Let $K$ be a field and let $x_1, \ldots, x_n$ be indeterminates over $K$. Prove that $K$ is algebraically closed in $K(x_1, \ldots, x_n)$.

Definition: Let $F$ be a field extension of $K$. We say that $K$ is algebraically closed in $F$ if every element of $F$ that is algebraic over $K$ is already in $K$.

My attempt:

We need to show for any $u \in K(x_1,...,x_n)$ there is $p(x) \in K[x]$ such that $p(u)=0,$ where $x_1,...,x_n$ are indeterminates over $K$. Let $u \in K(x_1,...,x_n)$, then we have $u=\displaystyle \frac{f(x_1,...,x_n)}{g(x_1,...,x_n)}$ where $g(x_1,...,x_n) \neq 0$. We note that $f(x_1,...,x_n), g(x_1,...,x_n) \in K[x_1,...,x_n].$ I can not go along from here, I can not find $f(x) \in K[x]$ satisfy $p(u)=0$.

Thanks for any help or hints

  • 5
    $\begingroup$ How do you define "$k$ is algebraically closed in $K$ for fields $k\subset K$?" $\endgroup$ Dec 4, 2019 at 20:08
  • 2
    $\begingroup$ Maybe: every $y \in K$ is either in $k$ or transcendent over $k$? $\endgroup$
    – Aphelli
    Dec 4, 2019 at 20:14
  • 1
    $\begingroup$ I just updated it with the definition. $\endgroup$
    – Ahmed
    Dec 4, 2019 at 21:01
  • 2
    $\begingroup$ @DietrichBurde, it’s just a problem with OP’s command of the language. I think it’s clear that he’s talking about the relative algebraic closure of the small field in the bigger. $\endgroup$
    – Lubin
    Dec 4, 2019 at 21:45

1 Answer 1


The result you need is that $K[x_1, \dotsc, x_n]$ is a UFD.

You already arrive at $p(u) = 0$, where $u = \frac{f}{g}$.

Writing $p(x) = x^d + a_{d - 1}x^{d - 1} + \dotsc + a_0$ with $a_i \in K$, we have: $$\left(\frac{f}{g}\right)^d + a_{d - 1}\left(\frac{f}{g}\right)^{d - 1} + \dotsc + a_0 = 0,$$ or: $$f^d + a_{d - 1}f^{d - 1}g + \dotsc + a_0g^d = 0.$$ This tells us that, as elements of $K[x_1, \dotsc, x_n]$, the polynomial $f^d$ is a multiple of $g$. Since $K[x_1, \dotsc, x_n]$ is a UFD, it follows that $g$ divides $f$.

Therefore $u = \frac{f}{g}$ is indeed a polynomial in $K[x_1, \dotsc, x_n]$. By comparing degree in the equation $p(u) = 0$, it's clear that $u$ must be of degree $0$, hence in $K$.

  • $\begingroup$ Why $$\left(\frac{f}{g}\right)^d + a_{d - 1}\left(\frac{f}{g}\right)^{d - 1} + \dotsc + a_0 ,$$ equal to 0 $\endgroup$
    – Ahmed
    Dec 5, 2019 at 1:42
  • $\begingroup$ It's your assumption $p(u)=0$. $\endgroup$
    – WhatsUp
    Dec 5, 2019 at 10:50

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