# Galois invariants of the function field of a projective variety defined over K

I apologize if this question is too simple but I cannot see why the following is true:

Let $$K$$ be a perfect field and $$\overline{K}$$ a fixed algebraic closure. Let $$X \subset \mathbb{P}^n(\overline{K})$$ be a projective variety defined over $$K$$ (The definition I'm using of 'defined over $$K$$' is that the homogeneous ideal $$I(X)$$ can be generated by homogeneous elements in $$K[X_0, \cdots, X_n]$$). Then I understand that the absolute Galois group $$G = \text{Gal}(\overline{K}/K)$$ acts on $$X$$ in the obvious manner: $$\sigma \cdot [x_0: \cdots : x_n] = [\sigma x_0 : \cdots : \sigma x_n]$$ and that the invariants $$X^G$$ are precisely the $$K$$-rational points $$X(K)$$.

Further, $$G$$ also acts on the function field $$\overline{K}(X)$$ by acting on coefficients. My question is: Why is $$\overline{K}(X)^G = K(X)$$?

I could only make the start: Let $$\frac{f}{g}$$ be an element of $$\overline{K}(X)^G$$, so that $$f,g$$ are homogeneous polynomials of the same degree in $$\overline{K}[X_0, \cdots, X_n]$$ and $$g$$ is not in $$I(X)$$. Then $$\sigma \cdot \frac{f}{g} = \frac{f}{g}$$ implies that $$\sigma f \cdot g - f \cdot \sigma g \in I(X)$$. Now I'm not sure how to proceed. To descend to the ground field, I know I have to involve Galois cohomology somehow but the map $$\sigma \rightarrow \sigma f \cdot g - f \cdot \sigma g$$ is not a $$1$$-cocycle.

Any help will be appreciated.

• You can assume that the denominator is invariant. You will get a cocycle. – Roland Feb 28 at 7:44