§1
Fields
The following are examples of fields.
- The field of rational numbers \(\mathbb{Q}\).
- The field of real numbers \(\mathbb{R}\).
- The field of complex numbers \(\mathbb{C}\).
- For any prime \(p\), the ring \(\mathbb{Z}/p\mathbb{Z}\) of integers modulo \(p\) is a field, denoted \(\mathbb{F}_p\).
- For any prime \(p\) and positive integer \(n\), there exists a unique (up to isomorphism) finite field of order \(p^n\), denoted \(\mathbb{F}_{p^n}\).
🇮🇳 Hinglish Explanation
Field kya hota hai? (Proper Definition) Ek set \(F\) jisme do operations hote hain: addition (\(+\)) aur multiplication (\(\times\)). Inhe kuch rules follow karne hote hain:
- Closure & Commutativity: \(a+b = b+a\) aur \(a \times b = b \times a\). Dono ka result \(F\) mein hi aana chahiye.
- Identities: Ek \(0\) (additive identity) aur ek \(1\) (multiplicative identity) exist karta hai jahan \(1 \neq 0\).
- Inverses: Har element \(a\) ka negative \(-a\) hota hai (yani \(a + (-a) = 0\)), aur har non-zero element \(a\) ka inverse \(a^{-1}\) hota hai (yani \(a \times a^{-1} = 1\)).
- Distributivity: \(a \times (b + c) = (a \times b) + (a \times c)\).
Simple terms mein: Ek aisa mathematical system jahan jod, ghata, guna, aur bhag (zero se nahi) perfectly kaam kare aur result wapas usi system mein aaye.
- ℚ (Rational): Sabhi fractions. Field hai.
- 𝔽_p (Finite): Modulo prime. Jaise 𝔽₂={0,1} mein 1+1=0.
Show that the following are not fields.
- The ring of integers \(\mathbb{Z}\).
- The polynomial ring \(\mathbb{Q}[x]\) in one variable over \(\mathbb{Q}\).
- The polynomial ring \(\mathbb{R}[x,y]\) in two variables over \(\mathbb{R}\).
🇮🇳 Hinglish Solution
Ye fields kyun nahi hain?
- ℤ: 2∈ℤ hai, lekin 2 ka inverse ½ ℤ mein nahi. → Field nahi.
- ℚ[x]: Polynomial ring — x ka inverse koi polynomial nahi. → Field nahi.
- ℝ[x,y]: Same reason, x ka inverse exist nahi karta. → Field nahi.
English Solution
Claim: None of \(\mathbb Z,\mathbb Q[x],\mathbb R[x,y]\) is a field.
Proof: In a field, every nonzero element must have a multiplicative inverse.
In \(\mathbb Z\), \(2\neq0\), but \(2x=1\) gives \(x=\frac12\notin\mathbb Z\). Hence \(2\) has no inverse.
In \(\mathbb Q[x]\), if \(x f(x)=1\), then \[ \deg(xf)=1+\deg f\ge1, \] which is impossible since \(\deg 1=0\). Hence \(x\) has no inverse.
Similarly, in \(\mathbb R[x,y]\), if \(xg(x,y)=1\), then \[ \deg(xg)=1+\deg g\ge1, \] which is impossible.
Therefore, \[ \boxed{\mathbb Z,\ \mathbb Q[x],\ \mathbb R[x,y]\text{ are not fields}.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Idea: Field mein har nonzero element ka multiplicative inverse hona chahiye.
\(\mathbb Z\) mein \(2\neq0\), lekin \[ 2x=1\Rightarrow x=\frac12\notin\mathbb Z. \] Isliye \(2\) ka inverse \(\mathbb Z\) mein nahi hai.
\(\mathbb Q[x]\) mein \(x\) ka inverse maan lo \(f(x)\). Tab \[ xf(x)=1. \] Lekin \[ \deg(xf)=1+\deg f\ge1, \] jabki \(\deg1=0\). Contradiction.
Bilkul same argument \(\mathbb R[x,y]\) mein bhi \(x\) ke liye apply hota hai.
Isliye \[ \boxed{\mathbb Z,\mathbb Q[x],\mathbb R[x,y]\text{ fields nahi hain}.} \]
Show that any finite integral domain is a field.
