§9
Galois Extensions
The inclusion-reversing correspondence in §8 is not necessarily a bijection between all intermediate extensions and all subgroups. A necessary condition is that the fixed field of \(\mathrm{Gal}(K/F)\) equals \(F\), i.e.\ \(K^{\mathrm{Gal}(K/F)} = F\).
Let \(K\) be an algebraic extension of a field \(F\). We say that \(K/F\) is a Galois extension if the fixed field of the Galois group \(\mathrm{Gal}(K/F)\) is equal to \(F\), that is, we have \[ K^{\mathrm{Gal}(K/F)} = F. \]
🇮🇳 Hinglish Explanation
Galois extension K/F: K^{Gal(K/F)} = F.
Matlab: Galois group ka fixed field exactly F hai — koi bigger nahi.
- ℂ/ℝ: K^{Gal} = ℝ ✓ Galois
- ℚ(√2,√3)/ℚ: K^{Gal} = ℚ ✓ Galois
- ℚ(∛3)/ℚ: Gal={id}, K^{Gal}=ℚ(∛3)≠ℚ ✗ NOT Galois
Note that any field automorphism \(\sigma\) of \(K\) can be viewed as a group homomorphism from \(K^\times\) to itself, where \(K^\times\) denotes the multiplicative group of non-zero elements of \(K\).
Let \(G\) be a group, and let \(K\) be a field. A character of \(G\) over \(K\) is a group homomorphism from \(G\) to the multiplicative group \(K^\times\) of non-zero elements of \(K\). The set of all characters from \(G\) to \(K\) is denoted by \(\mathrm{Hom}(G, K^\times)\).
🇮🇳 Hinglish Explanation
Character of G over K: Group homomorphism χ:G→K×.
Field automorphisms are characters of (L×,·) over L. Dedekind's lemma isi pe based hai.
Let \(\sigma_1,\ldots,\sigma_n\) be distinct characters of a group \(G\) over a field \(K\). Then the characters \(\sigma_1,\ldots,\sigma_n\) are linearly independent over \(K\): if \[ \sum_{i=1}^n c_i \sigma_i = 0 \] for some \(c_1,\ldots,c_n \in K\), then \(c_1 = c_2 = \cdots = c_n = 0\).
🇮🇳 Hinglish Explanation
Dedekind's Lemma: Distinct characters linearly independent over K.
Matlab: alag automorphisms genuinely different hain — linear combination cancel nahi kar sakti.
The following result is a consequence of Dedekind's lemma.
Let \(K/F\) be a finite extension. Then the Galois group \(\mathrm{Gal}(K/F)\) is a finite group, and its order is at most the degree of the extension \(K/F\), that is, \[ |\mathrm{Gal}(K/F)| \le [K:F]. \]
🇮🇳 Hinglish Explanation
K/F finite → |Gal(K/F)| ≤ [K:F]. Dedekind se automorphisms linearly independent.
Let \(G\) be a finite subgroup of the group of automorphisms of a field \(K\). Then the order of \(G\) is equal to the degree of the extension \(K/K^G\), that is, \[ |G| = [K : K^G]. \] Consequently, the extension \(K/K^G\) is finite and Galois, and has \(G\) as its Galois group, that is, \(G = \mathrm{Gal}(K/K^G)\).
🇮🇳 Hinglish Explanation
Artin's Theorem: G finite Aut(K) → |G|=[K:K^G], aur K/K^G Galois with Gal=G.
A finite extension \(K/F\) is Galois if and only if the order of the Galois group \(\mathrm{Gal}(K/F)\) is equal to the degree of the extension \(K/F\), that is, \[ |\mathrm{Gal}(K/F)| = [K:F]. \]
🇮🇳 Hinglish Explanation
Finite K/F Galois ↔ |Gal(K/F)| = [K:F]. Practical test: order compare karo.
Let \(K/F\) be a field extension, and let \(\alpha \in K\) be algebraic over \(F\). Then the order of the Galois group \(\mathrm{Gal}(F(\alpha)/F)\) is equal to the number of distinct roots of the minimal polynomial of \(\alpha\) over \(F\) in \(F(\alpha)\). Consequently, the extension \(F(\alpha)/F\) is Galois if and only if the number of distinct roots of the minimal polynomial of \(\alpha\) over \(F\) in \(F(\alpha)\) is equal to the degree of the extension \(F(\alpha)/F\).
