§4
Generated Subfields
Let \(L/K\) be a field extension and \(S \subseteq L\). The field generated by \(S\) over \(K\), written \(K(S)\), is the smallest subfield of \(L\) containing both \(K\) and \(S\). We say \(L/K\) is generated by \(S\) if \(L = K(S)\). The ring generated by \(S\) over \(K\), written \(K[S]\), is the smallest subring of \(L\) containing \(K\) and \(S\).
🇮🇳 Hinglish Explanation
K(S): S ke elements ko K mein adjoin karo → smallest field jo K aur S dono contain kare.
K[S]: Smallest ring jo K aur S contain kare (sirf polynomial expressions).
Difference: K[S] mein sirf polynomials. K(S) mein unke ratios bhi. K(S) ⊇ K[S] hamesha.
Note that the field generated by \(S\) over \(K\) is the intersection of all subfields of \(L\) that contain both \(K\) and \(S\). Similarly, the ring generated by \(S\) over \(K\) is the intersection of all subrings of \(L\) that contain both \(K\) and \(S\).
🇮🇳 Hinglish Explanation
K(S) = ∩{F : F ek subfield hai L ka, K⊆F, S⊆F} — sabhi aise fields ka intersection.
Ye well-defined hai: subfields ka intersection bhi subfield hoti hai (verify karo — close under +, ×, inverses).
An extension \(L/K\) is finitely generated if there exists a finite subset \(S \subset L\) such that \(L = K(S)\).
🇮🇳 Hinglish Explanation
Finitely generated L/K: Agar L = K(α₁,...,αₙ) kuch finite elements se.
Example: ℚ(√2,√3)/ℚ finitely generated — generators: {√2, √3}.
Contrast: ℚ̄/ℚ algebraic hai but infinitely generated (har degree ke liye naya element chahiye).
Let \(k\) be a field. Consider the field extension \(k(x_1,x_2,\ldots,x_n)/k\), where \(k(x_1,x_2,\ldots,x_n)\) is the field of rational functions in \(n\) variables over \(k\). Show that \(k(x_1,x_2,\ldots,x_n)\) is generated by the elements \(x_1,x_2,\ldots,x_n\) over \(k\), that is, \[ k(x_1,x_2,\ldots,x_n) = k(\{x_1,x_2,\ldots,x_n\}). \] Also show that the polynomial ring \(k[x_1,\ldots,x_n]\) is generated by \(x_1,\ldots,x_n\) over \(k\), that is, \[ k[x_1,\ldots,x_n] = k[\{x_1,\ldots,x_n\}]. \]
🇮🇳 Hinglish Solution
k(x₁,...,xₙ) by definition woh smallest field hai jo k aur x₁,...,xₙ contain kare. k({x₁,...,xₙ}) bhi exactly wahi hai. Dono same definition se same cheez hain.
Similarly k[{x₁,...,xₙ}] = k[x₁,...,xₙ] polynomial ring by definition.
English Solution
Claim: \[ k(x_1,\ldots,x_n)=k(\{x_1,\ldots,x_n\}) \] and \[ k[x_1,\ldots,x_n]=k[\{x_1,\ldots,x_n\}]. \]
Proof: Let \[ S=\{x_1,\ldots,x_n\}. \]
By definition, \(k[S]\) is the smallest subring containing \(k\) and \(S\). Therefore it contains every polynomial expression in \(x_1,\ldots,x_n\). Hence \[ k[S]=k[x_1,\ldots,x_n]. \]
Similarly, \(k(S)\) is the smallest subfield containing \(k\) and \(S\). It contains all polynomials and their quotients, so \[ k(S)=k(x_1,\ldots,x_n). \]
Therefore, \[ \boxed{k[S]=k[x_1,\ldots,x_n],\qquad k(S)=k(x_1,\ldots,x_n).} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Idea: \(k[x_1,\ldots,x_n]\) aur \(k(x_1,\ldots,x_n)\) bas notation ka difference nahi hai; ek ring generated by elements hai aur doosra field generated by elements.
Agar \[ S=\{x_1,\ldots,x_n\}, \] to \(k[S]\) smallest subring hai jo \(k\) aur \(S\) ko contain karta hai.
Isliye usmein exactly polynomial expressions milte hain: \[ k[S]=k[x_1,\ldots,x_n]. \]
Similarly, \(k(S)\) smallest subfield hai. Field hone ki wajah se polynomial ke quotients bhi include honge.
Therefore: \[ \boxed{k(S)=k(x_1,\ldots,x_n)}. \]
The field of rational functions \(k(x_1,x_2,\ldots,x_n)\) in \(n\) variables over \(k\) is the field of fractions of the integral domain \(k[x_1,\ldots,x_n]\). Moreover, \(k(x_1,x_2,\ldots,x_n)\) is generated by \(x_1,x_2,\ldots,x_n\) over \(k\), and hence the extension \(k(x_1,x_2,\ldots,x_n)/k\) is finitely generated. Furthermore, the ring \(k[x_1,\ldots,x_n]\) is generated by \(x_1,\ldots,x_n\) over \(k\).
🇮🇳 Hinglish Explanation
k(x₁,...,xₙ) = field of fractions of k[x₁,...,xₙ]. Extension k(x₁,...,xₙ)/k finitely generated (n generators).
Let \(L/K\) be a field extension, and let \(S\) be a subset of \(L\). Show that the field generated by \(S\) over \(K\) is equal to the set of all elements of \(L\) that can be expressed as a rational function in the elements of \(S\) with coefficients in \(K\). In other words, show that \[ K(S) = \bigcup_{\substack{S_f \subset S \\ |S_f| < \infty}} K(S_f). \] Also show that the ring generated by \(S\) over \(K\) is equal to the set of all elements of \(L\) that can be expressed as a polynomial in the elements of \(S\) with coefficients in \(K\). In other words, show that \[ K[S] = \bigcup_{\substack{S_f \subset S \\ |S_f| < \infty}} K[S_f]. \]
🇮🇳 Hinglish Solution
K(S) = ∪ K(Sf) over all finite Sf ⊂ S:
Koi bhi element of K(S) kuch finite elements of S pe depend karta hai (rational function mein finitely many variables). Isliye K(S) = union of K(Sf) over finite subsets Sf.
Right side ek field hai (directed union of fields). Smallest field containing S → equality.
