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Let $E:F$ be a field extension and $a\in K$. Then, $$a\text{ is algebraic over F }\iff F(a)=F[a]$$ Proof: Let's take the evaluation epimorphism $\epsilon:F[x]\longrightarrow F[a],f(x)\mapsto \epsilon(f(x)):=f(a)$. We have: $a\text{ is algebraic over F }\iff \exists f(x)\in F[x]:f(a)=0\iff \ker \epsilon\neq (0)$. Also, $F[x]/\langle \ker \epsilon \rangle \cong F[a]$ (from 1st Isomorphism Theorem). But, $F[x]$ is a PID, so, $F[a]$ is a domain $ \iff \ker \epsilon$ prime ideal $\iff \ker \epsilon$ is a maximal ideal $\iff F[a]$ is a field. But, the quotient field, $Q(F[a])=F(a)$ is the smallest field which containts $F[a]$. So, $F[a]=F(a)$.

Now my questions are:

  1. Can we move in this syllogism conversely (are we ok for the $\Leftarrow$ direction)?
  2. If yes, we proove this direction with another way?
  3. And the most important: What $F[a]=F(a)$ means in fact? How could these sets be equal, from the time that we now that the second one contains equivalevalent classes in the form $\frac{f(a)}{g(a)},\ g(a)\neq 0$?

Thank you in advance.

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    $\begingroup$ 3. $F[a] = F(a)$ means that because of the algebraic relations of powers of a, (For example $f(a) = 0$ or -1) all the inverses of polynomials are indeed polynomials. $\endgroup$ – B.A Mar 28 '17 at 23:14
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    $\begingroup$ You should be OK for the opposite direction. If $F(a)=F[a]$, then $\ker\epsilon$ is a maximal ideal in $F[x]$, so is non-zero. $\endgroup$ – John Gowers Mar 28 '17 at 23:20
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Yes, the statement is an "if and only if".

I can understand why you might think that $F[a] \neq F(a)$, but here's an example that will convince you otherwise. Consider the element $$ \frac 1 {\sqrt{2} - 1} \in \mathbb Q (\sqrt{2}).$$ As written, this element looks like it is not in $\mathbb Q[\sqrt{2}]$, because it is a fraction.

But this is an illusion! By "rationalising the denominator", you can write it as $$ \frac 1 {\sqrt{2} - 1} = \frac {\sqrt{2} + 1} {(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac {\sqrt{2} + 1} {2 - 1} = \sqrt{2} + 1.$$ So this element is in $\mathbb Q[\sqrt{2}]$.


Now, I'll give you an alternative proof that $f(a)/g(a)$ is in $ F[a]$ if $a$ is algebraic over $F$:

Since $a$ is algebraic, it has a minimal polynomial $m(x) \in F[x]$. The minimal polynomial $m(x)$ is irreducible, so the greatest common divisor of $m(x)$ and $g(x)$ is $1$. Therefore, by Euclid's algorithm, there exist polynomials $u(x) , v(x) \in F[x]$ such that $$u(x)g(x) + v(x)m(x) = 1.$$ Substituting $x = a$ into this equation, and using the fact that $m(a) = 0$, we find that $$\frac 1 {g(a)} = u(a),$$ and hence, $$\frac{f(a)}{g(a)} = f(a)u(a),$$ which is now manifestly an element of $F[a]$.

On the other hand, if $a$ is transcendental over $F$, then you can easily check that $F(a)$ is isomorphic to the field of rational functions $F(t)$, whereas $F[a]$ is isomorphic to the ring of polynomials $F[t]$, and these are different rings.

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  • $\begingroup$ Thank you for your answer. So, you say to think $\frac 1 {\sqrt{2} - 1}$ as an number in $\frac{1}{\sqrt{2} - 1} \in \mathbb{Q} (\sqrt{2})$, and with the rationalisation this number belongs in $\mathbb{Q} [\sqrt{2}]$? But, I can not understand how that the element $\frac{f(a)}{g(a)}:=[(f(a),g(a)]_{\sim}$ which is a set (!) is equal with an element like $f(a)$. How can this happen? (in my mind it's an equivalence class and a single element) $\endgroup$ – Chris Mar 28 '17 at 23:31
  • $\begingroup$ Sure, what I really mean is that $\frac 1 {\sqrt 2 - 1} = \frac{\sqrt{2} + 1} 1$. Or to put it more obstrusely, $[(1, \sqrt 2 - 1)]_{\sim} = [(\sqrt{2} + 1, 1)]_{\sim}$. $\endgroup$ – Kenny Wong Mar 28 '17 at 23:33
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    $\begingroup$ So we identify an element $f(a) \in F[a]$ with the element $f(a)/1 \in F(a)$, i.e. with the equivalence class $[(f(a),1)]$. And the claim is that this identification is an isomorphism. $\endgroup$ – Kenny Wong Mar 28 '17 at 23:34
  • $\begingroup$ So, do we say that $f(a)=\frac{f(a)}{1}$ under this identification? So, is it useful to think these cases like we think "operations" (roughly speaking) in $\mathbb{Q}$? $\endgroup$ – Chris Mar 28 '17 at 23:39
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    $\begingroup$ Yes. By the way, I tend to think about these things as fractions, rather than as sets. I say this because we all have very good intuition about how fractions behave and can be manipulated, because of the hours of pain we had to suffer in primary school. All of this intuition works in $F(a)$. This is because the field axioms, and the definition of $F(a)$, were designed to mimic our intuition from primary school. $\endgroup$ – Kenny Wong Mar 28 '17 at 23:42
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I may be missing something, but I don’t see any proof of the “$\Leftarrow$” direction.

Here it is: If $\alpha$ is not algebraic over $F$, then it’s transcendental, and the evaluation $F[T]\to F[\alpha]$ has zero kernel (that, in effect, is the definition of transcendentality) and is onto by construction. So $F[\alpha]$ is isomorphic to a polynomial ring, which is not a field. But by definition, $F(\alpha)$ is a field. So $F[\alpha]\ne F(\alpha)$.

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