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1) Show that x^2 +2 is irreducible in Z5 [x]
given this hint: if ax+b is a factor of x^2+2, then so is
x+a^-1b = a^-1(ax+b)

2) Show that x^2+3 is irreducible in Z5 [x]

(Z is the integers and Z5 is Z mod 5)

any help would be greatly appreciated!

2006-10-31 18:47:49 · 1 answers · asked by jjodom1010 3 in Science & Mathematics Mathematics

1 answers

5 is a prime number, so the ring of integers mod 5 has no zero divisors. Ergo, if (ax-b) is a factor of a polynomial P(x), then the congruence relation P(x)≡0 has a solution b/a. So all that is necessary to show that x²+2 is irreducible in Z5 is to show that x²+2≡0 has no solutions in Z5. This is easy, just note that this implies x²≡3 mod 5, but:
0²≡0 mod 5
1²≡1 mod 5
2²≡4 mod 5
3²≡4 mod 5
4²≡1 mod 5

so x²≡3 mod 5 has no solutions, and x²+2 is irreducible.

For the second one, you can use exactly the same method.

Edit: corrected minor typographical error.

2006-10-31 19:42:39 · answer #1 · answered by Pascal 7 · 2 0

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