English Solution
Claim: \[ K(S)=\bigcup_{\substack{S_f\subseteq S\\|S_f|<\infty}}K(S_f), \] and similarly \[ K[S]=\bigcup_{\substack{S_f\subseteq S\\|S_f|<\infty}}K[S_f]. \]
Proof: Let \[ U=\bigcup_{\substack{S_f\subseteq S\\|S_f|<\infty}}K(S_f). \]
Every \(K(S_f)\subseteq K(S)\), so \[ U\subseteq K(S). \]
Conversely, \(U\) is a field containing \(K\) and every element of \(S\). Hence, by the minimality of \(K(S)\), \[ K(S)\subseteq U. \]
Thus \[ \boxed{K(S)=U}. \]
The same argument using only addition and multiplication gives \[ \boxed{ K[S]=\bigcup_{\substack{S_f\subseteq S\\|S_f|<\infty}}K[S_f]. } \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Idea: Kisi element ko finite operations se banaya jaata hai. Isliye chahe \(S\) infinite ho, kisi particular element ko banane ke liye sirf finite elements of \(S\) chahiye.
Har finite subset \(S_f\subseteq S\) ke liye \[ K(S_f)\subseteq K(S). \] Isliye union bhi \(K(S)\) mein contained hai.
Reverse direction mein union khud ek field hai jo \(K\) aur \(S\) dono ko contain karta hai. Isliye smallest generated field hone ki wajah se \[ K(S) \] union ke andar hoga.
Hence: \[ \boxed{ K(S)=\bigcup_{S_f\subseteq S,\;S_f\ finite}K(S_f). } \]
Ring ke liye same idea: \[ \boxed{ K[S]=\bigcup_{S_f\subseteq S,\;S_f\ finite}K[S_f]. } \]
Let \(L/K\) be a field extension, and let \(\alpha_1,\alpha_2,\ldots,\alpha_n \in L\). Then the field \(K(\{\alpha_1,\alpha_2,\ldots,\alpha_n\})\) is also denoted by \(K(\alpha_1,\alpha_2,\ldots,\alpha_n)\).
🇮🇳 Hinglish Explanation
Shorthand: K({α₁,...,αₙ}) = K(α₁,...,αₙ) — curly braces drop karo jab finite set ho.
Ye sirf notation simplification hai.
Show that \(\mathbb{Q}(\sqrt{2})\) and \(\mathbb{Q}(\sqrt{3})\) are not isomorphic. What can be said about \(\mathbb{Q}(\sqrt{2},\sqrt{6})\) and \(\mathbb{Q}(\sqrt{3},\sqrt{6})\)?
🇮🇳 Hinglish Solution
ℚ(√2) ≇ ℚ(√3): Assume σ:ℚ(√2)→ℚ(√3) isomorphism ho. σ(√2)²=σ(2)=2. Toh σ(√2)=±√2. But ±√2 ∉ ℚ(√3) (kyunki ℚ(√3)={a+b√3} aur √2 irrational over ℚ(√3)). Contradiction.
ℚ(√2,√6) vs ℚ(√3,√6): Dono ≅ ℚ(√2,√3) kyunki √6=√2·√3. Map √2↦√3, √6↦√6 dono ek isomorphism deta hai.
English Solution
Part 1: Suppose \[ \varphi:\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3) \] is an isomorphism. Since \(\varphi\) fixes \(\mathbb Q\), \[ \varphi(\sqrt2)^2=2. \]
Write \[ \varphi(\sqrt2)=a+b\sqrt3,\qquad a,b\in\mathbb Q. \] Then \[ (a+b\sqrt3)^2 =a^2+3b^2+2ab\sqrt3=2. \] Hence \[ 2ab=0. \] If \(a=0\), then \(3b^2=2\), impossible in \(\mathbb Q\). If \(b=0\), then \(a^2=2\), also impossible.
Therefore \[ \boxed{\mathbb Q(\sqrt2)\not\cong\mathbb Q(\sqrt3).} \]
Part 2: Since \[ \sqrt3=\frac{\sqrt6}{\sqrt2}, \] we have \[ \mathbb Q(\sqrt3,\sqrt6) \subseteq\mathbb Q(\sqrt2,\sqrt6). \] Similarly, \[ \sqrt2=\frac{\sqrt6}{\sqrt3}, \] so the reverse inclusion holds.
Therefore \[ \boxed{\mathbb Q(\sqrt2,\sqrt6)=\mathbb Q(\sqrt3,\sqrt6).} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Part 1: Assume karo ki \[ \mathbb Q(\sqrt2)\cong\mathbb Q(\sqrt3). \]
Isomorphism rationals ko fix karega. Isliye \(\sqrt2\) ka image aisa element hoga jiska square \(2\) ho.
Write: \[ \varphi(\sqrt2)=a+b\sqrt3. \] Then \[ (a+b\sqrt3)^2=2. \]
Compare karne par: \[ 2ab=0. \]
Agar \(a=0\), to \(3b^2=2\), impossible. Agar \(b=0\), to \(a^2=2\), impossible.
Contradiction. Therefore dono fields isomorphic nahi hain.
Part 2:
\[ \sqrt3=\frac{\sqrt6}{\sqrt2} \] so first field second ke andar hai. Similarly, \[ \sqrt2=\frac{\sqrt6}{\sqrt3}. \]
Hence dono inclusions milti hain: \[ \boxed{\mathbb Q(\sqrt2,\sqrt6)=\mathbb Q(\sqrt3,\sqrt6)}. \]
Show that \(\mathbb{Q}(\sqrt{2}) = \mathbb{Q}[\{\sqrt{2}\}]\).
🇮🇳 Hinglish Solution
ℚ[√2] = {a+b√2 : a,b∈ℚ} — ring clearly.
Inverse exist karta hai: (a+b√2)⁻¹ = (a−b√2)/(a²−2b²). a²−2b²≠0 kyunki √2 irrational (agar a²=2b² hota toh √2=a/b∈ℚ — impossible). Toh inverse ℚ[√2] mein hai.
∴ ℚ[√2] ek field hai = ℚ(√2).
English Solution
Claim: \[ \mathbb Q(\sqrt2)=\mathbb Q[\sqrt2]. \]
Proof: Every element of \(\mathbb Q[\sqrt2]\) has the form \[ a+b\sqrt2,\qquad a,b\in\mathbb Q. \]
For \(a+b\sqrt2\neq0\), \[ (a+b\sqrt2)^{-1} =\frac{a-b\sqrt2}{a^2-2b^2}. \] Since \(a^2-2b^2\neq0\), this inverse belongs to \(\mathbb Q[\sqrt2]\).
Thus every nonzero element of \(\mathbb Q[\sqrt2]\) has an inverse in the same ring. Hence \(\mathbb Q[\sqrt2]\) is a field.
Therefore, \[ \boxed{\mathbb Q(\sqrt2)=\mathbb Q[\sqrt2].} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
\(\mathbb Q[\sqrt2]\) ka general element: \[ a+b\sqrt2. \]
Nonzero element ka inverse nikaalne ke liye conjugate multiply karo: \[ (a+b\sqrt2)(a-b\sqrt2)=a^2-2b^2. \]
Therefore \[ (a+b\sqrt2)^{-1} =\frac{a-b\sqrt2}{a^2-2b^2}. \]
Ye expression again \(\mathbb Q[\sqrt2]\) mein hai.
