Skip to content
Part CCSIR NET December 2023

Ideals, quotient rings, prime & maximal ideals, CRT

Let n∈Zn \in \mathbb{Z} be such that n ≡ 1 (mod 7) and n ≡ 4 (mod 15). Which of the following statements are true?

  1. A.n ≡ 1 (mod 3).✓ Correct
  2. B.n ≡ 1 (mod 35).
  3. C.n ≡ 1 (mod 21).✓ Correct
  4. D.n ≡ 1 (mod 5).

The idea, in plain words

A statement about a composite modulus is really several statements at once, one for each prime factor. Break both conditions into their prime pieces, see which pieces you actually know, and only claims built from known pieces survive.

Solution

n ≡ 4 (mod 15) gives n ≡ 1 (mod 3) and n ≡ 4 (mod 5).

With n ≡ 1 (mod 7), CRT gives n ≡ 1 (mod 21); n ≢ 1 (mod 5) kills the mod-35 claim.

Why each option is right or wrong

Checked against 6 symbolic computations
  • A.Correct

    True.

    Split the modulus: 15 = 3 × 5

    so n ≡ 4 (mod 15) gives n ≡ 4 ≡ 1 (mod 3) and n ≡ 4 (mod 5).

    SymPy verifies on the witness n = 64

    which satisfies both hypotheses (64 mod 7 = 1, 64 mod 15 = 4) and gives 64 mod 3 = 1.

    The trigger: whenever a composite modulus appears, factor it and work prime by prime.

  • B.WrongExecution slip

    False.

    35 = 5 × 7, and although n ≡ 1 (mod 7) holds, the mod-5 part does not: n ≡ 4 (mod 5), not 1.

    The witness settles it — 64 mod 35 = 29, not 1.

    This option is what you get by combining the modulus 7 with the modulus 15 without splitting off the 5 and checking it.

  • C.Correct

    True.

    21 = 3 × 7, and both parts hold: n ≡ 1 (mod 3) from (A) and n ≡ 1 (mod 7) by hypothesis.

    By CRT, n ≡ 1 (mod 21).

    SymPy confirms 64 mod 21 = 1.

    Note the contrast with (B): 21 works because both of its prime factors give residue 1, and 35 fails because one of them does not.

  • D.WrongExecution slip

    False. n ≡ 4 (mod 15) reduces to n ≡ 4 (mod 5), not 1 — the witness gives 64 mod 5 = 4.

    Reducing '4 mod 15' to '1 mod 5' is the slip; 4 and 1 are only congruent modulo 3.

2 of these 2 wrong options correspond to a named reasoning error — the same errors recur across subjects.

Related counterexample: Every prime ideal is maximal

More on this topic

The chapter behind this: Ideals, quotients and the Chinese remainder theorem — free to read

From Rings and Fields › Ideals, quotient rings, prime & maximal ideals, CRT

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

ShareWhatsAppTelegram