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I've just started to work myself into the topic of irreducibility of polynomials and I would like to show the following:

Proof that $x^4+1$ is irreducible over $\mathbb{Q}$

I've managed to come up with a proof that seems reasonable to me but I'm not sure if it is actually correct. Could someone tell me if I made any beginner's mistakes?

My proof: Clearly $x^4+1$ has no root in $\mathbb{Q}$, so it can only be decomposed into two polynomials of degree 2. Therefore there must be $a, b \in \mathbb{Q}$ s.t. $(x^4+1) = (x^2+ax+1)(x^2+bx+1)$. If we multiply this out we get $x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1$. If we now equate the coefficients we get that $a+b = 0$ and therefore $a = -b$ and that $ab+2=0$ and therefore that $-(a^2)=2$, which is not possible in $\mathbb{Q}$.

Any feedback is appreciated!

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    Why could it not be $(x^2+ax-1)(x^2+bx-1)$? – 2017-01-13
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    How do you know that the constant terms of the two quadratic factors have to be $1$? I think you need to invoke Gauß's lemma explicitly and work over $\mathbb Z$ to do that. (And even then, it would seem that @lulu's objection still applies). – 2017-01-13
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    For a different approach you can use a substitution $x\mapsto y+1$ to obtain a polynomial to which Eisenstein's criterion can be applied – 2017-01-13

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Your proof is almost fine - you just have to write more generally $$ x^4+1=(x^2+ax+b)(x^2+cx+d), $$ where we obtain of course $bd=1$, which over the integers only leaves the possibilities $b=d=1$, or $b=d=-1$. In both cases, the associated Diophantine equations have no solution, and this is easy to see (we obtain $c=−a$, and $a^2 =b+d=±2$, which is not possible).

Furthermore, the polynomial $x^4+1$ is the $8$-th cyclotomic polynomial, and hence irreducible by the theorem of your algebra lecture. Of course, there are several other methods to prove this directly, e.g., writing $f(x)=x^4+1$, and applying Eisenstein to $$ f(x+1)=x^4+4x^3+6x^2+4x+2. $$

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For this polynomial, it can simplified, since we know its factorisation over $\mathbf R$: $$x^4+1=(x^2+1)^2-2x^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1),$$ and this factorisation does not have rational coefficients. So by the uniqueness of factorisation, it cannot have another factorisation over $\mathbf Q$.

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    Nice idea to show it like this. To avoid potential over-generalization of this method by some reader, it might be worth stressing that this works because we know this factorization over the reals is the factorization into *irreducible* polynomials (over the reals). Just writing down some decomposition into two factors over the reals that is not one over the rationals does in general not suffice. – 2017-01-13
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    You're quite right. I just started from the initial observation of the O.P. – 2017-01-13
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Your approach in principle works fine you just tried to take a short-cut that is not available, yet is not really needed. Instead:

$$(x^4+1) = (x^2+ax+c)(x^2+bx+d)$$

We get

$$x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd.$$

You get $cd=1$ and thus $c=d^{-1}$. You still get $a=-b$ (from the third power), and then $ad +bc$ becomes $ad-ac$ and further $a(c^{-1}-c)$.

Thus, $a=0$ or $c= c^{-1}$ (that is, $c = \pm 1$).

In the former case, you get $c+ c^{-1}= c+d = 0$ which is impossible. In the latter case, you already dealt with $c=d=1$ and $c= d= -1$ is essentially the same, it yields the impossible $a^2 = -2$.

That said, the Eisenstein approach mentioned in a comment is more elegant, if you have it available.