So \(\mathbb Q[\sqrt2]\) already a field hai. Aur \(\mathbb Q(\sqrt2)\) smallest field hai containing \(\sqrt2\).
Hence: \[ \boxed{\mathbb Q(\sqrt2)=\mathbb Q[\sqrt2]}. \]
Show that \(\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt{2}+\sqrt{3})\).
🇮🇳 Hinglish Solution
β = √2+√3. Goal: ℚ(β) = ℚ(√2,√3).
⊆ obvious: β ∈ ℚ(√2,√3).
⊇: 1/β = 1/(√2+√3) = (√3−√2)/((√3)²−(√2)²) = (√3−√2)/1 = √3−√2.
β + 1/β = 2√3 → √3 = (β+1/β)/2 ∈ ℚ(β). Similarly √2 = β−√3 ∈ ℚ(β). Done.
English Solution
Claim: \[ \mathbb Q(\sqrt2,\sqrt3)=\mathbb Q(\sqrt2+\sqrt3). \]
Let \[ u=\sqrt2+\sqrt3. \] Clearly, \[ \mathbb Q(u)\subseteq\mathbb Q(\sqrt2,\sqrt3). \]
Also, \[ (\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1, \] so \[ u^{-1}=\sqrt3-\sqrt2. \]
Hence \[ \sqrt3=\frac{u+u^{-1}}2,\qquad \sqrt2=\frac{u-u^{-1}}2. \] Thus \[ \sqrt2,\sqrt3\in\mathbb Q(u). \]
Therefore \[ \mathbb Q(\sqrt2,\sqrt3)\subseteq\mathbb Q(u). \]
Hence \[ \boxed{\mathbb Q(\sqrt2,\sqrt3)=\mathbb Q(\sqrt2+\sqrt3).} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Let \[ u=\sqrt2+\sqrt3. \]
Clearly \(u\) first field mein hai, so \[ \mathbb Q(u)\subseteq\mathbb Q(\sqrt2,\sqrt3). \]
Ab \[ u^{-1}=\sqrt3-\sqrt2 \] because \[ (\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=1. \]
Therefore: \[ \sqrt3=\frac{u+u^{-1}}2, \qquad \sqrt2=\frac{u-u^{-1}}2. \]
So \(\sqrt2,\sqrt3\in\mathbb Q(u)\).
Thus reverse inclusion bhi mil gayi.
\[ \boxed{\mathbb Q(\sqrt2,\sqrt3)=\mathbb Q(\sqrt2+\sqrt3)}. \]
Show that \(\mathbb{Q}(\sqrt{2},\omega) = \mathbb{Q}(\sqrt{2}+\omega)\), where \(\omega\) is a primitive cube root of unity in \(\mathbb{C}\).
🇮🇳 Hinglish Solution
ω = e^{2πi/3} = (−1+i√3)/2, satisfies ω²+ω+1=0.
β = √2+ω. β ∈ ℚ(√2,ω) clearly.
Reverse: β−√2=ω... lekin √2 bhi recover karna hai. β² = 2+2√2·ω+ω². ω²=−1−ω. Toh β²=2+2√2·ω−1−ω = 1+(2√2−1)ω. Phir equations solve karke ω aur √2 recover hote hain from β.
English Solution
Claim: \[ \mathbb Q(\sqrt2,\omega)=\mathbb Q(\sqrt2+\omega), \] where \(\omega^2+\omega+1=0\).
Let \[ u=\sqrt2+\omega. \] Clearly, \[ \mathbb Q(u)\subseteq\mathbb Q(\sqrt2,\omega). \]
Since \[ \omega=u-\sqrt2, \] substitute in \(\omega^2+\omega+1=0\): \[ (u-\sqrt2)^2+(u-\sqrt2)+1=0. \] After simplification, \[ (2u+1)\sqrt2=u^2+u+3. \]
Thus \[ \sqrt2=\frac{u^2+u+3}{2u+1}\in\mathbb Q(u). \] Therefore \[ \omega=u-\sqrt2\in\mathbb Q(u). \]
Hence \[ \mathbb Q(\sqrt2,\omega)\subseteq\mathbb Q(u). \]
Therefore, \[ \boxed{\mathbb Q(\sqrt2,\omega)=\mathbb Q(\sqrt2+\omega).} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Let \[ u=\sqrt2+\omega. \]
Clearly \[ \mathbb Q(u)\subseteq\mathbb Q(\sqrt2,\omega). \]
Kyuki \[ \omega=u-\sqrt2 \] aur \[ \omega^2+\omega+1=0, \] substitute karne par: \[ (2u+1)\sqrt2=u^2+u+3. \]
Hence \[ \sqrt2=\frac{u^2+u+3}{2u+1}\in\mathbb Q(u). \]
Then \[ \omega=u-\sqrt2\in\mathbb Q(u). \]
So reverse inclusion bhi mil gayi.
\[ \boxed{\mathbb Q(\sqrt2,\omega)=\mathbb Q(\sqrt2+\omega)}. \]
Let \(n\) be a positive integer, and let \(\zeta_n\) denote a primitive \(n\)-th root of unity in \(\mathbb{C}\). Then the extension \(\mathbb{Q}(\zeta_n)\) of \(\mathbb{Q}\), generated by \(\zeta_n\), is called the cyclotomic extension of \(\mathbb{Q}\) obtained by adjoining the \(n\)-th roots of unity.
🇮🇳 Hinglish Explanation
ζₙ = e^{2πi/n} = primitive n-th root of unity. ζₙⁿ=1 aur koi chhota positive power nahi deta 1.
Cyclotomic extension ℚ(ζₙ)/ℚ: ζₙ adjoin karo ℚ mein.
- n=4: ζ₄=i, ℚ(i)/ℚ degree 2
- n=6: ζ₆=e^{iπ/3}, degree 2
- n=5: degree 4
Determine the degree of \(\mathbb{Q}(\zeta_n)/\mathbb{Q}\) for \(n = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\).
🇮🇳 Hinglish Solution
[ℚ(ζₙ):ℚ] = φ(n) (Euler's totient).
| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| φ(n) | 1 | 1 | 2 | 2 | 4 | 2 | 6 | 4 | 6 | 4 |
English Solution
Claim: \[ [\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n). \]
The minimal polynomial of \(\zeta_n\) over \(\mathbb Q\) is the \(n\)-th cyclotomic polynomial \(\Phi_n(X)\).
Since \[ \deg\Phi_n(X)=\varphi(n), \] we get \[ \boxed{[\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n).} \]
For \(n=1,\ldots,10\): \[ \begin{array}{c|cccccccccc} n&1&2&3&4&5&6&7&8&9&10\\ \hline \varphi(n)&1&1&2&2&4&2&6&4&6&4 \end{array} \]
Hence the required degrees are \[ \boxed{1,1,2,2,4,2,6,4,6,4}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Important formula:
Primitive \(n\)-th root of unity \(\zeta_n\) ka minimal polynomial cyclotomic polynomial \(\Phi_n(X)\) hota hai.
