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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 | + | 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> |
| + | 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> | ||
--[[User:Bcaulkin|Bcaulkin]] 22:38, 8 April 2009 (UTC) | --[[User:Bcaulkin|Bcaulkin]] 22:38, 8 April 2009 (UTC) | ||
Revision as of 17:40, 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)
I agree for parts b and d, however, 0 is contained in any $ Z/Z_n $ 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?
I do know Einstein's explicitly says p cannot divide any $ a_n $ so 2, 3, 5, 7 cannot be used for e if we multiply by 14.
--Bcaulkin 22:38, 8 April 2009 (UTC)
