Track ideals through a quotient

Try

In ℤ/12ℤ, the ideal generated by [4] is ❴[0],[4],[8]❵. Its full inverse image in ℤ is 4ℤ, not merely the three displayed integers. Every representative differing by a multiple of 12 must return with its residue class.

This example suggests a general rule: ideals in a quotient R/I should correspond to ideals in R that contain I. The containment is essential. The image of 8ℤ modulo 12 is also generated by [4], because 8ℤ+12ℤ=4ℤ. Passing to the quotient has forgotten the distinction between these two original ideals.

Throughout, rings are commutative with identity and ring maps preserve identity. An ideal need not be proper; the zero ring is allowed as a quotient. Prime and maximal ideals, however, are proper. Let π:R→R/I denote the quotient map r↦r+I.

Predict

Predict all ideals of ℤ/12ℤ. For an ideal dℤ of ℤ to contain 12ℤ, which divisibility condition must d satisfy? Then predict the inclusion direction: does a larger divisor d give a larger or smaller ideal?

Next compare sums, intersections and products of quotient ideals. Sums and intersections should match the corresponding operations upstairs. For products, pause: if J and K contain I, must JK contain I? Try J=4ℤ, K=6ℤ and I=12ℤ. The product is 24ℤ, which does not contain 12ℤ. The expression JK/I is therefore not defined as a quotient of these additive groups.

Experiment

Every ideal of ℤ is dℤ for a unique d≥0. Containing 12ℤ forces d to be a positive divisor of 12. Use this fact and the proof below to justify that the table is exhaustive, rather than merely a list of examples. Write [a] for the residue of a modulo 12.

For a full lattice description, draw an upward arrow for each immediate strict inclusion. The arrows are 12ℤ→6ℤ, 12ℤ→4ℤ, 6ℤ→3ℤ, 6ℤ→2ℤ, 4ℤ→2ℤ, 3ℤ→ℤ and 2ℤ→ℤ. Use the same arrows for the corresponding quotient ideals. This supplies every inclusion by following paths; ideals in different branches need not be comparable.

Predict the sum, intersection and product of the quotient ideals generated by [4] and [6].

Check all three operations

Upstairs, 4ℤ+6ℤ=2ℤ and 4ℤ∩6ℤ=12ℤ. Therefore the quotient sum is generated by [2] and the intersection is zero. The quotient product is also zero because every product is a multiple of 24, hence zero modulo 12. Its full preimage is 24ℤ+12ℤ=12ℤ. We had to add I back to the product upstairs; this correction is part of the theorem.

Understand

Construct the two maps and prove they are inverse

For an ideal L of R/I, define its inverse image π⁻¹(L)=❴r∈R:π(r)∈L❵. Differences stay in this set because π is additive and L is an additive subgroup. If r belongs to it and a∈R, then π(ar)=π(a)π(r) belongs to L by absorption. Thus π⁻¹(L) is an ideal. It contains I because every element of I maps to zero.

Conversely, let J be an ideal of R containing I. Its image π(J), written J/I, is an ideal in R/I. It is closed under differences. For absorption, every scalar in R/I has a representative a∈R, and (a+I)(j+I)=aj+I belongs to π(J). This is the step using surjectivity of π.

If r belongs to π⁻¹(π(J)), then π(r)=π(j) for some j∈J. Hence r−j∈I⊆J, so r∈J. The reverse inclusion is immediate. Therefore π⁻¹(π(J))=J. Also π(π⁻¹(L))=L: one inclusion follows from the definition and the other from taking a representative of each element of L. We have proved inverse bijections, not merely maps in one direction.

For an arbitrary ideal J not assumed to contain I, the same calculation gives π⁻¹(π(J))=J+I. This explains exactly what was lost in the initial 8ℤ example.

Inclusion, sums and intersections

Both image and inverse image preserve inclusion. Because the maps are inverse on the specified ideals, they reflect inclusion as well: J/I⊆K/I if and only if J⊆K, provided I⊆J,K.