Uski degree: \[ \deg\Phi_n=\varphi(n). \]
Isliye directly: \[ \boxed{[\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n)}. \]
For \(n=1,\ldots,10\): \[ 1,1,2,2,4,2,6,4,6,4. \]
§5
Degree of Field Extensions
If \(L/K\) is a field extension, then \(L\) can be viewed as a vector space over \(K\), where scalar multiplication is given by the field multiplication.
Show that the map \(\mathbb{C} \times \mathbb{C} \to \mathbb{C},\ (a,b) \mapsto \bar{a}b\), where \(\bar{a}\) denotes the complex conjugate, defines a \(\mathbb{C}\)-linear structure on \(\mathbb{C}\) making it a vector space over \(\mathbb{C}\).
🇮🇳 Hinglish Solution
Map (a,b)↦ā·b pe ℂ-vector space structure check karo:
- Scalar multiplication: a·b defined as ā·b. Distributive: ā(b+c)=āb+āc ✓
- 1·b = 1̄·b = 1·b = b ✓ (1̄=1)
- (a₁a₂)·b = ā₁ā₂·b vs a₁·(a₂·b) = ā₁·(ā₂·b) = ā₁ā₂·b ✓ (since conjugate is multiplicative)
So ℂ is a 1-dimensional ℂ-vector space under this twisted action (basis: {1}).
English Solution
Claim: Define a scalar multiplication on \(\mathbb C\) by \[ a\cdot b=\overline a\,b, \qquad a,b\in\mathbb C. \] Then \(\mathbb C\) is a \(\mathbb C\)-vector space.
Proof: For any \(a,b,c\in\mathbb C\), \[ a\cdot(b+c)=\overline a(b+c)=\overline a b+\overline a c=a\cdot b+a\cdot c, \] and \[ (a+b)\cdot c=(\overline a+\overline b)c=\overline a c+\overline b c=a\cdot c+b\cdot c. \] Also, \[ (ab)\cdot c=\overline{ab}\,c=\overline a\,\overline b\,c=a\cdot (b\cdot c), \] while \[ 1\cdot c=\overline1\,c=c. \] Hence all vector-space axioms hold, and therefore \[ \boxed{\mathbb C\text{ is a }\mathbb C\text{-vector space under }a\cdot b=\overline a b.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Yahan scalar multiplication define hai: \[ a\cdot b=\overline a\,b, \qquad a,b\in\mathbb C. \] Humein vector-space axioms check karne hain.
Step 1: Addition ke respect mein, \[ a\cdot(b+c)=\overline a(b+c)=\overline a b+\overline a c=a\cdot b+a\cdot c. \] Isliye distributivity correct hai.
Step 2: Scalar addition ke respect mein, \[ (a+b)\cdot c=(\overline a+\overline b)c=\overline a c+\overline b c=a\cdot c+b\cdot c. \]
Step 3: Scalar multiplication ke respect mein, \[ (ab)\cdot c=\overline{ab}\,c=\overline a\,\overline b\,c=a\cdot(b\cdot c). \]
Step 4: Identity element: \[ 1\cdot c=\overline1\,c=c. \]
Sab axioms satisfy hote hain. Hence: \[ \boxed{\mathbb C\text{ is a }\mathbb C\text{-vector space under }a\cdot b=\overline a b.} \] \(\blacksquare\)
Let \(L/K\) be a field extension. We say \(L\) is a finite extension of \(K\) if \(L\) is finite-dimensional over \(K\). The degree is \([L:K] = \dim_K L\). If \(L\) is not finite-dimensional we say the extension is infinite.
🇮🇳 Hinglish Explanation
[L:K] = degree = dim_K(L) = L ko K-vector space ki tarah consider karo, uska dimension.
- [ℂ:ℝ] = 2. Basis {1,i}. ℂ = {a+bi : a,b∈ℝ}.
- [ℚ(√2):ℚ] = 2. Basis {1,√2}.
- [ℝ:ℚ] = ∞ (uncountable basis).
Show that \([L:K] = 1\) if and only if \(L = K\).
🇮🇳 Hinglish Solution
[L:K]=1 means dim_K(L)=1. Basis {u} for some u∈L. Har element c·u form ka hai (c∈K). Agar u=1 (which we can assume), toh L={c·1 : c∈K}=K. Toh L=K.
Ulta: L=K toh obvious dim=1.
English Solution
Claim: \[ [L:K]=1\iff L=K. \]
Proof:
(\(\Leftarrow\)) If \(L=K\), then \(\{1\}\) is a basis of \(L\) over \(K\). Hence \[ [L:K]=1. \]
(\(\Rightarrow\)) Suppose \[ [L:K]=1. \] Let \(\{v\}\) be a basis of \(L\) over \(K\). Since \(1\in L\), \[ 1=cv \] for some \(c\in K\). Thus \[ v=c^{-1}\in K. \] Every \(x\in L\) has the form \[ x=av,\qquad a\in K, \] so \(x\in K\). Therefore \(L\subseteq K\).
Since \(K\subseteq L\), we get \[ \boxed{L=K}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Direction 1: Agar \(L=K\), to \(\{1\}\) basis hai. Hence \[ [L:K]=1. \]
Direction 2: Suppose \[ [L:K]=1. \] To \(L\) ka ek nonzero element \(v\) basis hoga.
Kyuki \(1\in L\), \[ 1=cv \] for some \(c\in K\). Therefore \(v\in K\).
Ab har \(x\in L\): \[ x=av,\qquad a\in K. \] Since \(v\in K\), \(x\in K\).
Thus \[ L\subseteq K. \] Already \(K\subseteq L\), so \[ \boxed{L=K}. \]
Show that \([k(x):k] = \infty\) for any field \(k\).
🇮🇳 Hinglish Explanation
k(x) mein elements: 1, x, x², x³, ... Consider karo in k ke upar. Koi finite linear combination c₀+c₁x+...+cₙxⁿ=0 (cᵢ∈k) sirf tab hoti hai jab sab cᵢ=0 (polynomial over field has finitely many roots). Toh {1,x,x²,...} infinite linearly independent set → dim=∞ → [k(x):k]=∞.
Show that \(k(t)/k(t^2)\) is a finite extension of degree \(2\).
🇮🇳 Hinglish Solution
t satisfies X²−t²=0 over k(t²) → min poly degree ≤ 2 → [k(t):k(t²)] ≤ 2.
t ∉ k(t²) (elements of k(t²) have only even powers). Toh [k(t):k(t²)] > 1.
∴ [k(t):k(t²)] = 2. Basis: {1, t}.
