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In the field extension $F_p(x)/F_p$, where $F_p(x)$ is the field of fractions of polynomials over $F_p$, is it by definition that $x \in F_p(x)$ is not algebraic? In other words, should I claim that $x$ is indeterminate when I make this field extension, otherwise somehow it is no longer the field of fractions. In other words, if $x \in F_p$ perhaps $F_p(x)$ is a smaller field than the field of fractions.

Further, I read in an answer here: Example of infinite field of characteristic $p\neq 0$ that any element in $F_p(x)$ that involves $x$ will not be algebraic over $F_p$, but I don't see how.

Mike
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    That $x$ is not algebraic is obvious: Evaluating a polynomial in $x$ at $x$ just gives you that polynomial back; it is only $0$ if it was $0$. The harder question is why other non-constant rational functions $a \in F_p\left(x\right)$ are not algebraic. To show that, write $a$ as $u/v$ for two coprime polynomials $u$ and $v$, and try to get a contradiction from assuming that $a$ is algebraic. – darij grinberg Feb 08 '19 at 20:15
  • @darijgrinberg so in this case $x$ is assumed not to be an element of $F_p$, right? Otherwise evaluating a polynomial at $x$ would yield some element in $F_p$, and perhaps that element is zero. Am I missing the point of the word "indeterminate"? Is it supposed to mean 'placeholder' i.e. 'don't think of this as an element in any field'?

    For the second part, Let $\frac{u}{v} \in F_p(t)$ be algebraic, then suppose $a_0+a_1\frac{u}{v} +a_2 (\frac{u}{v})^2 + ... + a_n (\frac{u}{v} )^n=0$. Playing around with this I haven't been able to get a contradiction (I assume it should come from 'coprime')

     – Mike Feb 08 '19 at 20:22
  • Nope, $x$ is an indeterminate -- the one from the polynomial ring $F_p\left[x\right]$. – darij grinberg Feb 08 '19 at 20:30
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    And as for the contradiction, try multiplying by $v^n$ and checking which of the resulting addends are divisible by $u$. – darij grinberg Feb 08 '19 at 20:31
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    @darijgrinberg

    So $a_0 v^n + ... + a_nu^n=0$. Since $0$ is divisible by $u$, the left-hand side is divisible by $u$, hence every term on the sum is divisible by $u$, hence $v^n$ is divisible by $u$, therefore $v$ is divisible by $u$, contradiction.

    How does this look?

     – Mike Feb 08 '19 at 20:40
  • Almost. "$v^n$ divisible by $u$" does not yield "$v$ divisible by $u$", but it still contradicts the assumption that $u$ and $v$ are coprime (okay, I forgot to say that you have to make that assumption). – darij grinberg Feb 08 '19 at 20:43
  • @darijgrinberg Oh ok, that makes sense. Thanks! – Mike Feb 08 '19 at 20:45

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