Galois group

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#1 h Mar 11, 2025, 5:19 AM

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Let $K$ be a field. Find the Galois groups

$a) \text{Gal}(K(x), K)$

$b) \text{Gal}(K(x,y), K)$

This post has been edited 3 times. Last edited by ILOVEMYFAMILY, Mar 11, 2025, 5:21 AM

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#2 h Mar 11, 2025, 5:35 PM • 1 Y

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I decided to describe in detail the solution of the first one. The second problem is solved almost similarly (try it yourself)

We consider automorphisms of the field $K(x)$ that leave the field $K$ fixed

An automorphism $\sigma$ of $K(x)$ is a bijective mapping of $K(x)$ onto itself that preserves all its operations. Since $\sigma$ fixes $K$ , all elements $a \in K$ remain fixed. Since $K(x)$ is generated by $K$ and an element $x$ that is transcendental over $K$ , the action of $\sigma$ is completely determined by where $\sigma$ maps $x$ . Let $\sigma(x) = f(x)$ , where $f(x) \in K(x)$ , that is, $f(x)$ is some rational function. For $\sigma$ to be an automorphism, the field generated by $K$ and $\sigma(x)$ must equal the whole of $K(x)$ . We know that $\sigma$ is invertible, so there is an automorphism $\tau = \sigma^{-1}$ such that $\tau(f(x)) = x$ , therefore $f(x)$ must be a rational function such that there exists an inverse rational function $g(x) \in K(x)$ that satisfies $f(g(x)) = x$ and $g(f(x)) = x$

The field $K(x)$ can be viewed as a field of functions on the projective line $\mathbb{P}^1_K$ over $K$ . The automorphisms of $K(x)$ that fix $K$ correspond to the biregular automorphisms of $\mathbb{P}^1_K$ . We know well that the group of automorphisms of the projective line $\mathbb{P}^1_K$ over any field $K$ is isomorphic to the group of projective linear transformations $\mathrm{PGL}(2, K)$ (this follows from the fact that each automorphism is uniquely determined by the matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ with nonzero determinant, and matrices that differ by a scalar factor define the same automorphism), and therefore $f(x)$ must be a fractional linear function

This is not so hard to check. Let $\sigma(x) = \frac{ax + b}{cx + d}$ , then for any element $P(x)/Q(x) \in K(x)$ the automorphism $\sigma$ acts

$\sigma\left(\frac{P(x)}{Q(x)}\right) = \frac{P(\sigma(x))}{Q(\sigma(x))} = \frac{P\left(\frac{ax + b}{cx + d}\right)}{Q\left(\frac{ax + b}{cx + d}\right)}$
Since $P$ and $Q$ are polynomials with coefficients from $K$ , and $\sigma$ fixes $K$ , we get a rational function from $K(x)$ (bijectivity follows from the fact that the fractional-linear transformation is invertible, that is, the inverse mapping is given by the matrix inverse to $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ up to a scalar). And the fact that $K(\sigma(x)) = K(x)$ is easy to verify yourself

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#3 h Mar 12, 2025, 3:00 PM

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still stuck at $\text{Gal}(K(x,y),K)$ .

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#4 h Mar 12, 2025, 7:39 PM

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ILOVEMYFAMILY wrote:

still stuck at $\text{Gal}(K(x,y),K)$ .

What did you try? Can you show me your attempt? I described everything in great detail above

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ishan.panpaliya

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ishan.panpaliya

#5 h Mar 12, 2025, 8:25 PM

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@Tip_pay

Can you explain a little more why this statement is true? I am not familiar with projective coordinates.

Quote:

The field $K(x)$ can be viewed as a field of functions on the projective line $\mathbb{P}^1_K$ over $K$

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#7 h Mar 23, 2025, 1:49 AM

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Bump bump

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