English Solution
Claim: \[ [k(t):k(t^2)]=2. \]
Proof: The element \(t\) satisfies \[ X^2-t^2=0, \] where \(t^2\in k(t^2)\). Hence \[ [k(t):k(t^2)]\le2. \]
We show \(t\notin k(t^2)\). If \[ t=\frac{f(t^2)}{g(t^2)}, \] then \[ t\,g(t^2)=f(t^2). \] The left side contains only odd powers of \(t\), while the right side contains only even powers of \(t\). This is impossible for nonzero \(f,g\).
Thus \(t\notin k(t^2)\), so the minimal polynomial of \(t\) has degree \(2\).
Therefore, \[ \boxed{[k(t):k(t^2)]=2.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
\(t\) satisfy karta hai: \[ X^2-t^2=0. \] Isliye degree at most \(2\).
Ab prove karo ki \(t\notin k(t^2)\).
Agar \(t\in k(t^2)\), to \[ t=\frac{f(t^2)}{g(t^2)}. \] Cross multiplication: \[ tg(t^2)=f(t^2). \]
Left side mein sirf odd powers of \(t\) aayengi, aur right side mein sirf even powers.
Nonzero polynomial simultaneously odd aur even powers ka nahi ho sakta.
Hence contradiction. So minimal polynomial degree \(2\) hai.
\[ \boxed{[k(t):k(t^2)]=2}. \]
Determine the degrees of the extensions: \[ \mathbb{Q}(\sqrt{2})/\mathbb{Q},\quad \mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q},\quad \mathbb{Q}(\sqrt{2},\sqrt[3]{3})/\mathbb{Q},\quad \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})/\mathbb{Q}. \]
🇮🇳 Hinglish Solution
Tower Law lagao: [L:F]=[L:K]·[K:F].
- [ℚ(√2):ℚ] = 2 (min poly X²−2 irreducible).
- [ℚ(√2,√3):ℚ] = [ℚ(√2,√3):ℚ(√2)]·[ℚ(√2):ℚ] = 2·2 = 4.
- [ℚ(√2,∛3):ℚ] = 2·3 = 6 (√2 degree 2, ∛3 degree 3; coprime → product).
- [ℚ(√2,√3,√5):ℚ] = 2·2·2 = 8.
English Solution
Using the tower law:
1. Since \(X^2-2\) is irreducible over \(\mathbb Q\), \[ \boxed{[\mathbb Q(\sqrt2):\mathbb Q]=2}. \]
2. Since \(\sqrt3\notin\mathbb Q(\sqrt2)\), \[ [\mathbb Q(\sqrt2,\sqrt3):\mathbb Q(\sqrt2)]=2. \] Hence \[ \boxed{[\mathbb Q(\sqrt2,\sqrt3):\mathbb Q]=4}. \]
3. \(X^3-3\) is irreducible over \(\mathbb Q\) by Eisenstein's criterion, so \[ [\mathbb Q(\sqrt[3]3):\mathbb Q]=3. \] Since \(2\) and \(3\) are coprime, \[ \boxed{[\mathbb Q(\sqrt2,\sqrt[3]3):\mathbb Q]=6}. \]
4. Adjoining \(\sqrt2,\sqrt3,\sqrt5\) successively doubles the degree: \[ 2\cdot2\cdot2=8. \] Therefore, \[ \boxed{[\mathbb Q(\sqrt2,\sqrt3,\sqrt5):\mathbb Q]=8}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Step 1: \[ X^2-2 \] irreducible hai, so \[ [\mathbb Q(\sqrt2):\mathbb Q]=2. \]
Step 2: \(\sqrt3\notin\mathbb Q(\sqrt2)\), hence ek aur factor \(2\) milega: \[ [\mathbb Q(\sqrt2,\sqrt3):\mathbb Q]=4. \]
Step 3: \[ X^3-3 \] Eisenstein at \(3\) se irreducible hai. Hence degree \(3\).
Degrees \(2\) aur \(3\) coprime hain, so combined degree: \[ 2\cdot3=6. \]
Step 4: \(\sqrt2,\sqrt3,\sqrt5\) ko successively adjoin karne par har baar degree \(2\) double hoti hai: \[ 2\cdot2\cdot2=8. \]
Final answers: \[ \boxed{2,\ 4,\ 6,\ 8}. \]
Determine the degrees over \(\mathbb{Q}\): \(\mathbb{Q}(i),\ \mathbb{Q}(\omega),\ \mathbb{Q}(\zeta_n)\).
🇮🇳 Hinglish Solution
- [ℚ(i):ℚ] = 2, min poly X²+1.
- [ℚ(ω):ℚ] = 2, min poly X²+X+1.
- [ℚ(ζₙ):ℚ] = φ(n).
English Solution
Since \(i\) satisfies \(X^2+1\), which is irreducible over \(\mathbb Q\), \[ \boxed{[\mathbb Q(i):\mathbb Q]=2}. \]
Since \(\omega\) satisfies \[ X^2+X+1=0, \] and this polynomial is irreducible over \(\mathbb Q\), \[ \boxed{[\mathbb Q(\omega):\mathbb Q]=2}. \]
For a primitive \(n\)-th root of unity \(\zeta_n\), \[ \boxed{[\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n)}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
\(i\) ka minimal polynomial: \[ X^2+1. \] So \[ [\mathbb Q(i):\mathbb Q]=2. \]
\(\omega\) ke liye: \[ \omega^2+\omega+1=0, \] so \[ [\mathbb Q(\omega):\mathbb Q]=2. \]
General primitive \(n\)-th root ke liye: \[ \boxed{[\mathbb Q(\zeta_n):\mathbb Q]=\varphi(n)}. \]
Let \(L/K\) be a finite extension of degree \(n\). Show that any element of \(L\) is a root of a polynomial of degree at most \(n\) with coefficients in \(K\).
🇮🇳 Hinglish Solution
Agar [L:K]=n, toh {1,α,α²,...,αⁿ} mein n+1 elements hain. n-dimensional space mein n+1 elements linearly dependent hote hain. Toh ∃ c₀,...,cₙ ∈ K (not all zero): c₀+c₁α+...+cₙαⁿ=0. Ye degree-n polynomial hai jisme α root hai.
English Solution
Claim: If \([L:K]=n\), then every \(\alpha\in L\) is a root of a nonzero polynomial in \(K[X]\) of degree at most \(n\).
Proof: Since \(L\) is an \(n\)-dimensional vector space over \(K\), the \(n+1\) elements \[ 1,\alpha,\alpha^2,\ldots,\alpha^n \] are linearly dependent.
Hence there exist \(c_0,\ldots,c_n\in K\), not all zero, such that \[ c_0+c_1\alpha+\cdots+c_n\alpha^n=0. \]
Let \[ f(X)=c_0+c_1X+\cdots+c_nX^n. \] Then \(f(X)\neq0\), \(\deg f\le n\), and \[ f(\alpha)=0. \]
Therefore, \[ \boxed{\alpha\text{ is algebraic over }K\text{ of degree at most }n.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
\(L\) ki dimension \(n\) hai.
