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| ``Re: Proof''
by mathcam on 2007-12-28 14:25:58 |
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| > No. The polynomial may well be irreducible (e.g. x^3-15x-4;
> can you factorise it in Q[x]?).
> It's true that f can be always be factorised as you write,
> but the real numbers a_j are not rational unless n is not a
> power of 2.
> Jussi
No, but you can factor it in R[x]. The point is that the term "irreducible" is relative to a given ring. When you say that a polynomial has real coefficients (as it does in the entry casus irreducibilis), the most natural place to consider its irreducibility is in the ring R[x]. If I gave you the polynomial
pi * x^2 - e*x +3,
it's essentially nonsense to talk about whether it's irreducible or not in Q[x], since it's not an element of Q[x]! Thus the only way the term "irreducible" in that entry can be taken is to mean "irreducible in R[x]", which can't be the case if it has three real roots.
There are a couple of resolutions to this. You could insist that the coefficients be rational, in which case your use of the word irreducible is fine. This is perhaps the historically best solution for the "casus irredcubilis" entry since presumably this terminology was invented only for the case of integer/rational coefficients. On the other hand, the phenomenon extends to real coefficients, so it would be a slight disservice to unnecessarily leave them out. The other option would be to remove the word irreducible.
Cam |
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