🇮🇳 Hinglish Solution
Integral domain: Ring jisme zero divisors na hon — yaani a·b=0 ⟹ a=0 ya b=0.
Proof idea: Finite domain mein, koi a≠0 lo. Map φ(x)=ax injective hai (ax=ay ⟹ x=y, zero divisor nahi). Finite set pe injective = surjective. Toh koi b hai jisme ab=1. Isliye a invertible → domain ek field hai.
English Solution
Claim: Every finite integral domain \(R\) is a field.
Proof: Let \(a\in R\), \(a\neq0\). Define \[ \varphi_a:R\to R,\qquad \varphi_a(x)=ax. \]
If \(\varphi_a(x)=\varphi_a(y)\), then \[ a(x-y)=0. \] Since \(R\) is an integral domain and \(a\neq0\), \[ x-y=0\implies x=y. \] Thus \(\varphi_a\) is injective.
Since \(R\) is finite, \(\varphi_a\) is also surjective. Hence there exists \(b\in R\) such that \[ \varphi_a(b)=ab=1. \]
Thus every nonzero element of \(R\) has an inverse. Therefore, \[ \boxed{R\text{ is a field}.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Idea: Humein prove karna hai ki finite integral domain ke har nonzero element ka inverse exist karta hai.
Kisi \(a\neq0\) ko lo aur map define karo: \[ \varphi_a(x)=ax. \]
Agar \[ ax=ay, \] to \[ a(x-y)=0. \] Integral domain mein zero-product property hoti hai. Aur \(a\neq0\), isliye \[ x-y=0\Rightarrow x=y. \] So \(\varphi_a\) injective hai.
Ab important point: \(R\) finite hai. Finite set mein injective map \(R\to R\) automatically surjective hota hai.
Isliye \(1\) bhi map ka output hoga. Matlab koi \(b\in R\) hoga jiske liye \[ ab=1. \]
To har nonzero \(a\) ka inverse mil gaya.
Hence \[ \boxed{R\text{ is a field}.} \]
Show that any finite field has order \(p^n\) for some prime \(p\) and positive integer \(n\).
🇮🇳 Hinglish Solution
Characteristic of a field: Woh smallest n jisme 1+1+...+1 (n times) = 0. Agar koi aisa n nahi → char=0.
Steps: Finite field F ka char p (prime) hona chahiye (composite hota toh zero divisors aate). Toh 𝔽_p = {0,1,...,p-1} ⊆ F. F ek finite-dimensional vector space hai 𝔽_p ke upar, dimension n maan lo. Toh |F| = p^n.
English Solution
Claim: If \(F\) is a finite field, then \[ |F|=p^n \] for some prime \(p\) and positive integer \(n\).
Proof: The characteristic of \(F\) is a positive integer \(p\). Since \(F\) is a field, \(p\) must be prime.
Thus \(F\) contains the prime field \[ \mathbb F_p. \] Hence \(F\) is a vector space over \(\mathbb F_p\).
Let \[ n=\dim_{\mathbb F_p}F. \] Choosing a basis of \(F\), every element of \(F\) has a unique representation using \(n\) coefficients from \(\mathbb F_p\).
There are \(p\) choices for each coefficient, so \[ |F|=p^n. \]
Therefore, \[ \boxed{|F|=p^n.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Idea: Finite field \(F\) ke andar ek prime field \(\mathbb F_p\) hota hai.
Field ki characteristic \(p\) hoti hai, aur field ki characteristic hamesha prime hoti hai.
Isliye \[ \mathbb F_p\subseteq F. \]
Ab \(F\) ko \(\mathbb F_p\) ke upar vector space samjho. Agar \[ \dim_{\mathbb F_p}F=n, \] to har element ko basis ke \(n\) coefficients se uniquely represent kiya ja sakta hai.
Har coefficient ke \(p\) choices hain. Isliye total choices: \[ p^n. \]
Hence \[ \boxed{|F|=p^n}. \]
For any field \(k\), the ring of polynomials in one variable \(x\) over \(k\) is denoted \(k[x]\). The field of rational functions \(k(x)\) consists of all elements of the form \(\dfrac{f(x)}{g(x)}\) where \(f(x), g(x) \in k[x]\) and \(g(x) \neq 0\). It is the field of fractions of the integral domain \(k[x]\).
