The map φ: ℤ[x] → ℤ defined by φ(p(x)) = p(0) sends every polynomial to its constant term. What is ker(φ)?
AThe zero polynomial only
BAll polynomials with integer coefficients
CAll polynomials with zero constant term — i.e., multiples of x
DAll polynomials with no x² term
The kernel consists of all elements sent to 0 by φ. Here φ(p(x)) = p(0), so ker(φ) = {p(x) : p(0) = 0} — exactly the polynomials whose constant term is 0, i.e., polynomials of the form a₁x + a₂x² + ... This is the ideal (x) ⊆ ℤ[x] generated by x. Notice that this set is closed under addition and absorbs multiplication by any element of ℤ[x] — the two key properties of an ideal — as predicted by the theorem that kernels of ring homomorphisms are ideals.
Question 2 Multiple Choice
Why is the kernel of a ring homomorphism φ: R → S an ideal of R, rather than merely a subring?
ABecause φ preserves addition, the kernel is closed under subtraction, which is the definition of an ideal
BThe absorption property — if φ(i) = 0 and r ∈ R, then φ(ri) = φ(r)φ(i) = φ(r)·0 = 0, so ri ∈ ker(φ)
CIdeals are exactly the subrings that contain the identity element
DKernels must be ideals by convention, since ideals generalize normal subgroups
The absorption property of ideals is forced by the multiplicative condition of ring homomorphisms. If i ∈ ker(φ) (so φ(i) = 0) and r is any element of R, then φ(ri) = φ(r)φ(i) = φ(r)·0 = 0, so ri ∈ ker(φ). This holds for both left and right multiplication. Being closed under addition follows from the additive condition, but absorption is the additional property that makes ker(φ) an ideal, not just a subring. A subring only needs to be closed under the ring operations among its own elements; an ideal must absorb multiplication from outside.
Question 3 True / False
Every ideal I of a ring R is the kernel of some ring homomorphism from R.
TTrue
FFalse
Answer: True
Given any ideal I ⊆ R, the natural quotient map φ: R → R/I defined by φ(r) = r + I is a ring homomorphism (this will be proved when you study quotient rings), and ker(φ) = {r : r + I = 0 + I} = I. So ideals are not merely a superset of kernels — they are exactly the kernels, and every ideal arises as the kernel of the quotient map. This perfect correspondence ('ideals = kernels') is the analogue of 'normal subgroups = kernels of group homomorphisms' and is one of the central structural facts of ring theory.
Question 4 True / False
A ring homomorphism φ: R → S mainly needs to satisfy φ(a + b) = φ(a) + φ(b) — the multiplicative condition φ(ab) = φ(a)φ(b) follows automatically from the additive one.
TTrue
FFalse
Answer: False
The multiplicative condition does NOT follow from the additive condition. A map that preserves addition is a group homomorphism on the additive group (R, +), but a ring has additional structure — multiplication — and preserving addition says nothing about how the map interacts with it. Consider the zero map φ(r) = 0 for all r: it preserves addition but φ(ab) = 0 ≠ φ(a)φ(b) = 0·0 = 0... actually this one works. Better: take φ: ℤ → ℤ × ℤ sending n to (n, 0). This preserves addition but φ(1·1) = (1,0) while φ(1)φ(1) = (1,0)(1,0) = (1,0) — actually also fine. The point is that both conditions must be checked independently; neither implies the other in general.
Question 5 Short Answer
State the First Isomorphism Theorem for rings and explain what it reveals about the relationship between homomorphisms, kernels, and ideals.
Think about your answer, then reveal below.
Model answer: The First Isomorphism Theorem states: if φ: R → S is a ring homomorphism, then R/ker(φ) ≅ im(φ). It reveals that homomorphisms, ideals, and quotient rings are three descriptions of the same phenomenon. Every homomorphism factors through a quotient ring — you divide R by the 'information φ forgets' (the kernel, which is an ideal) to get something isomorphic to the image. Conversely, every ideal I arises as a kernel (of the quotient map R → R/I), and every quotient ring is an image. The theorem unifies these concepts: studying surjective homomorphisms out of R is the same as studying ideals of R.
The First Isomorphism Theorem is a cornerstone of abstract algebra because it converts the purely structural question ('what maps are there from R?') into the combinatorial question ('what ideals does R have?'). It means that to understand all possible ring images of R (up to isomorphism), you need only understand all ideals of R — a much more manageable task.