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

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

• How do you define "$k$ is algebraically closed in $K$ for fields $k\subset K$?" Dec 4, 2019 at 20:08
• Maybe: every $y \in K$ is either in $k$ or transcendent over $k$? Dec 4, 2019 at 20:14
• I just updated it with the definition. Dec 4, 2019 at 21:01
• @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. Dec 4, 2019 at 21:45

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