Talk:Casus irreducibilis

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Mathematics rating: Start Class Low Priority  Field: Algebra

[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(xr1)(xr2)(xr3) 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)
With your recent edits all my points have been satisfactorily addressed. Thanks.  --Lambiam 15:07, 1 May 2008 (UTC)