? What does the map kill? The theorem then packages those answers into an isomorphism. This is why the theorem is called a 'structure theorem' — it provides a blueprint for algebraic identification.