Re: degree 61 polynomial

On Thu, 19 Feb 2009 21:05:00 -0800 (PST), Mariano Suárez-Alvarez
<mariano.suarezalvarez@xxxxxxxxx> wrote:

On Feb 20, 2:52 am, quasi <qu...@xxxxxxxx> wrote:
On Thu, 19 Feb 2009 23:00:18 -0500, quasi <qu...@xxxxxxxx> wrote:
Prove or disprove:

If an irreducible polynomial f in Q[x] is such that the sum of some
nonempty proper subset of the roots of f is zero, then the sum of all
the roots of f is zero.

To see the relation to tommy1729's original problem, assume the claim
of the above problem. Let's call it "the hypothetical claim".

Now consider tommy's question:

suppose a student has an irreducible integer polynomial
of degree 61.

he or she claims that polynomial has 5 zeros such that :

x1 + x2 + x3 + x4 + x5 = 3

and dares you to prove or disproof it without using a computer.

could you ?

Let's also assume that tommy meant a _monic_ irreducible integer
polynomial of degree 61.

With these assumptions, the answer to tommy's question is "yes" -- the
student's claim can be disproved -- no such polynomial exists.

Suppose otherwise.

Let y_n = 5*x_n - 3, for n = 1, ..., 61.

Then y_1, ..., y_61 are the roots of the monic integer polynomial

   g(y) = (5^61)*f((y+3)/5)

Since the fields Q(x_1) and Q(y_1) are equal, and since x1 has degree
61 over Q, the degree of y1 over Q is also 61. It follows that g is
irreducible in Q[y].

Urgh. There is no need to invoke fields and degrees
for this! Just suppose that g is reducible, and
deduce (by a change of variables inverse to the
one that got you g from f) that f itself is reducible.

Yes I saw that reasoning as well, but I thought the explanation would
appear cleaner using field degrees.

quasi
.

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