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I agree for parts b and d, however, 0 is contained in any <math>Z/Z_n</math> so using p=3 for parts a, and c fails because 0 is a zero.  However I've missed a few classes, so perhaps we don't consider 0?<br>
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Einstein's explicitly says p cannot divide any <math>a_n</math> so p cannot be 3 for parts a, c or e<br>
I do know Einstein's explicitly says p cannot divide any <math>a_n</math> so 2, 3, 5, 7 cannot be used for e if we multiply by 14.<br>
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--[[User:Bcaulkin|Bcaulkin]] 22:38, 8 April 2009 (UTC)
 
--[[User:Bcaulkin|Bcaulkin]] 22:38, 8 April 2009 (UTC)
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According to theorem 17.4 in the book says that the leading coefficient cannot be divisible by p, but all the others must be.  In addition the last coefficient cannot be divisible by p^2.
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My fault, I misread it.  Thanks!

Latest revision as of 18:50, 8 April 2009


Are the following irreducible over Q?

  • a) $ x^5 + 9x^4 + 12x^2 + 6 $
  • b) $ x^4 + x + 1 $
  • c) $ x^4 + 3x^2 + 3 $
  • d) $ x^5 + 5x^2 + 1 $
  • e) $ (5/2)x^5 + (9/2)x^4 + 15x^3 + (3/7)x^2 + 6x + (3/14) $

a.) Look at Eisenstein's with p = 3.
b.) A polynomial is irreducible in Q if there's a p such that f(x) mod p is irreducible. Look at p = 2.
c.) See part a.
d.) See part b.
e.) Multiply by 14 then see part a.
--Jniederh 22:12, 8 April 2009 (UTC)


Einstein's explicitly says p cannot divide any $ a_n $ so p cannot be 3 for parts a, c or e
--Bcaulkin 22:38, 8 April 2009 (UTC)


According to theorem 17.4 in the book says that the leading coefficient cannot be divisible by p, but all the others must be. In addition the last coefficient cannot be divisible by p^2.


My fault, I misread it. Thanks!

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Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

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