# Algebraic Closure of $\mathbb{Q}$ in $\mathbb{Q}(x)$

Defining algebraic closure of $$F$$ in $$E$$ as the set of elements in $$E$$ that are algebraic over $$F$$, I am trying to understand the following statement from Fraleigh's textbook:

"$$\mathbb{Q}$$ is the algebraic closure of $$\mathbb{Q}$$ in $$\mathbb{Q}(x)$$."

From my understanding, I need to show that $$\forall \alpha \in \mathbb{Q}(x)$$, $$\exists f(x) \in \mathbb{Q}[x]$$, $$f(\alpha) = 0$$, where $$\mathbb{Q}(x)$$ is the field of quotients of $$\mathbb{Q}[x]$$. How can this be done? For example, how can I find a polynomial in $$\mathbb{Q}[x]$$ that when evaluated on $$x^2+1$$ equals to zero?

• @ancientmathematician Isn't that what I wrote in the original question? If not, my apologies Commented Oct 12, 2021 at 6:39
• You are misunderstanding the statement. You want to prove that if an element of $f\in \mathbb{Q}(x)$ is algebraic over $\mathbb{Q}$, then $f\in\mathbb{Q}$. In other words, given $f\in\mathbb{Q}(x)$, if there exists a nonzero $P\in \mathbb{Q}[T]$ such that $P(f(x))=0$, then $f\in\mathbb{Q}$. Commented Oct 12, 2021 at 6:51
• @GreginGre Right, that makes it very clear. Was going in the wrong direction Commented Oct 12, 2021 at 7:03
• @JadenPark I am sorry, I didn't think I had posted that comment, but it got saved by mistake. Commented Oct 12, 2021 at 8:12
• @ancientmathematician no worries :) Commented Oct 12, 2021 at 8:19

Fraleigh says that the algebraic closure of $$\mathbb Q$$ in $$\mathbb Q(x)$$ is $$\mathbb Q$$. This means that for all rational $$\alpha$$ there is a non constant polynomial with rational coefficients which kills $$\alpha$$. This is certainly true, take $$f(x) = x - \alpha$$. Also it means that these are the only elements of $$\mathbb Q(x)$$ which are algebraic over $$\mathbb Q$$. To see that this is true, assume $$f/g$$ is algebraic over $$\mathbb Q$$. The ring $$\mathbb Q$$ is a unique factorization domain, so by dividing out common factors we may assume that $$f$$ and $$g$$ are coprime. There is an equation of the form \begin{align}(f/g)^n + a_{n-1}(f/g)^{n-1}+…+a_0 =0\end{align} with rational $$a_k$$. Multiplying this equation by $$g^n$$ yields \begin{align}f^n+ a_{n-1}gf^{n-1}+…+a_0 g^n=0\end{align} which shows that $$g$$ divides $$f$$. But since they are coprime, this can only be true if $$g$$ is a unit. Thus $$f/g$$ is an element of $$\mathbb Q[x]$$ which is algebraic over $$\mathbb Q$$. There are no non-constant polynomials with this property, so we conclude that $$f/g$$ is a rational number.