🇮🇳 Hinglish Explanation
|Gal(F(α)/F)| = # distinct roots of min poly inside F(α). Galois iff ye number = deg.
Determine the Galois group of the extension \(\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q}\) and its subgroups, and determine the corresponding intermediate extensions of \(\mathbb{Q}(\sqrt[3]{3})/\mathbb{Q}\).
🇮🇳 Hinglish Solution
ℚ(∛3)/ℚ: degree 3, |Gal|=1. NOT Galois. Intermediate: sirf ℚ aur ℚ(∛3) (prime degree). Subgroups: sirf {id}. Correspondence bijective nahi.
English Solution
Let \[ K=\mathbb Q(\sqrt[3]3). \] Since \(X^3-3\) is irreducible over \(\mathbb Q\), \[ [K:\mathbb Q]=3. \]
The other two roots of \(X^3-3\) are non-real, so they are not in \(K\subset\mathbb R\). Hence \[ \boxed{\operatorname{Gal}(K/\mathbb Q)=\{\mathrm{id}\}.} \]
Since the degree \(3\) is prime, the only intermediate fields are \[ \mathbb Q,\qquad K. \]
Thus there are two intermediate fields but only one subgroup of the Galois group. Therefore the Galois correspondence is not exact.
\(\blacksquare\)🇮🇳 Hinglish Solution
\[ K=\mathbb Q(\sqrt[3]3). \]
Degree: \[ [K:\mathbb Q]=3. \]
Polynomial \[ X^3-3 \] ke other two roots complex hain, aur \(K\) real field hai.
So only one root field mein hai. Therefore: \[ \operatorname{Gal}(K/\mathbb Q)=\{1\}. \]
Degree prime \(3\) hone ki wajah se intermediate fields only: \[ \mathbb Q,\quad K. \]
So one subgroup but two intermediate fields. Hence correspondence fail karti hai because extension Galois nahi hai.
Similar problem as the previous exercise, but for the extension \(k(t)/k(t^2)\), or for the extension \(k(t)/k(t^p)\) for any prime \(p\) and any field \(k\) of characteristic \(p\).
🇮🇳 Hinglish Solution
k(t)/k(t²): degree 2, Gal≅ℤ/2, Galois. k(t)/k(t^p), char p: inseparable (X−t)^p, |Gal|=1≠p, NOT Galois.
English Solution
Case 1: \(\operatorname{char}k\neq2\).
We have \[ [k(t):k(t^2)]=2 \] and \[ \operatorname{Gal}(k(t)/k(t^2)) =\{\mathrm{id},t\mapsto-t\}. \] Hence the extension is Galois.
The two subgroups correspond to \[ k(t),\qquad k(t^2). \]
Case 2: \(\operatorname{char}k=p\).
We have \[ [k(t):k(t^p)]=p>1, \] but \[ \operatorname{Gal}(k(t)/k(t^p)) =\{\mathrm{id}\}. \] Hence the extension is not Galois.
Since \(p\) is prime, the only intermediate fields are \[ k(t^p),\qquad k(t). \]
Thus the Galois correspondence fails in characteristic \(p\).
\(\blacksquare\)🇮🇳 Hinglish Solution
Characteristic \(\neq2\):
\[ [k(t):k(t^2)]=2, \] aur nontrivial automorphism \[ t\mapsto -t. \] Isliye \[ \operatorname{Gal}(k(t)/k(t^2))=\{\mathrm{id},t\mapsto -t\}\cong C_2. \] So extension Galois hai.
Characteristic \(p\):
Agar \[ F=k(t^p)\subseteq k(t), \] then \[ [k(t):F]=p, \] but every \(F\)-automorphism must fix \(t\), because \[ \sigma(t)^p=\sigma(t^p)=t^p, \] thus \(\sigma(t)=t\). So only identity automorphism survives:
\[ \operatorname{Gal}(k(t)/k(t^p))=\{\mathrm{id}\}. \]
Isliye extension purely inseparable hai aur Galois nahi hai.
Show that the only automorphism of \(\mathbb{Q}\) is the identity.
🇮🇳 Hinglish Solution
Aut(ℚ)={id}: koi σ, σ(1)=1 → σ(n)=n for n∈ℤ, phir σ(p/q)=p/q. Toh σ=id.
