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Math Help - Show that x^3-2 is irreducible in Z/19Z [x]/<x^2-2>

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    Show that x^3-2 is irreducible in Z/19Z [x]/<x^2-2>

    Please show that x^3-2 is irreducible in Z/19Z [x]/<x^2-2>. There HAS to be a better way than plugging and chugging 361 polynomials.

    Thanks!

    Julian
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    Quote Originally Posted by aznmaven View Post
    Please show that x^3-2 is irreducible in Z/19Z [x]/<x^2-2>. There HAS to be a better way than plugging and chugging 361 polynomials.

    Thanks!

    Julian
    It is sufficient to show t^3 - 2 has no roots in \mathbb{Z}_{19}[x]/(x^2-2). Let \alpha = x + (x^2-2). Then we see that \alpha^2 = 2.

    Now let a+b\alpha be an element in this field with a,b\in \mathbb{Z}_{19}. Now show that (a+b\alpha)^3 - 2 =0 is impossible by expanding out this the expression and using the fact \alpha^2 = 2, and comparing coefficient.
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