Isliye \(n+1\) vectors \[ 1,\alpha,\alpha^2,\ldots,\alpha^n \] linearly dependent honge.
So kuch coefficients \(c_i\in K\), not all zero, milenge such that \[ c_0+c_1\alpha+\cdots+c_n\alpha^n=0. \]
Ab polynomial define karo: \[ f(X)=c_0+c_1X+\cdots+c_nX^n. \]
Then \[ f(\alpha)=0 \] and \[ \deg f\le n. \]
Hence every \(\alpha\in L\) algebraic hai with degree at most \(n\).
Now let's look at some examples of finite extensions.
Consider \(\mathbb{Q}(\sqrt{2})/\mathbb{Q}\). Any element can be written \(a + b\sqrt{2}\), \(a,b \in \mathbb{Q}\). The set \(B = \{1, \sqrt{2}\}\) is a basis, so \([\mathbb{Q}(\sqrt{2}):\mathbb{Q}] = 2\).
🇮🇳 Hinglish Explanation
ℚ(√2) mein elements: √2 hona chahiye. √2 satisfies X²−2=0 → [ℚ(√2):ℚ] ≤ 2.
Basis {1,√2}: Independent? a+b√2=0 → b=0 (√2 irrational) → a=0. Toh linearly independent.
Spanning: Har element a+b√2 form ka — by construction ℚ(√2)=ℚ[√2] (Exercise 1.24).
∴ [ℚ(√2):ℚ] = 2. ✓
Let \(L/K\), \(K/F\) be field extensions. If both are finite then \(L/F\) is finite and \[ [L:F] = [L:K][K:F]. \]
🇮🇳 Hinglish Explanation
Tower Law: Agar K ek F ka extension hai aur L ek K ka extension, toh [L:F]=[L:K]·[K:F].
Intuition: Degrees multiply hoti hain! K ka F ke upar dimension \(n\) hai, L ka K ke upar dimension \(m\) hai, toh total dimension \(m \times n\) hogi.
If either \([L:K]\) or \([K:F]\) is infinite then \([L:F]\) is also infinite (an infinite \(K\)-basis of \(L\) is also an infinite \(F\)-basis).
Let \(\{u_1,\ldots,u_m\}\) be a basis of \(L/K\) and \(\{v_1,\ldots,v_n\}\) a basis of \(K/F\). We claim \(B = \{u_i v_j : 1 \le i \le m,\ 1 \le j \le n\}\) is a basis of \(L/F\).
Spanning. For any \(\alpha \in L\), write \(\alpha = \sum_i c_i u_i\) with \(c_i \in K\), then write each \(c_i = \sum_j d_{ij} v_j\) with \(d_{ij} \in F\). Then \(\alpha = \sum_{i,j} d_{ij} u_i v_j\).
Independence. Suppose \(\sum_{i,j} e_{ij} u_i v_j = 0\). Grouping: \(\sum_i \bigl(\sum_j e_{ij} v_j\bigr) u_i = 0\). Since \(\{u_i\}\) is linearly independent over \(K\), each \(\sum_j e_{ij} v_j = 0\). Since \(\{v_j\}\) is linearly independent over \(F\), all \(e_{ij} = 0\). Hence \([L:F] = |B| = mn = [L:K][K:F]\). \(\square\)
🇮🇳 Hinglish Explanation
3-step proof:
- Infinite case handle karo: Dono mein se agar ek ∞ hai → total ∞. Clear.
- Basis banao: {uᵢvⱼ : 1≤i≤m, 1≤j≤n} — mn elements. Claim: ye F ke upar L ka basis hai.
- Spanning: α∈L → α=∑cᵢuᵢ (K-coefficients) → cᵢ=∑dᵢⱼvⱼ (F-coefficients) → α=∑dᵢⱼuᵢvⱼ. ✓
- Independence: ∑eᵢⱼuᵢvⱼ=0 → group as ∑(∑eᵢⱼvⱼ)uᵢ=0 → each inner sum=0 (uᵢ independent over K) → each eᵢⱼ=0 (vⱼ independent over F). ✓
If an extension has prime degree, show that it has no proper intermediate extensions.
🇮🇳 Hinglish Solution
Agar [L:K]=p (prime) aur M intermediate hai (K⊆M⊆L).
Tower Law: p=[L:K]=[L:M]·[M:K]. p prime toh ya [M:K]=1→M=K, ya [L:M]=1→M=L.
Koi proper intermediate possible nahi. ✓
English Solution
Claim: If \[ [L:K]=p \] where \(p\) is prime, then there is no proper intermediate field between \(K\) and \(L\).
Proof: Let \[ K\subseteq M\subseteq L. \] By the tower law, \[ [L:K]=[L:M][M:K]. \] Therefore, \[ p=[L:M][M:K]. \]
Since \(p\) is prime, either \[ [M:K]=1 \] or \[ [L:M]=1. \]
Thus \[ M=K\quad\text{or}\quad M=L. \]
Hence there is no proper intermediate field.
\[ \boxed{\text{The only intermediate fields are }K\text{ and }L.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Intermediate field lo: \[ K\subseteq M\subseteq L. \]
Tower law: \[ [L:K]=[L:M][M:K]. \]
Agar \[ [L:K]=p \] prime hai, to \(p\) ko do positive integers ke product ke form mein sirf trivial way mein likh sakte hain.
So either \[ [M:K]=1\Rightarrow M=K, \] or \[ [L:M]=1\Rightarrow M=L. \]
Hence beech mein koi proper field nahi hai.
\[ \boxed{M=K\text{ or }M=L.} \]
Let \(L/K\) be a field extension and \(\alpha_1,\ldots,\alpha_n \in L\). Then \(K(\alpha_1,\ldots,\alpha_n)/K\) is finite if and only if each \(K(\alpha_i)/K\) is finite. Moreover, if finite, \[ [K(\alpha_1,\ldots,\alpha_n):K] \le \prod_{i=1}^n [K(\alpha_i):K]. \]
🇮🇳 Hinglish Explanation
α₁,...,αₙ ek ek karke adjoin karo: K ⊆ K(α₁) ⊆ K(α₁,α₂) ⊆ ... ⊆ K(α₁,...,αₙ).
Tower Law se: total degree = product of each step degree.
Har step: [K(...,αᵢ):K(...,αᵢ₋₁)] ≤ [K(αᵢ):K] kyunki αᵢ ka min poly over bigger field ka degree ≤ min poly over K.
∴ [K(α₁,...,αₙ):K] ≤ ∏[K(αᵢ):K].
§6
Algebraic Elements and Algebraic Extensions
Let \(L/K\) be a field extension, \(\alpha \in L\).
- Show that multiplication-by-\(\alpha\) is a \(K\)-linear map \(L \to L\).