English Solution
Claim: The only field automorphism of \(\mathbb Q\) is the identity.
Proof: Let \[ \sigma:\mathbb Q\to\mathbb Q \] be a field automorphism.
Since \(\sigma(1)=1\), \[ \sigma(n)=n \] for every \(n\in\mathbb Z\).
For \(p/q\in\mathbb Q\), \(q\neq0\), \[ q\cdot\frac pq=p. \] Applying \(\sigma\), \[ q\,\sigma\left(\frac pq\right)=p. \] Hence \[ \sigma\left(\frac pq\right)=\frac pq. \]
Therefore every rational number is fixed, so \[ \boxed{\sigma=\mathrm{id}_{\mathbb Q}}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Field automorphism \(\sigma:\mathbb Q\to\mathbb Q\) lo.
Sabse pehle: \[ \sigma(1)=1. \]
Addition se: \[ \sigma(n)=n \] for all integers \(n\).
Ab rational number: \[ \frac pq. \] Since \[ q\frac pq=p, \] apply \(\sigma\): \[ q\sigma\left(\frac pq\right)=p. \]
Hence: \[ \sigma\left(\frac pq\right)=\frac pq. \]
Har rational fixed hai. Therefore: \[ \boxed{\sigma=\mathrm{id}}. \]
Show that the only automorphism of \(\mathbb{R}\) is the identity.
🇮🇳 Hinglish Solution
Aut(ℝ)={id}: σ order-preserving (squares → positives). Dense ℚ pe id → continuous extension id on ℝ.
English Solution
Claim: The only field automorphism of \(\mathbb R\) is the identity.
Proof: Let \[ \sigma:\mathbb R\to\mathbb R \] be an automorphism.
By Exercise 1.80, \(\sigma\) fixes \(\mathbb Q\) pointwise.
If \(x>0\), then \(x=y^2\) for some \(y\in\mathbb R\). Hence \[ \sigma(x)=\sigma(y)^2>0. \] Thus \(\sigma\) preserves positivity and therefore order.
If \(q Since \(\mathbb Q\) is dense in \(\mathbb R\),
\[
\sigma(r)=r
\]
for every \(r\in\mathbb R\). Therefore,
\[
\boxed{\sigma=\mathrm{id}_{\mathbb R}}.
\]
\(\blacksquare\)
🇮🇳 Hinglish Solution
Let \[ \sigma:\mathbb R\to\mathbb R \] be an automorphism.
Sabse pehle Exercise 1.80 se \(\mathbb Q\) pointwise fixed hai.
Agar \(x>0\), then \[ x=y^2 \] for some \(y\). Hence \[ \sigma(x)=\sigma(y)^2>0. \]
So positive numbers positive hi rahenge. Therefore order preserve hota hai.
Ab rational numbers \(\mathbb R\) mein dense hain. Kisi real \(r\) ke neeche ke rationals aur upar ke rationals same position mein rahenge.
Isliye \(\sigma(r)\) aur \(r\) different nahi ho sakte.
Hence: \[ \boxed{\sigma=\mathrm{id}_{\mathbb R}}. \]
Let \(F\) be a field of characteristic not equal to \(2\). Let \(K\) be an extension of \(F\) of degree two. Prove that \(K = F(\alpha)\) for some \(\alpha \in K\) with \(\alpha^2 \in F\). Show that \(K\) is Galois over \(F\). What can be said about degree two extensions of a field of characteristic \(2\)?
🇮🇳 Hinglish Solution
Char≠2, [K:F]=2 → K=F(α), α²∈F (complete the square / min poly X²−d). Non-id auto: α↦−α → Galois. Char 2: Artin–Schreier type X²+X+a possible.
English Solution
Claim: If \(\operatorname{char}F\neq2\) and \([K:F]=2\), then \(K/F\) is Galois.
Let \[ K=F(\alpha), \] where the minimal polynomial is \[ X^2+bX+c. \]
Put \[ \beta=\alpha+\frac b2. \] Then \[ \beta^2=\frac{b^2-4c}{4}\in F. \]
Thus the minimal polynomial of \(\beta\) is \[ X^2-\beta^2. \] Its roots are \[ \beta,\qquad-\beta, \] both of which belong to \(K\).
