Let $p$ be a prime and $n=p+1$.
Then prove or disprove whether $x^{2p}+nx+n$ is irreducible over $\mathbb{Q}[\sqrt[n]{p}]$.
This question appeared in our quiz today. But I couldn't even prove that the polynomial is irreducible over $\mathbb{Q}[x]$. Provided that I can prove that the polynomial, which I will call $f$ from now on, is irreducible over $\mathbb{Q}[x]$, we will have
$$[\mathbb{Q}(\alpha,\sqrt[n]{p}):\mathbb{Q}]\le [\mathbb{Q}(\alpha):\mathbb{Q}][\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}]=2p(p+1)$$
,where $\alpha$ is a root of $f$. But we also have $p+1=[\mathbb{Q}(\sqrt[n]{p}):\mathbb{Q}]\mid[\mathbb{Q}(\alpha,\sqrt[n]{p}):\mathbb{Q}]$
, similarly we also have $2p\mid [\mathbb{Q}(\alpha,\sqrt[n]{p}):\mathbb{Q}]$.Thus either
$[\mathbb{Q}(\alpha,\sqrt[n]{p}):\mathbb{Q}]=p(p+1)$ or $2p(p+1)$. Maybe some further pondering will actually show us that the polynomial is indeed irreducible over the field $\mathbb{Q}[\sqrt[n]{p}]$.
But I am yet unable to prove that the polynomial is irreducible over $\mathbb{Q}[x]$. I can't see any immediate way to apply Eisenstein.
Best Answer
Not really a quiz-appropriate solution, but...
Irreducibility over $\mathbb{Q}$:
Let $g(x)$ be a factor of $f(x) = x^{2p}+nx+n$ of degree $d$. Every root $\alpha$ of $g(x)$ satisfies $\alpha^{2p}+n\alpha+n = 0$, so $$\alpha^{2p} = n(-1-\alpha).$$ Multiplying over all roots of $g$ gives $$ g(0)^{2p} = \prod_{g(\alpha)=0} \alpha^{2p} = \prod_{g(\alpha)=0} n(-1-\alpha) = n^d g(-1).$$ Also, $g(-1)$ divides $f(-1) = (-1)^{2p}+n(-1)+n = 1$, and by the above equation $g(-1)$ is positive, so $g(-1) = 1$. So we have $$g(0)^{2p} = n^d.$$
If $p\not\mid d$, then $n$ is a perfect $p$th power. But this gives a contradiction, since $n = p+1$ is too small to be a $p$th power.
If $p\mid d$ and $g(x)$ is a nontrivial factor, then $d = p$. So $g(0)^{2p} = n^p$, which means $n$ is a perfect square. But then $p= m^2-1 = (m-1)(m+1)$ for some integer $m$. This contradicts the primality of $p$, unless $m=2$ and $p=3$.
So we have proven that the polynomial is irreducible for $p \ne 3$. For $p=3$, the above reasoning gives that $d=3$, $g(-1) = 1$, and $g(0) = \pm 2$ for each factor $g(x)$, which reduces the possible factorizations to two cases: $$x^6 + 4x+4 = (x^3 + ax^2+ax + 2)\cdot(x^3 + bx^2 + bx + 2)$$ or $$x^6 + 4x+4 = (x^3 + cx^2 + (c-4)x - 2)\cdot(x^3 + dx^2 + (d-4)x - 2).$$ Each of these cases can be shown to be impossible by comparing the coefficients for the $x^5$ and $x^1$ terms on both sides.
Irreducibility over $\mathbb{Q}(\sqrt[n]{p})$:
Consider an algebraic field extension $L/\mathbb{Q}$, and let $\sigma$ be any automorphism of $L$. Note that if $f(x) \in \mathbb{Q}[x]$ factors into two monic irreducible factors over $L[x]$ as $$ f(x) = g_1(x)\cdot g_2(x),$$ then we can get another factorization of $f(x)$ by applying $\sigma$ coefficientwise to both sides: $$ f(x) = \sigma(f(x)) = \sigma(g_1(x))\cdot\sigma(g_2(x)).$$ Since the irreducible factorization is unique up to order, $\sigma$ permutes the $g_i(x)$. If moreover $L$ is a Galois extension of $\mathbb{Q}$, then the subgroup of automorphisms that fix $g_1(x)$ and $g_2(x)$ has index 1 or index 2 in the Galois group of $L/\mathbb{Q}$. By the fundamental theorem of Galois theory, this implies that the coefficients of $g_1(x)$ and $g_2(x)$ either all lie in $\mathbb{Q}$ or all lie in the same quadratic subfield of $L$. If $f(x)$ is irreducible over $\mathbb{Q}$, then only the second case is possible.
Applying to this problem: Let $K= \mathbb{Q}(\sqrt[n]{p})$ and let $L$ be the Galois closure of $K$, i.e. $L$ is the splitting field of $x^n - p$. In your post you basically showed that $$[\mathbb{Q}(\alpha, \sqrt[n]{p}) : K] = p \quad\text{ or }\quad [\mathbb{Q}(\alpha, \sqrt[n]{p}) : K]=2p,$$ i.e. $f(x)$ splits into two factors of degree $p$ or is irreducible over $K$.
The same basic idea applies to $L$ as well. The case $p=2$ is straightforward; for $p\ne 2$ we have $\sqrt{p} \in K$, so $x^n-p$ can be factored as $$x^n-p = (x^{n/2}-\sqrt{p})(x^{n/2}+\sqrt{p})$$ over $K$. The degree of the splitting field of each factor can only have prime factors that are at most $\frac{n}{2} = \frac{p+1}{2} < p$, so the degree of $L$ is relatively prime to $p$. Hence by the same reasoning as in your post and by the argument above, $f(x)$ splits into two factors of degree $p$ or is irreducible over $L$.
Thus if $f(x)$ factors over $L$, it splits into two factors with coefficients in a quadratic subfield of $L$. If moreover $f(x)$ factors over $K$, then the quadratic field must be a subfield of $K$ as well, i.e. it must be $\mathbb{Q}(\sqrt{p})$, the only quadratic subfield of $K$.
In summary, $f(x)$ is reducible over $K$ if and only if it splits into two conjugate factors of degree $p$ over $\mathbb{Q}(\sqrt{p})$.
In that case we could then write $$f(x) = (a(x) + b(x)\sqrt{p})\cdot(a(x) - b(x)\sqrt{p}) = a(x)^2 - p \cdot b(x)^2.$$ This says that $f(x)$ would need to be perfect square modulo $p$. But $f'(x) \equiv n \pmod{p}$ is relatively prime to $f(x)$, so $f(x)$ is squarefree modulo $p$. Hence $f(x)$ is not reducible over $K$.