For such J,K, (J+K)/I=(J/I)+(K/I) by taking sums of representatives. For intersections, the forward containment is immediate. If a class belongs to both J/I and K/I, choose j∈J and k∈K representing it. Then j−k∈I⊆K, so j∈K as well; the class has a representative in J∩K. Therefore (J∩K)/I=(J/I)∩(K/I). The same representative argument or the order correspondence handles arbitrary intersections. An empty intersection is the whole ring on both sides.

Products need the kernel added back

Recall that JK consists of finite sums of products jk, with j∈J and k∈K. Applying π to such sums proves π(JK)=π(J)π(K), including both containments: every product of quotient representatives lifts to one of those finite sums.

But the full inverse image of π(JK) is JK+I. Consequently the correct quotient formula is

(J/I)(K/I)=(JK+I)/I.

Only when I⊆JK can we shorten the right side to JK/I. Products and intersections are different operations; neither equality of their quotient images nor a familiar integer example permits replacing one by the other without proof.

Prime ideals, maximal ideals and the third isomorphism theorem

Let J contain I. The map R/I→R/J sending r+I to r+J is well-defined because I⊆J. It is an onto unital homomorphism with kernel J/I. The first ring isomorphism theorem gives

(R/I)/(J/I)≅R/J.

This is the third isomorphism theorem, with explicit map and kernel. The inverse correspondence also shows J/I is proper exactly when J is proper. Using the existing domain and field quotient criteria, J/I is prime exactly when J is prime, and J/I is maximal exactly when J is maximal. Alternatively, maximality follows from the inclusion-preserving bijection on intermediate ideals.

If I=R, then R/I is the zero ring. There is one ideal on either side of the correspondence, and neither side has a prime or maximal ideal in this interval. If J=R, the third theorem still holds: both resulting quotient rings are zero. Properness is required for prime and maximal claims, not for the correspondence or isomorphism itself.

What changes for a general ring map?

For a surjective unital homomorphism φ:R→S, the identical proof identifies ideals of S with ideals of R containing ker φ. The image of any ideal J is then an ideal of S, but its inverse image is J+ker φ. Images become one-to-one only after restricting to ideals containing the kernel.

For a map that is not onto, inverse images of ideals are still ideals, but images of ideals need not be ideals of the codomain. For the inclusion ℤ→ℚ, the image of 2ℤ is not an ideal of ℚ: multiply its element 2 by 1/2 to obtain 1 outside it. The image is an ideal of the image subring ℤ; this is not the same ambient ring. Also contraction of a maximal ideal need not be maximal without surjectivity: the maximal ideal 0 of the field ℚ contracts to 0 in ℤ, which is not maximal. Contraction of a prime ideal does remain prime for unital maps, by pulling back a zero-product condition and retaining properness.

Apply

Change the ambient ring to ℚ[t] and the kernel ideal to I=(t³). Find every ideal of A=ℚ[t]/(t³), identify its prime and maximal ideals, and compute the square and cube of the ideal generated by the class of t. Give a completeness argument, not only a list of powers.

Check the polynomial quotient

The polynomial ring over a field is a PID. Ideals containing (t³) are generated, after monic normalization, by divisors of t³: 1,t,t²,t³. Thus A has exactly four ideals: A, (t)/(t³), (t²)/(t³) and zero, forming a chain. The quotient by (t)/(t³) is ℚ, so that ideal is both prime and maximal. The quotient by (t²)/(t³) has the nonzero class of t with square zero, so it is not a domain. A itself also has nonzero zero divisors, so its zero ideal is not prime. The whole ideal is not proper.

Writing J for the ideal generated by the class of t, J² is generated by the class of t² and is nonzero, whereas J³=0. For example, J²≠J even though both are nonzero ideals. The third isomorphism theorem identifies A/J² with ℚ[t]/(t²); it does not turn this quotient into a field.

As a separate check, explain why replacing I=(t³) by I=(t) collapses the interval of ideals to just the whole ring and (t), giving the field ℚ. A larger kernel leaves fewer distinctions, and the correspondence describes precisely which ideals survive.

Continue

Return to ideals and quotients for absorption and finite-sum products, or Euclidean domains for the principal-ideal arguments used in the examples.

Continue to quadratic norms and factorization to distinguish irreducible elements from prime elements through a concrete ring. The later Noetherian and Hilbert-basis lesson will reuse ideal correspondence to prove finite-generation results for quotient rings.