Part CCSIR NET December 2024a-product-of-fields-is-not-a-field

A product of fields is not a field

Let $f \in \mathbb{R}[x]$ be a product of distinct monic irreducible polynomials $P_{1}, P_{2}, \dots, P_{n}$ , where n ≥ 2. Let (f) denote the ideal generated by f in the ring $\mathbb{R}[x]$ . Which of the following statements are true?

A. $\mathbb{R}[x]/(f)$ is a field.
B. $\mathbb{R}[x]/(f)$ is a finite dimensional $\mathbb{R}-$ vector space.✓ Correct
C. $\mathbb{R}[x]/(f)$ is a direct sum of fields, each of which is isomorphic to $\mathbb{R}$ or $\mathbb{C}$ .✓ Correct
D.There are no non-zero elements $u \in \mathbb{R}[x]/(f)$ such that u^m = 0 for some m ≥ 1.✓ Correct

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests property not inherited.

See pricing

50 are analysed free — try those first.

The trap it tests

Property not inherited

A property assumed to pass to subobjects, quotients, or through a chain. It does not.

Drill statements like this

Related counterexample: Every prime ideal is maximal

A product of fields is not a field

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