# Let $\alpha$ be algebraic over $\mathbb{Q}$ with $[\mathbb{Q}(\alpha) : \mathbb{Q}]=2$ and let $F=\mathbb{Q}(\alpha)$.

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 $$\frac{d}{2}$$.

• Clue: Consider the splitting field Jun 7, 2019 at 21:01

## 2 Answers

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.

• What about the second part? Jun 10, 2019 at 17:44
• 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. 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 $$e=[\Q(\alpha,\beta):\Q(\alpha)] (why?). On the other hand $$[\Q(\beta):\Q]=d$$, so we have $$d\mid 2e$$. Since $$d$$ is odd, we conclude that $$d\mid e$$, a contradiction.