- If \(L/K\) is finite, its characteristic polynomial is monic of degree \([L:K]\) and vanishes at \(\alpha\).
- If \(\alpha\) is algebraic over \(K\) and \(L = K(\alpha)\), show the characteristic polynomial equals the minimal polynomial of \(\alpha\).
🇮🇳 Hinglish Solution
(a) Linear map: mα(cx+y)=α(cx+y)=cαx+αy=c·mα(x)+mα(y). ✓
(b) Characteristic polynomial: mα ek n×n matrix hai (n=[L:K]). Char poly monic degree n, aur Cayley-Hamilton se mα satisfies its own char poly → α satisfies it.
(c) L=K(α) case: Basis {1,α,...,αⁿ⁻¹}. mα ki matrix companion matrix hai min poly ki. Char poly = min poly (both monic, same degree, same α root, min poly divides char poly, degrees equal → same).
English Solution
(a) Define \[ \mu_\alpha:L\to L,\qquad \mu_\alpha(x)=\alpha x. \] For \(x,y\in L\) and \(c\in K\), \[ \mu_\alpha(x+y)=\mu_\alpha(x)+\mu_\alpha(y), \] and \[ \mu_\alpha(cx)=c\mu_\alpha(x). \] Hence \(\mu_\alpha\) is \(K\)-linear.
(b) If \([L:K]=n\), then \(\mu_\alpha\) is represented by an \(n\times n\) matrix \(A\) over \(K\). By Cayley-Hamilton, \[ \chi_A(A)=0. \] Applying this to \(1\in L\), \[ \chi_A(\alpha)=0. \] Thus \(\alpha\) satisfies a monic polynomial of degree \(n\) over \(K\).
(c) If \(L=K(\alpha)\) and \[ m(X) \] is the minimal polynomial of \(\alpha\), then \[ \deg m=[L:K]=n. \] Since \(m(X)\mid\chi_A(X)\) and both are monic of degree \(n\), \[ \boxed{m(X)=\chi_A(X)}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
(a) Map: \[ \mu_\alpha(x)=\alpha x. \] Addition aur scalar multiplication dono preserve hote hain, so \(\mu_\alpha\) linear hai.
(b) Finite-dimensional case mein matrix \(A\) lo. Cayley-Hamilton: \[ \chi_A(A)=0. \] Isko \(1\) par apply karne se: \[ \chi_A(\alpha)=0. \]
(c) Agar \(L=K(\alpha)\), to minimal polynomial \(m(X)\) ki degree \[ [L:K]=n \] hai.
Since \[ m(X)\mid\chi_A(X) \] aur dono monic degree \(n\) ke hain, \[ \boxed{m(X)=\chi_A(X)}. \]
An element \(\alpha \in L\) is algebraic over \(K\) if there exists a nonzero \(f(x) \in K[x]\) with \(f(\alpha) = 0\). Otherwise \(\alpha\) is transcendental over \(K\).
🇮🇳 Hinglish Explanation
Algebraic over K: α ∈ L algebraic hai agar koi nonzero polynomial f(x)∈K[x] exist kare jisme f(α)=0.
Transcendental: Koi aisa polynomial nahi.
- √2 algebraic over ℚ: X²−2=0 ✓
- i algebraic over ℚ: X²+1=0 ✓
- π transcendental over ℚ (Lindemann 1882) ✓
- x ∈ k(x) transcendental over k (definition se) ✓
Is every algebraic extension finitely generated? Prove or give a counterexample.
🇮🇳 Hinglish Solution
Answer: NO. Algebraic extensions finitely generated nahi hoti necessarily.
Counter-example: ℚ̄ = {α∈ℂ : α algebraic over ℚ}. Ye algebraic extension hai ℚ ka (by definition). Lekin infinitely generated — ∛2, ⁴√2, ⁵√2, ... sab chahiye, koi finite set kafi nahi.
English Solution
Claim: An algebraic extension need not be finitely generated.
Example: Consider the algebraic closure \(\overline{\mathbb Q}\) of \(\mathbb Q\).
Suppose \[ \overline{\mathbb Q} =\mathbb Q(\alpha_1,\ldots,\alpha_m) \] for finitely many algebraic elements \(\alpha_i\).
Then the extension would be finite: \[ [\overline{\mathbb Q}:\mathbb Q]=N \] for some \(N\).
But for every \(n\ge1\), \[ X^n-2 \] is irreducible over \(\mathbb Q\) by Eisenstein's criterion. Hence \[ [\mathbb Q(\sqrt[n]{2}):\mathbb Q]=n. \]
Since \(\overline{\mathbb Q}\) contains such fields for arbitrarily large \(n\), its degree over \(\mathbb Q\) cannot be finite.
Therefore, \[ \boxed{\overline{\mathbb Q}/\mathbb Q\text{ is algebraic but not finitely generated}.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Counterexample: Algebraic closure \[ \overline{\mathbb Q} \] lo.
Iska har element \(\mathbb Q\) par algebraic hai, so extension algebraic hai.
Agar ye finitely generated hota, maan lo \[ \overline{\mathbb Q} =\mathbb Q(\alpha_1,\ldots,\alpha_m), \] to extension finite hota.
Lekin har \(n\) ke liye \[ \sqrt[n]2 \] algebraic hai aur \[ [\mathbb Q(\sqrt[n]2):\mathbb Q]=n. \]
So arbitrarily large finite-degree subextensions exist karte hain. Ye finite extension ke against hai.
Hence: \[ \boxed{\overline{\mathbb Q}/\mathbb Q\text{ algebraic but not finitely generated}.} \]
Let \(L/K\) be a field extension, \(\alpha \in L\). The following are equivalent:
- \(K(\alpha)/K\) is finite.
- \(\alpha\) is algebraic over \(K\).
Moreover, \(\alpha\) is algebraic over \(K\) iff \(\alpha^n\) is algebraic over \(K\) for some positive integer \(n\).
🇮🇳 Hinglish Explanation
Equivalence: K(α)/K finite ⟺ α algebraic over K.
⟹: Agar K(α)/K finite of degree n, toh {1,α,...,αⁿ} linearly dependent over K → koi polynomial relation → α algebraic.
⟸: Agar α algebraic, min poly degree n, toh {1,α,...,αⁿ⁻¹} spans K(α) over K → [K(α):K]=n finite.
Bonus: α algebraic ⟺ αⁿ algebraic for some n≥1 (reason: finite extension K(α)/K implies K(αⁿ)/K finite implies αⁿ algebraic; converse: K(αⁿ) finite → K(α) finite over K(αⁿ) → tower).
If \(\alpha \in L\) is algebraic over \(K\), its minimal polynomial over \(K\) is the unique monic polynomial \(f(x) \in K[x]\) of least degree with \(f(\alpha) = 0\).
