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- Jun 22, 2012

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*Let F be a field and let [TEX] A \ne 0 [/TEX] be an ideal of F[x]. Then a uniquely determined monic polynomial h exists exists in F[x] such that A = (h).*

**Theorem 1.**The beginning of the proof reads as follows:

*Because [TEX] A \ne 0 [/TEX], it contains non-zero polynomials and hence contains monic polynomials (being an ideal) ... ... etc. etc.*

**Proof:**BUT! why must A contain monic polynomials??

Help with this matter would be appreciated!

Peter

[This has also been posted on MHF]