Let $\alpha$ be algebraic over $\mathbb{Q}$ with $[\mathbb{Q}(\alpha) : \mathbb{Q}]=2$ and let $F=\mathbb{Q}(\alpha)$. Suppose that $f(x) \in \mathbb{Q}[x]$ is irreducible of degree d.

(i) If d is odd, show that $f(x)$ remains irreducible in $F[x]$.

(ii) If d is even, show that $f(x)$ either remains irreducible in $F[x]$ or $f(x)$ is the product of two irreducible polynomials in $F[x]$ of degree $\frac{d}{2}$.

The first part has been answered and I think I have a solution for the second part. Are there any issues with this?

(ii) Suppose that $f$ is not irreducible over F. Now we know that $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}(\beta)] \leq 2$. If $[\mathbb{Q}(\alpha,\beta): \mathbb{Q}(\beta)] = 2$ then $[F(\beta):F]= [\mathbb{Q}(\alpha,\beta): \mathbb{Q}(\alpha)] = \frac{[\mathbb{Q}(\alpha,\beta): \mathbb{Q}]}{[\mathbb{Q}(\alpha): \mathbb{Q}]}=\frac{2d}{2} =d$. So $f$ is irreducible over F which is a contradiciton. Hence $[F(\beta):\mathbb{Q}(\beta)]=1$ and $[F(\beta):\mathbb{Q}]=d$. It follows that $[F(\beta):F]= [\mathbb{Q}(\alpha,\beta): \mathbb{Q}(\alpha)] = \frac{d}{2}$ which means $deg(m_{\beta, F}) = \frac{d}{2}$. Since this is true for any root of $f$ we have that $f$ must be the product of two irreducible polynomials in $F[x]$ of degree $\frac{d}{2}$.

  • 1
    $\begingroup$ Clue: Consider the splitting field $\endgroup$ Jun 7, 2019 at 21:01

2 Answers 2


Let $g(x) \in F[x]$ be an irreducible factor of $f(x)$ in $F[x]$ and $\beta \in \mathbb{C}$ be a root of $g(x)$.

Since $g(x)$ is irreducible in $F[x]$, we have $m_{\beta, F}(x) = g(x)$, and hence $$\deg\left( g(x) \right) = \left[ F(\beta) : F \right] = \left[ \mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\alpha) \right] \, \text{.}$$

Therefore, we have $$2 \deg\left( g(x) \right) = \left[ \mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\alpha) \right] \left[ \mathbb{Q}(\alpha) : \mathbb{Q} \right] = \left[ \mathbb{Q}(\alpha, \beta) : \mathbb{Q} \right] = \left[ \mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\beta) \right] \left[ \mathbb{Q}(\beta) : \mathbb{Q} \right] \, \text{.}$$

Now, note that $m_{\beta, \mathbb{Q}}(x) = f(x)$ since $\beta$ is a root of $f(x)$ and $f(x)$ is irreducible in $\mathbb{Q}[x]$. It follows that $\left[ \mathbb{Q}(\beta) : \mathbb{Q} \right] = \deg\left( m_{\beta, \mathbb{Q}}(x) \right) = d$, and hence $$2 \deg\left( g(x) \right) = d \left[ \mathbb{Q}(\alpha, \beta) : \mathbb{Q}(\beta) \right] = d \, \deg\left( m_{\alpha, \mathbb{Q}(\beta)}(x) \right) \, \text{.}$$

Finally, note that $m_{\alpha, \mathbb{Q}(\beta)}(x)$ is a factor of $m_{\alpha, \mathbb{Q}}(x)$, which has degree $2$. Thus, $\deg\left( m_{\alpha, \mathbb{Q}(\beta)}(x) \right) \in \lbrace 1, 2 \rbrace$, and the desired result is proved.

  • $\begingroup$ What about the second part? $\endgroup$
    – wasatar
    Jun 10, 2019 at 17:44
  • $\begingroup$ My post answers both parts. We have $2 \deg\left( g(x) \right) = d \, \deg\left( m_{\alpha, \mathbb{Q}(\beta)}(x) \right)$ so $\deg\left( g(x) \right) = \frac{d}{2}$ if $\deg\left( m_{\alpha, \mathbb{Q}(\beta)}(x) \right) = 1$ and $\deg\left( g(x) \right) = d$ if $\deg\left( m_{\alpha, \mathbb{Q}(\beta)}(x) \right) = 2$. Note that the first case obviously can't happen if $d$ is odd. $\endgroup$
    – v_lentin
    Jun 10, 2019 at 18:04

Let $\beta$ be a root of $f(x)$ in an extension field of $F=\mathbb{Q}(\alpha)$. Then we have $$\def\Q{\mathbb{Q}} [\Q(\alpha,\beta):\Q]=[\Q(\alpha,\beta):\Q(\alpha)][\Q(\alpha):\Q] =[\Q(\alpha,\beta):\Q(\beta)][\Q(\beta):\Q] $$ If $f$ is reducible over $F$, then $e=[\Q(\alpha,\beta):\Q(\alpha)]<d$ (why?). On the other hand $[\Q(\beta):\Q]=d$, so we have $d\mid 2e$. Since $d$ is odd, we conclude that $d\mid e$, a contradiction.


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