Talk:Casus irreducibilis
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[edit] Something wrong?
Take the cubic polynomial x3 − 3x2 + 2x. Its discriminant is equal to 4. According to the article, this mean we are in casus irreducibilis. However, the polynomial factors as x(x − 1)(x − 2).
Furthermore, what is the meaning of real field in the section Formal statement and proof?
- If the meaning is that F is the field of real numbers, and the three real roots are r1, r2, and r3, the polynomial can be written in the form a(x − r1)(x − r2)(x − r3) and so is by definition not irreducible over F. Also, then why not simply use ℝ instead of F?
- If F is a formally real field, then what does this have to do with polynomials having real roots?
--Lambiam 14:37, 1 May 2008 (UTC)
- Hi Lambiam, let me respond to your points.
- The polynomial has to be irreducible over F (or, in the case of the lead, Q). I see that this condition had been left out of the discriminant condition, even though it was mentioned in the previous paragraph. I have added it to the second paragraph as well, since that is an obvious source of confusion.
- By "real field" I meant "formally real field". Thanks for catching the fact that real field is a disambiguation page. I think the terminology "real field" is quite established for what we are calling a "formally real field", and that the only reason there is a disambiguation page at all is because of a certain laziness in linking.
At any rate, any real field is a subfield of the real numbers (although the real numbers are not needed to define a real field).In this context, "real root" means "root contained in the real closure". I will try to clarify this in the article. silly rabbit (talk) 14:55, 1 May 2008 (UTC)
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- With your recent edits all my points have been satisfactorily addressed. Thanks. --Lambiam 15:07, 1 May 2008 (UTC)