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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!

atlantic0cean
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  • Should be the algebraic closure of the field of rational fractions with coefficients in $\mathbf Q$. It can't be isomorphic to $\overline{\mathbf Q}$ as it has transcendence degree $1$ over $\mathbf Q$. – Bernard Dec 14 '20 at 11:24
  • @Bernard Thank you! But there are yet some problems I can't understand. What does 'transcendence degree' mean? Why can't they be isomorphic if $\overline{\mathbb{Q}(x)}$ has transcendence degree 1 over $\mathbb{Q}$? – atlantic0cean Dec 14 '20 at 11:28
  • You can define an algebraic closure of a field $K$ per se. It will be a field $C$ which is algebraic over $K$, and such that all non-constant polynomials with coefficients in $K$ have a root in $C$. One shows that such a $C$ exists, and that all such $C$ are $K$-isomorphic. – Andreas Caranti Dec 14 '20 at 11:30
  • Roughly speaking, it is the number of algebraically independent generators. For instance, over any field $K$, $F(X,Y)$ has transcendence degree $2$. The algebraic closure of $\mathbf Q$, being algebraic, has transcendence degree $0$. – Bernard Dec 14 '20 at 11:32
  • @Bernard I think I've understood intuitively! Thank you! – atlantic0cean Dec 14 '20 at 11:37
  • The title does not reflect the question – lhf Dec 14 '20 at 12:34
  • By definition, X is transcendental over $Q$, wheras $\bar Q$ is algebraic . – nguyen quang do Dec 14 '20 at 20:31

2 Answers2

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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.

Tom Sharpe
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  • I think you've given a great explanation about what does $\overline{\mathbb{Q}(x)}$ look like! Thanks a lot! – atlantic0cean Dec 14 '20 at 11:45
  • Simply speaking, $\overline{\Bbb Q(x)}$ has a transcendent element over its prime field $\Bbb Q$ while $\overline{\Bbb Q}$ doesn't have. – Berci Dec 14 '20 at 11:45
  • Thanks! If you think I've answered your question, please 'Accept' this answer. – Tom Sharpe Dec 14 '20 at 14:00
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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 $\mathbb Q$.

lhf
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