Hence the minimal polynomial splits completely in \(K\), so \(K/F\) is normal and separable.
Therefore, \[ \boxed{K/F\text{ is Galois}.} \]
Characteristic \(2\): A quadratic extension may be purely inseparable, in which case it is not Galois. A separable quadratic extension of Artin-Schreier type is Galois.
\(\blacksquare\)🇮🇳 Hinglish Solution
Assume: \[ \operatorname{char}F\neq2,\qquad[K:F]=2. \]
Let \[ K=F(\alpha) \] and minimal polynomial: \[ X^2+bX+c. \]
Complete the square: \[ \beta=\alpha+\frac b2. \]
Then: \[ \beta^2=\frac{b^2-4c}{4}\in F. \]
So minimal polynomial: \[ X^2-\beta^2. \]
Roots: \[ \beta,\quad-\beta. \] Dono \(K\) mein hain.
Hence polynomial splits completely. So extension normal and separable hai.
Therefore: \[ \boxed{K/F\text{ is Galois}.} \]
Characteristic \(2\) mein purely inseparable quadratic extension possible hai, jo Galois nahi hoti.
Show that there exist automorphisms \(\sigma,\tau\) of \(k(x)\) such that \(\sigma(x) = 1/x\) and \(\tau(x) = 1 - 1/x\). Describe the subgroup of \(\mathrm{Gal}(k(x)/k)\) generated by \(\sigma,\tau\).
🇮🇳 Hinglish Solution
σ:x↦1/x, τ:x↦1−1/x. Ye PGL₂ / anharmonic group generate karte hain — order 6 ≅ S₃ subgroup of Gal(k(x)/k).
English Solution
Define \[ \sigma(x)=\frac1x,\qquad \tau(x)=1-\frac1x. \]
Then \[ \sigma^2(x)=x, \] so \[ \sigma^2=\mathrm{id}. \]
Also, \[ \tau^2(x)=\frac1{1-x}, \] and \[ \tau^3(x)=x. \] Hence \[ \tau^3=\mathrm{id}. \]
Further, \[ \sigma\tau(x)=\frac{x}{x-1}, \] and \[ (\sigma\tau)^2(x)=x. \] Thus \[ (\sigma\tau)^2=\mathrm{id}. \]
Therefore, \[ \langle\sigma,\tau\rangle = \langle\sigma,\tau\mid \sigma^2=\tau^3=(\sigma\tau)^2=1\rangle \cong S_3. \]
Hence \[ \boxed{\langle\sigma,\tau\rangle\cong S_3.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Do automorphisms define kiye hain: \[ \sigma(x)=\frac1x, \qquad \tau(x)=1-\frac1x. \]
First: \[ \sigma^2(x)=x. \] So order \(2\).
Second: \[ \tau^2(x)=\frac1{1-x}, \qquad \tau^3(x)=x. \] So order \(3\).
Also: \[ (\sigma\tau)^2(x)=x. \]
Ye exactly \(S_3\) ke standard relations hain: \[ \sigma^2=\tau^3=(\sigma\tau)^2=1. \]
Hence: \[ \boxed{\langle\sigma,\tau\rangle\cong S_3}. \]
§11
Normal Extensions
Let \(K\) be an extension of \(F\). Show that \(K\) is a splitting field over \(F\) for a set \(\{f_1,\ldots,f_n\}\) of polynomials in \(F[x]\) if and only if \(K\) is a splitting field over \(F\) for the polynomial \(f_1 \cdots f_n\).
🇮🇳 Hinglish Solution
Splitting field of {f₁,...,fₙ} = splitting field of product f₁⋯fₙ — same roots adjoin karne padte hain.
English Solution
Claim: \(K\) is a splitting field for \(\{f_1,\ldots,f_n\}\) over \(F\) if and only if \(K\) is a splitting field for \[ f=f_1f_2\cdots f_n. \]
The roots of \(f\) are exactly the union of the roots of \(f_1,\ldots,f_n\).
Therefore \(f\) splits over \(K\) if and only if every \(f_i\) splits over \(K\).
Also, the field generated by all roots of \(f\) is the same as the field generated by all roots of \(f_1,\ldots,f_n\).
