$\bar{\mathbb{Q}}$ is not isomorphic to $\overline{\mathbb{Q}(x)}$? Here the over line means algebraic closure. That is to say, if $E/K$ is a field extension, then $\bar{K}$ denotes the algebraic closure of $K$ in $E$. It is easy to imagine $\bar{\mathbb{Q}}$: all (complex) roots of rational polynomials. However, what is $\overline{\mathbb{Q}(x)}$ by definition exactly? It is confusing if I still use the definition "all roots of something"...
Thank you in advance for your contribution and help!
 A: The algebraic closure of a field $K$ is the smallest algebraically closed field $\bar{K}$ containing $K$; that is, $\bar{K}$ is algebraically closed, contains $K$, and if $E$ is any algebraically closed field such that $K\hookrightarrow E$, then also $\bar{K}\hookrightarrow E$. This definition can be applied with $K=\mathbb{Q}$ or with $K=\mathbb{Q}(x)$, the field of rational functions in 1 variable over $\mathbb{Q}$. You can think of $\overline{\mathbb{Q}(x)}$ as solutions in $y$ to equations of the form
$$f_n(x)y^n+\ldots+f_1(x)y+f_0(x) = 0,$$
where $f_i(x)$ are rational functions in $x$ over $\mathbb{Q}$. This means that the solution $y$ will also be a function in $x$, but not necessarily a rational one.
There is a theorem that states that an algebraically closed field is determined up to isomorphism by its transcendence degree over its prime subfield. The prime subfield of $K$ is the smallest subfield $k\hookrightarrow K$. The transcendence degree of $K/k$ is the cardinality of the largest subset of $K$ which is algebraically independent over $k$. In both your cases, the prime subfield is $\mathbb{Q}$. $\bar{\mathbb{Q}}$ is by definition algebraic over $\mathbb{Q}$, so it's transcendence degree is 0. $\overline{\mathbb{Q}(x)}$ is algebraic over $\mathbb{Q}(x)$, which clearly has transcendence degree 1 over $\mathbb{Q}$. Thus the transcendence degree of $\overline{\mathbb{Q}(x)}/\mathbb{Q}$ is still 1. Therefore, $\bar{\mathbb{Q}}$ and $\overline{\mathbb{Q}(x)}$ cannot be isomorphic.
A: The answer to the question in the title is no because $\mathbb{Q}(x)$ cannot be embedded into $\overline{\mathbb{Q}}$:
If $\alpha \in \overline{\mathbb{Q}}$, then $\mathbb Q(\alpha)$ has finite dimension over $\mathbb Q$. But $\mathbb{Q}(x)$ has infinite dimension over $\mathbb Q$.