🇮🇳 Hinglish Explanation
k[x] = polynomial ring. Elements jaise: 3x²+2x−1.
k(x) = rational function field. Elements: f(x)/g(x) jahan g≠0. Jaise (x²+1)/(x−3).
Analogy: k[x] waise hai jaise ℤ (integer ring), aur k(x) waise hai jaise ℚ (rationals). k[x] se k(x) waise banta hai jaise ℤ se ℚ — field of fractions.
For any field \(k\), the ring of formal power series in one variable \(x\) over \(k\) is denoted \(k[[x]]\). The field of formal Laurent series \(k((x))\) consists of generalized formal power series \[ \sum_{n \ge n_0} a_n x^n \] for some integer \(n_0\), where \(a_n \in k\) for all \(n \ge n_0\) and \(a_{n_0} \neq 0\). It is the field of fractions of \(k[[x]]\).
🇮🇳 Hinglish Explanation
k[[x]]: Formal power series — ∑ aₙxⁿ, n≥0. "Formal" matlab convergence ki chinta nahi.
k((x)): Laurent series — kuch negative powers bhi ok. Jaise x⁻²+3+x+x²+
Kaise banta hai: k[[x]] se k((x)) waise banta hai jaise k[x] se k(x) — field of fractions.
§2
Field Homomorphisms and Isomorphisms
Let \(K\) and \(L\) be fields. A field homomorphism \(\sigma: K \to L\) preserves the field operations: for all \(a, b \in K\),
- \(\sigma(a+b) = \sigma(a) + \sigma(b)\),
- \(\sigma(ab) = \sigma(a)\sigma(b)\),
- \(\sigma(1_K) = 1_L\).
A homomorphism \(\sigma: K \to L\) is an isomorphism if there exists a homomorphism \(\tau: L \to K\) with \(\tau \circ \sigma = \mathrm{id}_K\) and \(\sigma \circ \tau = \mathrm{id}_L\).
🇮🇳 Hinglish Explanation
Field homomorphism σ:K→L: Ek map jo field structure preserve kare.
- σ(a+b) = σ(a)+σ(b) — addition preserve karo
- σ(ab) = σ(a)σ(b) — multiplication preserve karo
- σ(1_K) = 1_L — identity preserve karo
Isomorphism: Agar inverse map τ bhi exist kare → dono fields essentially same structure ke hain, sirf naam alag.
Show that if \(\sigma: K \to L\) is a field homomorphism, then \(\sigma\) is injective. Is \(\sigma\) necessarily surjective? If yes, prove it. If not, give a counterexample.
🇮🇳 Hinglish Solution
Injectivity (hamesha): ker(σ) = {a∈K : σ(a)=0} ek ideal hai K ka. K field hai toh K ke sirf ideals hain {0} aur K. ker(σ)≠K (kyunki σ(1)=1≠0). Toh ker(σ)={0} → σ injective.
Surjectivity nahi (necessarily): Counter-example: Inclusion ℚ ↪ ℝ ek field homomorphism hai — lekin surjective nahi (π∈ℝ, π∉ℚ).
English Solution
Claim: Every field homomorphism \(\sigma:K\to L\) is injective, but it need not be surjective.
Proof: The kernel of \(\sigma\) is an ideal of the field \(K\). Hence \[ \ker\sigma=\{0\}\quad\text{or}\quad K. \] Since \[ \sigma(1_K)=1_L\neq0, \] we have \(\ker\sigma\neq K\). Thus \[ \ker\sigma=\{0\}, \] so \(\sigma\) is injective.
It need not be surjective. For example, the inclusion \[ \mathbb Q\hookrightarrow\mathbb R \] is a field homomorphism but is not surjective, since \(\sqrt2\notin\mathbb Q\).
Hence \[ \boxed{\text{Every field homomorphism is injective, but not necessarily surjective.}} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Idea: Field homomorphism ka kernel field ka ideal hota hai.
Field ke sirf do ideals hote hain: \[ \{0\}\quad\text{aur}\quad K. \]
Lekin \[ \sigma(1)=1\neq0, \] isliye kernel poora \(K\) nahi ho sakta.
Therefore \[ \ker\sigma=\{0\}. \] Aur zero kernel ka matlab map injective hai.
Lekin surjective hona zaroori nahi hai. Example: \[ \mathbb Q\hookrightarrow\mathbb R. \] \(\sqrt2\) real hai but rational nahi hai.