Hence the two splitting-field conditions are equivalent.
\[ \boxed{K\text{ is a splitting field for }\{f_1,\ldots,f_n\} \iff K\text{ is a splitting field for }f.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Product polynomial lo: \[ f=f_1f_2\cdots f_n. \]
Field mein product zero tabhi hota hai jab kisi factor ka value zero ho.
So \(f\) ke roots exactly \(f_1,\ldots,f_n\) ke roots ka union hain.
Therefore: \[ f\text{ splits} \iff f_1,\ldots,f_n\text{ sab split hote hain}. \]
Roots se generated field bhi same rahega.
Hence: \[ \boxed{ K\text{ is splitting field for all }f_i \iff K\text{ is splitting field for }f. } \]
Let \(K\) be a splitting field of a set \(S\) of polynomials over \(F\). If \(L\) is a subfield of \(K\) containing \(F\) for which each \(f \in S\) splits over \(L\), show that \(L = K\).
🇮🇳 Hinglish Solution
Minimal splitting field: agar L⊆K pe already sab polynomials split, toh generators (roots) L mein → L=K.
English Solution
Claim: If \(K\) is a splitting field of \(S\subseteq F[X]\) over \(F\), and every polynomial in \(S\) splits over \(L\), where \[ F\subseteq L\subseteq K, \] then \[ L=K. \]
Since \(K\) is a splitting field, it is generated over \(F\) by all roots of the polynomials in \(S\).
But every \(f\in S\) splits over \(L\). Hence all its roots in \(K\) already belong to \(L\).
Therefore \(K\subseteq L\).
Since \(L\subseteq K\), we obtain \[ \boxed{L=K}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Splitting field ki definition ke according \(K\) roots ko adjoin karke generate hota hai.
Agar \(L\) mein har polynomial \(f\in S\) already completely split hai, to uske saare roots \(L\) mein hain.
Therefore \(K\) ko generate karne wale saare roots \(L\) mein already present hain.
Hence: \[ K\subseteq L. \]
Given: \[ L\subseteq K. \]
Therefore: \[ \boxed{L=K}. \]
Provide an example of a field extension \(F \subseteq K \subseteq L\) such that \(L/K\) and \(K/F\) are normal, but \(L/F\) is not normal.
🇮🇳 Hinglish Solution
Classic: F=ℚ, K=ℚ(∛2), L=ℚ(∛2,ω). K/F not normal; wait — need L/K and K/F normal but L/F not. Better: F=ℚ, K=ℚ(√2), L=ℚ(√[4]{2}). K/F =ℚ(√2)/ℚ normal (degree 2). L/K=ℚ(√[4]{2})/ℚ(√2) normal degree 2. L/F not normal (min poly X⁴−2 doesn't split: missing i√[4]{2}).
English Solution
Example: Let \[ F=\mathbb Q, \qquad K=\mathbb Q(\sqrt2), \qquad L=\mathbb Q(\sqrt[4]2). \] Then \(K/F\) is quadratic, so it is normal. Also \(L/K\) is quadratic because \[ (\sqrt[4]2)^2=\sqrt2\in K, \] so \(L/K\) is normal as well.
However, \(L/F\) is not normal. Indeed, the polynomial \[ X^4-2 \] is irreducible over \(\mathbb Q\), it has \(\sqrt[4]2\in L\) as a root, and its remaining roots are \[ -\sqrt[4]2, \quad i\sqrt[4]2, \quad -i\sqrt[4]2, \] which do not lie in \(L\subseteq\mathbb R\). Hence \(X^4-2\) does not split over \(L\).
Therefore, \[ \boxed{K/F\text{ and }L/K\text{ are normal, but }L/F\text{ is not normal}.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Example: \[ \mathbb Q \subseteq \mathbb Q(\sqrt2) \subseteq \mathbb Q(\sqrt[4]2). \]
First extension: \(\mathbb Q(\sqrt2)/\mathbb Q\) quadratic hai, isliye normal hai.
Second extension: \[ (\sqrt[4]2)^2=\sqrt2\in \mathbb Q(\sqrt2), \] so \[ [L:K]=2. \] Ye bhi quadratic hai, isliye normal hai.
