Irreducible polynomial mod p^2

**robeuler** · November 19th 2010, 05:33 AM

Consider the polynomial $x^{2} + p$ modulo $p^{2}$ .

I would like to show that this polynomial is irreducible. Unfortunately I don't see how I can apply any of the usual tricks (Eisenstein, simple enumeration in the case of fields of small characteristic) might work.

**roninpro** · November 19th 2010, 06:54 AM

Suppose $x^2+p\pmod{p^2}$ is reducible. Then, $x^2+p\equiv 0\pmod{p^2}$ has a solution. By reducing the modulus, we have $x^2\equiv 0\pmod{p}$ , so $x$ is divisible by $p$ . Write $x=py$ . Putting this back into the original equation gives $(py)^2+p\equiv p^2y^2+p\equiv p \equiv 0\pmod{p^2}$ , which is certainly not true.

Therefore, $x^2+p\pmod{p^2}$ is irreducible.

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