🇮🇳 Hinglish Explanation
Minimal polynomial of α over K: Woh unique monic polynomial f(x)∈K[x] jo minimum degree ka ho aur f(α)=0 satisfy kare.
- √2 over ℚ: min poly = X²−2
- i over ℚ: min poly = X²+1
- ∛2 over ℚ: min poly = X³−2
Let \(L/K\) be a field extension and \(\alpha \in L\) algebraic over \(K\). Then:
- The minimal polynomial of \(\alpha\) is irreducible in \(K[x]\).
- Any \(f(x) \in K[x]\) with \(f(\alpha) = 0\) is divisible by the minimal polynomial.
- \([K(\alpha):K]\) equals the degree of the minimal polynomial.
- \(K(\alpha) = K[\alpha]\).
🇮🇳 Hinglish Explanation
4 fundamental properties of min poly:
- Irreducible: Agar f=g·h (reducible), toh g(α)=0 ya h(α)=0 → smaller degree polynomial → contradicts minimality.
- Divisibility: Koi bhi f(α)=0 wala polynomial min poly se divisible. (Division algo: f=q·(min poly)+r, deg r < deg min poly, r(α)=0 → r=0 by minimality.)
- Degree: [K(α):K] = deg(min poly). Basis {1,α,...,αⁿ⁻¹}.
- K(α)=K[α]: K[α] ≅ K[X]/(min poly). Min poly irreducible → (min poly) maximal ideal → quotient = field. Toh K[α] already field hai = K(α).
If \(L/K\) is a finite extension, then \(L\) is algebraic over \(K\) and finitely generated as a field extension of \(K\).
🇮🇳 Hinglish Explanation
Finite extension L/K → algebraic: Exercise 1.36 se, [L:K]=n → har α∈L, degree ≤ n polynomial satisfy karta hai → algebraic.
Finite extension → finitely generated: Koi K-basis {α₁,...,αₙ} lo → L=K(α₁,...,αₙ). ✓
Let \(L/K\) be a field extension and \(X \subseteq L\) with every element of \(X\) algebraic over \(K\). Then \(K(X)/K\) is algebraic. Moreover, if \(X\) is finite then \(K(X)/K\) is finite.
🇮🇳 Hinglish Explanation
Agar har element of X algebraic over K hai, toh K(X)/K algebraic hai.
Key idea: Koi β∈K(X) lo. β depend karta hai finite subset Sf={α₁,...,αₖ}⊆X pe. K(Sf)/K finite hai (Lemma 1.40 — finite many algebraic adjoined). β∈K(Sf) → β algebraic over K (Lemma 1.47). ✓
Let \(L/K\), \(K/F\) be algebraic extensions. Then \(L/F\) is also algebraic.
🇮🇳 Hinglish Explanation
Transitivity of algebraic extensions: L/K algebraic, K/F algebraic ⟹ L/F algebraic.
Proof: Koi α∈L lo. α satisfies f(X)=Xⁿ+cₙ₋₁Xⁿ⁻¹+...+c₀∈K[X] (α algebraic over K). Coefficients c₀,...,cₙ₋₁ ∈ K, algebraic over F. M=F(c₀,...,cₙ₋₁) finite over F. α algebraic over M (same polynomial, M⊇K coefficients). M(α)/M finite. Toh [M(α):F]=[M(α):M]·[M:F] finite → α algebraic over F. ✓
Let \(L/K\) be a field extension. The set of all elements of \(L\) algebraic over \(K\) forms a subfield of \(L\) — it is the largest algebraic extension of \(K\) contained in \(L\).
🇮🇳 Hinglish Explanation
Algebraic elements form a subfield:
Agar α, β algebraic over K → K(α,β) finite (Lemma 1.40) → sare elements of K(α,β) algebraic (Lemma 1.47) → α+β, α−β, αβ, α/β sab algebraic. ✓ So algebraic elements ek subfield banate hain.
The algebraic closure of \(K\) in \(L\) is the set of all elements of \(L\) algebraic over \(K\). The extension \(L/K\) is an algebraic extension if every element of \(L\) is algebraic over \(K\).
🇮🇳 Hinglish Explanation
Algebraic closure of K in L: L ke woh sare elements jo K ke upar algebraic hain.
Example: ℚ ka algebraic closure ℂ mein = ℚ̄ = {α∈ℂ : α satisfies some nonzero polynomial over ℚ} = "algebraic numbers".
Algebraic extension: L/K algebraic hai agar L = algebraic closure of K in L, yaani har element algebraic hai.
Let \(\overline{\mathbb{Q}}\) denote the algebraic closure of \(\mathbb{Q}\) in \(\mathbb{C}\). Show that \(\overline{\mathbb{Q}}\) contains an extension of \(\mathbb{Q}\) of degree \(n\) for every positive integer \(n\). Using this, show that \(\overline{\mathbb{Q}}/\mathbb{Q}\) is infinite.
🇮🇳 Hinglish Solution
ℚ̄ mein degree-n extension: Xⁿ−2 irreducible over ℚ (Eisenstein). Toh ℚ(∛[n]{2}) degree n extension hai ℚ ka. ∛[n]{2}∈ℚ̄ toh ℚ(∛[n]{2})⊆ℚ̄.
Toh ℚ̄ mein degree n ka extension exist karta hai for every n≥1 → [ℚ̄:ℚ] cannot be any finite number N (kyunki agar N hota, toh N+1 degree extension fit nahi hoti by Tower Law). ∴ [ℚ̄:ℚ]=∞.
English Solution
Claim: \(\overline{\mathbb Q}/\mathbb Q\) is an infinite extension.
For every \(n\ge1\), consider \[ f(X)=X^n-2. \] By Eisenstein's criterion at \(2\), \(f(X)\) is irreducible over \(\mathbb Q\).
Therefore, \[ [\mathbb Q(\sqrt[n]{2}):\mathbb Q]=n. \]
Since \(\sqrt[n]{2}\) is algebraic over \(\mathbb Q\), \[ \mathbb Q(\sqrt[n]{2})\subseteq\overline{\mathbb Q}. \]
Thus \(\overline{\mathbb Q}\) contains subfields of degree \(n\) for every \(n\). Hence its degree over \(\mathbb Q\) cannot be finite.
Therefore, \[ \boxed{[\overline{\mathbb Q}:\mathbb Q]=\infty}. \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Har \(n\ge1\) ke liye polynomial: \[ X^n-2. \]
Eisenstein at \(2\) se irreducible hai.
Isliye: \[ [\mathbb Q(\sqrt[n]2):\mathbb Q]=n. \]
Ye element algebraic hai, so ye field \(\overline{\mathbb Q}\) ke andar hai.
Matlab \(\overline{\mathbb Q}\) ke andar degree \(1,2,3,\ldots\) sab possible hain.
Isliye extension finite nahi ho sakta.
\[ \boxed{[\overline{\mathbb Q}:\mathbb Q]=\infty}. \]