But full extension: Polynomial \[ X^4-2 \] ke roots hain \[ \sqrt[4]2, -\sqrt[4]2, i\sqrt[4]2, -i\sqrt[4]2. \] Ismein \(\sqrt[4]2\in L\) hai, lekin baaki do non-real roots \(L\) mein nahi hain, kyunki \(L\subseteq\mathbb R\).
Isliye \(X^4-2\) full split nahi hota hai over \(L\). Thus: \[ \boxed{K/\mathbb Q\text{ and }L/K\text{ normal hain, but }L/\mathbb Q\text{ normal nahi hai}.} \]
Let \(K\) be an extension of a field \(F\). Assume that \(K\) is algebraically closed. Show that the algebraic closure of \(F\) in \(K\) is an algebraic closure of \(F\).
🇮🇳 Hinglish Solution
Algebraic closure of F in algebraically closed K = absolute algebraic closure of F (up to iso): every poly over F splits in K, roots algebraic over F.
English Solution
Claim: If \(K\) is algebraically closed and \(F\subseteq K\), then the algebraic closure \(A\) of \(F\) in \(K\) is an algebraic closure of \(F\).
Proof: By definition, \[ A=\{\alpha\in K:\alpha\text{ is algebraic over }F\}. \] Thus \(A/F\) is algebraic.
Let \[ g(X)\in A[X] \] be nonconstant. Since \(K\) is algebraically closed, \(g\) has a root \(\gamma\in K\).
The coefficients of \(g\) are algebraic over \(F\). Hence they generate a finite extension of \(F\). Since \(\gamma\) is algebraic over this finite extension, the tower law gives that \(\gamma\) is algebraic over \(F\).
Therefore \[ \gamma\in A. \] So every nonconstant polynomial in \(A[X]\) has a root in \(A\).
Hence \(A\) is algebraically closed and algebraic over \(F\). Therefore, \[ \boxed{A\text{ is an algebraic closure of }F.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Algebraic closure define karo: \[ A=\{\alpha\in K:\alpha\text{ is algebraic over }F\}. \]
Clearly \(A/F\) algebraic hai.
Ab \(g(X)\in A[X]\) nonconstant lo.
Since \(K\) algebraically closed hai, \(g\) ka koi root \[ \gamma\in K \] hoga.
Polynomial ke coefficients \(F\) par algebraic hain. Isliye coefficients se banne wala extension finite hai.
\(\gamma\) us finite extension ke upar algebraic hai, so tower law se \(\gamma\) \(F\) par bhi algebraic hai.
Hence: \[ \gamma\in A. \]
So \(A\) algebraically closed hai.
Therefore: \[ \boxed{A\text{ is an algebraic closure of }F.} \]
Provide an example of a field \(K\) such that \(\mathrm{Aut}(K)\) contains an element of infinite order.
🇮🇳 Hinglish Solution
Example: Aut(k(x)/k) contains σ: x↦x+1 of infinite order (σⁿ: x↦x+n ≠ id for n≠0).
English Solution
Example: Let \[ K=\mathbb Q(x). \] Define \[ \sigma(x)=x+1, \qquad \sigma|_{\mathbb Q}=\mathrm{id}. \]
This extends to an automorphism of \(\mathbb Q(x)\), whose inverse is \[ \sigma^{-1}(x)=x-1. \]
For \(n\ge1\), \[ \sigma^n(x)=x+n. \]
Since \[ x+n\neq x \] for every \(n\ge1\), \[ \sigma^n\neq\mathrm{id}. \]
Therefore \(\sigma\) has infinite order.
Hence, \[ \boxed{\operatorname{Aut}(\mathbb Q(x)) \text{ contains an element of infinite order}.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Simple example lo: \[ K=\mathbb Q(x). \]
Define: \[ \sigma(x)=x+1. \]
Iska inverse: \[ \sigma^{-1}(x)=x-1. \] So \(\sigma\) automorphism hai.
Ab powers calculate karo: \[ \sigma^2(x)=x+2, \] \[ \sigma^3(x)=x+3, \] and generally \[ \sigma^n(x)=x+n. \]
Kisi bhi \(n\ge1\) ke liye \[ x+n\neq x. \] Therefore \[ \sigma^n\neq\mathrm{id}. \]
So \(\sigma\) ki order finite nahi hai.
Hence: \[ \boxed{\sigma\text{ has infinite order in }\operatorname{Aut}(\mathbb Q(x)).} \]