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Determine whether f(x)=x^{4}+15x+7\, is irreducible over {\mathbb  {Q}}\, or not.

Mapping this polynomial to a new field by way of \sigma :{\mathbb  {Q}}[x]\rightarrow {\mathbb  {Z}}/(2)[x]\, gives f(x)\mapsto f^{\sigma }(x)=x^{4}+x+1\,.

Check all elements in the new field for roots (linear factors): f^{\sigma }(0)=f^{\sigma }(1)=1\, so there are no linear factors.

x^{2}\, is reducible. x^{2}+1\, has 1 as a root. x^{2}+x=x(x+1)\, The only irreducible quadratic in this field is x^{2}+x+1\,. Polynomial long division will show that it does not divide into x^{4}+x+1\,. Therefore f^{\sigma }\, is irreducible over {\mathbb  {Z}}/(2)\, and f\, is irreducible over {\mathbb  {Q}}\,.

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