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Suppose I am given a field $F$. Suppose also we have $a, b$ which are algebraic over $F$. In every book I have read, it was written that it is trivial to show that $F(a)(b)=F(b)(a)$. I tried to prove this equality algebraically using ring homomorphism but could not. How is this equality is usually shown?

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    $F[a]$ is the smallest ring containing $F$ and $a$, no? – 2017-01-11
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    Yes but here since we are talking about some abstract $a$ so $F(a)$ is actually the quotient $F[x]\over $.where $f$ is the irreducible polynomial which $a$ is its root – 2017-01-11
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    Yes, but you still have $$F(a) = F[a] = \{ Q \pmod f \mid Q \in F[X]\} = \{u_na^n+\dots+u_1a+u_0 \mid u_i \in F\}$$ where $n=deg(f)-1$ and $a=X \pmod f$. – 2017-01-11
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    @User666x: The statement doesn't make sense if you are talking about "some abstract $a$". What would it mean to say that $F(a)(b)=F(b)(a)$? What does $F(a)(b)$ even mean? What if the polynomial you intended to have as the minimal polynomial of $b$ becomes reducible over $F(a)$? Talking about field extensions like $F(a)(b)$ or $F(b)(a)$ only makes sense when you already have some larger field containing both $a$ and $b$. – 2017-01-11

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The fact that $a$ and $b$ are algebraic is irrelevant. We have a field $K$ with a subfield $F\subseteq K$ and two elements $a,b\in K$, and wish to show that $F(a)(b)=F(b)(a)$. By definition, $F(a)$ is the smallest subfield of $K$ containing $F$ and $a$. By definition again, $F(a)(b)$ is the smallest subfield of $K$ containing $F(a)$ and $b$. Since a subfield of $K$ contains $F(a)$ iff it contains both $F$ and $a$, $F(a)(b)$ is the smallest subfield of $K$ containing $F$, $a$, and $b$.

Similarly, $F(b)(a)$ is the smallest subfield of $K$ containing $F$, $b$, and $a$. Since "and" is commutative (that is, a subfield of $K$ contains $b$ and $a$ iff it contains $a$ and $b$), this is the same thing as $F(a)(b)$.

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    Nobody said we know what that field $K$ is – 2017-01-12
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    $a$ and $b$ are just symbols. Where did you bring this field $K$ from? – 2017-01-12
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    If $a$ and $b$ are "just symbols", then what on earth does $F(a)(b)$ mean? – 2017-01-12
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    We can build them abstractly using homomorphisms and quotient rings – 2017-01-12
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    You need to state precisely what your definition of $F(a)(b)$ is then. There is no standard definition of $F(a)(b)$ when $a$ and $b$ are not elements of a single field containing $F$.... – 2017-01-12
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Hint: prove it for rings first, namely $F[a][b]=F[b][a]$. Then use the fact that $a$ and $b$ are algebraic over $F$.

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we can use the theorem:

Theorem: Let be $K$ is extension field of $k$ and $\alpha_1,\alpha_2,...,\alpha_n \in K$ algebraic over $k$ and $\sigma\in S_n$ then:

$$k(\alpha_1)(\alpha_2)...(\alpha_n)=k(\alpha_1,\alpha_2,...,\alpha_n)= k(\alpha_{\sigma(1)},\alpha_{\sigma(2)},...,\alpha_{\sigma(n)})=k(\alpha_{\sigma(1)})(\alpha_{\sigma(2)})...(\alpha_{\sigma(n)})$$

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    Your "other way" is wrong: $\Bbb Q(\sqrt 2, \sqrt 3 - \sqrt 2) \neq \Bbb Q(\sqrt 2 + \sqrt 3 - \sqrt 2) =\Bbb Q(\sqrt 3)$. – 2017-01-12
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    ok, thank you very much ''Waston'', I edit the answer. – 2017-01-12