Can we use Eisenstein's Irreducibility Criterion to show that $x^4+1$ is not reducible in Q?

Question

As such:

Let $a(x)=x^4+1\in\mathbb{Q}\left[x\right]$. Then choose any prime $p$. By Eisenstein's Criterion, we see that $p\nmid 1$, $p\mid 0$ (since all coefficients of intermediate terms are 0), and $p^2\nmid 1$. Thus we conclude that $a(x)$ is not reducible in $\mathbb{Q}$.

Is this valid, or am I making some glaring omission?

My professor used the Rational Root Theorem, but it turned out to be a much longer process (with all of the testing for possible roots).

Edit -- since I have to have $p \mid 1$, I tried a different method, and substituted $x=\bar{x}+1$. Then $x^4+1=\bar x^4+4\bar x^3 + 6\bar x^2 + 4\bar x + 2$, and chose $p=2$. Then, I believe, it meets the criterion. Is this correct?

Answer 1

For applying Eisenstein on a polynomial $P = a_n x^n+ ... + a_1 x + a_0$ you should have $p | a_k \forall k <n$ and $p^2 \not|a_0 $ and $p\not| a_n$.

But in your case $P(X) = X^4+1$. But you can apply Eisenstein if you substitute $X = Y+1$. Then

$$P(X) = Y^4+4Y^3+6Y+4Y+2$$ and you can use $p=2$.

Answer 2

You can apply Gauss' lemma and then Eisenstein at $2$ to $a(x+1)=(x+1)^4+1=x^4+4x^3+6x^2+4x+2$ to conclude that $a$ is irreducible.