Is Eisenstein all powerful?
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From: quasi on 3 Aug 2010 14:04 Prove or disprove: If f in Z[x] with deg(f) > 1 is irreducible then there exists g in Z[x] with deg(g) >= 1 such that the polynomial f(g(x)) is irreducible by the Eisenstein criterion. quasi
From: quasi on 3 Aug 2010 14:34 On Tue, 03 Aug 2010 13:04:31 -0500, quasi <quasi(a)null.set> wrote: >Prove or disprove: > >If f in Z[x] with deg(f) > 1 is irreducible then there exists g in >Z[x] with deg(g) >= 1 such that the polynomial f(g(x)) is irreducible >by the Eisenstein criterion. To clarify ... It's immediate that if f(g(x)) is irreducible and deg(g)>=1, then f(x) is also irreducible. But the problem asks, for irreducible f with deg(f) > 1, is there always such a g for which the Eisenstein criterion could be _directly_ applied to prove the irreducibility of f(g(x)) (and thus indirectly establish the irreducibility of f). quasi
From: Robert Israel on 3 Aug 2010 17:27 quasi <quasi(a)null.set> writes: > On Tue, 03 Aug 2010 13:04:31 -0500, quasi <quasi(a)null.set> wrote: > > >Prove or disprove: > > > >If f in Z[x] with deg(f) > 1 is irreducible then there exists g in > >Z[x] with deg(g) >= 1 such that the polynomial f(g(x)) is irreducible > >by the Eisenstein criterion. > > To clarify ... > > It's immediate that if f(g(x)) is irreducible and deg(g)>=1, then f(x) > is also irreducible. > > But the problem asks, for irreducible f with deg(f) > 1, is there > always such a g for which the Eisenstein criterion could be _directly_ > applied to prove the irreducibility of f(g(x)) (and thus indirectly > establish the irreducibility of f). > > quasi Try f(x) = x^2 + 4. Suppose Eisenstein worked for g(x) = sum_{j=0}^n a_j x^j, f(g(x)) = (a_0^2 + 4) + sum_{k=1}^{2n} sum_{i=max(0,k-n)}^{min(k,n)} a_i a_{k-i} x^k Since a0^2 + 4 == 0 or 1 mod 4, p can't be 2. Now p does not divide a_n, but (considering the case k=2n-1) does divide a_{n-1}, and (similarly by looking at k=2n-2, ..., k=n) must divide a_{n-2}, ..., a_0. But then p can't divide a_0^2 + 4, contradiction. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: quasi on 4 Aug 2010 02:07 On Tue, 03 Aug 2010 16:27:52 -0500, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >quasi <quasi(a)null.set> writes: > >> On Tue, 03 Aug 2010 13:04:31 -0500, quasi <quasi(a)null.set> wrote: >> >> >Prove or disprove: >> > >> >If f in Z[x] with deg(f) > 1 is irreducible then there exists g in >> >Z[x] with deg(g) >= 1 such that the polynomial f(g(x)) is irreducible >> >by the Eisenstein criterion. >> >> To clarify ... >> >> It's immediate that if f(g(x)) is irreducible and deg(g)>=1, then f(x) >> is also irreducible. >> >> But the problem asks, for irreducible f with deg(f) > 1, is there >> always such a g for which the Eisenstein criterion could be _directly_ >> applied to prove the irreducibility of f(g(x)) (and thus indirectly >> establish the irreducibility of f). >> >> quasi > >Try f(x) = x^2 + 4. Suppose Eisenstein worked for >g(x) = sum_{j=0}^n a_j x^j, >f(g(x)) = (a_0^2 + 4) > + sum_{k=1}^{2n} sum_{i=max(0,k-n)}^{min(k,n)} a_i a_{k-i} x^k > >Since a0^2 + 4 == 0 or 1 mod 4, p can't be 2. >Now p does not divide a_n, but (considering the case k=2n-1) does divide >a_{n-1}, and (similarly by looking at k=2n-2, ..., k=n) must divide >a_{n-2}, ..., a_0. But then p can't divide a_0^2 + 4, contradiction. A nice, simple counterexample. Thanks. quasi
From: quasi on 4 Aug 2010 14:07
On Wed, 04 Aug 2010 01:07:43 -0500, quasi <quasi(a)null.set> wrote: >On Tue, 03 Aug 2010 16:27:52 -0500, Robert Israel ><israel(a)math.MyUniversitysInitials.ca> wrote: > >>quasi <quasi(a)null.set> writes: >> >>> On Tue, 03 Aug 2010 13:04:31 -0500, quasi <quasi(a)null.set> wrote: >>> >>> >Prove or disprove: >>> > >>> >If f in Z[x] with deg(f) > 1 is irreducible then there exists g in >>> >Z[x] with deg(g) >= 1 such that the polynomial f(g(x)) is irreducible >>> >by the Eisenstein criterion. >>> >>> To clarify ... >>> >>> It's immediate that if f(g(x)) is irreducible and deg(g)>=1, then f(x) >>> is also irreducible. >>> >>> But the problem asks, for irreducible f with deg(f) > 1, is there >>> always such a g for which the Eisenstein criterion could be _directly_ >>> applied to prove the irreducibility of f(g(x)) (and thus indirectly >>> establish the irreducibility of f). >>> >>> quasi >> >>Try f(x) = x^2 + 4. Suppose Eisenstein worked for >>g(x) = sum_{j=0}^n a_j x^j, >>f(g(x)) = (a_0^2 + 4) >> + sum_{k=1}^{2n} sum_{i=max(0,k-n)}^{min(k,n)} a_i a_{k-i} x^k >> >>Since a0^2 + 4 == 0 or 1 mod 4, p can't be 2. >>Now p does not divide a_n, but (considering the case k=2n-1) does divide >>a_{n-1}, and (similarly by looking at k=2n-2, ..., k=n) must divide >>a_{n-2}, ..., a_0. But then p can't divide a_0^2 + 4, contradiction. > >A nice, simple counterexample. > >Thanks. At this point, it appears that the Eisenstein criterion can only be regarded as a special case method. It works beautifully when it works, but its lack of general power is exposed by its apparent inability to establish the irreducibility of x^2 + 4. However, let's boost the power of Eisenstein and try again ... An amplified Eisenstein criterion: Let f in Z[x] with deg(f) = n > 1 be given by f(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_0 Suppose there is a prime p, and a positive integer k such that (1) p does not divide a_n (2) p^k divides a_i for i = 0 ... (n-1) (3) p^(k+1) does not divide a_0 Then f is irreducible. Proof: Omitted for now, but almost line by line (I think) the same as the proof of the ordinary Eisenstein criterion. I hope it's correct. With this amplified criterion, x^2 + 4 is proved irreducible without the need for transformation by composition (use p=2, k=2). So now we can repose the original question, replacing the ordinary Eisenstein criterion with the amplified version. quasi |