I need to show that $p(x)=x^4-2x^2-4$ is irreductible over $\mathbb Q.$ Here's what I've done:
Please tell me if it's correct
Over $\mathbb C,$ $x^4-2x^2-4\\=(x^2-1)^2-5\\=(x^2-1+\sqrt 5)(x^2-1-\sqrt 5)\\=(x+\sqrt{1+\sqrt 5})(x-\sqrt{1+\sqrt 5})(x+\sqrt{1-\sqrt 5})(x-\sqrt{1-\sqrt 5})...(1)$
Since $\mathbb C[x]$ is an UFD so representing $p(x)$ as $(1)$ is unique upto rearrangement of the irreducible factors and inserting associates. So if we want to express $p(x)$ as $f(x)g(x)$ then $f,g$ must be some or all products of $(x+\sqrt{1+\sqrt 5}),$$(x-\sqrt{1+\sqrt 5}),$$(x+\sqrt{1-\sqrt 5}),$$(x-\sqrt{1-\sqrt 5})$ multiplied by a rational. As long as not all of them are taken the product won't belong to $\mathbb Q[x].$ So $p(x)$ can only be factored in the trivial way over $\mathbb Q.$