$K$ is algebraically closed in $K(x_1, \ldots, x_n)$.

Question

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

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 $K$.