So: \[ \boxed{\text{Homomorphism injective hota hai, but necessarily surjective nahi.}} \]
Let \(\sigma: K \to L\) be a field homomorphism. Show that \(\sigma\) is an isomorphism if and only if \(\sigma\) is surjective.
🇮🇳 Hinglish Solution
Exercise 1.8 se σ hamesha injective hai. Agar σ surjective bhi hai toh bijective = isomorphism.
Inverse map τ=σ⁻¹ bhi field homomorphism hoga — verify karo: τ(a+b)=τ(a)+τ(b) etc. (σ ki properties use karke).
English Solution
Claim: A field homomorphism \(\sigma:K\to L\) is an isomorphism if and only if it is surjective.
Proof:
(\(\Rightarrow\)) If \(\sigma\) is an isomorphism, then it has an inverse. Hence \(\sigma\) is bijective, and therefore surjective.
(\(\Leftarrow\)) Suppose \(\sigma\) is surjective. By Exercise 1.8, every field homomorphism is injective. Hence \(\sigma\) is bijective.
Therefore \(\sigma^{-1}:L\to K\) exists. For \(a=\sigma(x)\), \(b=\sigma(y)\), \[ \sigma^{-1}(a+b)=x+y =\sigma^{-1}(a)+\sigma^{-1}(b), \] and similarly, \[ \sigma^{-1}(ab)=\sigma^{-1}(a)\sigma^{-1}(b). \] Also, \[ \sigma^{-1}(1_L)=1_K. \] Thus \(\sigma^{-1}\) is a field homomorphism.
Therefore \(\sigma\) is an isomorphism.
\[ \boxed{\sigma\text{ is an isomorphism }\iff \sigma\text{ is surjective}.} \] \(\blacksquare\)
🇮🇳 Hinglish Solution
Simple idea: Isomorphism ka matlab hi bijective homomorphism hai.
Agar \(\sigma\) isomorphism hai, to obviously surjective hai.
Ab agar \(\sigma\) surjective hai, to Exercise 1.8 ke according \(\sigma\) already injective hai.
Therefore \(\sigma\) bijective hai. Isliye inverse map exist karega: \[ \sigma^{-1}:L\to K. \]
Homomorphism ki properties se inverse bhi field homomorphism hota hai.
Hence: \[ \boxed{\sigma\text{ isomorphism }\iff\sigma\text{ surjective}.} \]
§3
Extension of Fields
Let \(K\) be a field. A field extension of \(K\) is a field \(L\) that contains \(K\) as a subfield. We write \(L/K\) to indicate that \(L\) is an extension of \(K\). We call \(K\) the base field and \(L\) the extension field.
🇮🇳 Hinglish Explanation
Field extension L/K: L aur K dono fields hain, aur K ⊆ L (K subfield hai L ka).
Notation L/K padho "L over K". K = base field (chhota), L = extension field (bada).
- ℝ/ℚ — reals over rationals
- ℂ/ℝ — complex over reals
- \(\mathbb{C}\) is an extension of \(\mathbb{R}\); \(\mathbb{R}\) is an extension of \(\mathbb{Q}\).
- For any field \(k\), \(k(x)/k\) and \(k((x))/k\) are field extensions.
🇮🇳 Hinglish Explanation
Classic tower: ℚ ⊂ ℝ ⊂ ℂ. Har step pe ek extension hai.
- k ⊂ k(x): x adjoin karo → rational functions milti hain
- k ⊂ k((x)): formal Laurent series
Let \(L/K\) be a field extension. An intermediate extension is a field \(M\) with \(K \subseteq M \subseteq L\). We say \(M/K\) is a subextension of \(L/K\).
🇮🇳 Hinglish Explanation
Intermediate extension: K ⊆ M ⊆ L. Teen fields ki chain.
M/K bhi ek extension hai (subextension of L/K).
Example: ℚ ⊆ ℚ(√2) ⊆ ℝ. Yahan ℚ(√2) intermediate hai ℝ/ℚ ka.
For any field \(k\), the field \(k(x^2)\) is an intermediate extension of \(k(x)/k\).
🇮🇳 Hinglish Explanation
k ⊆ k(x²) ⊆ k(x). Chain valid hai kyunki x²∈k(x).
k(x²) = {f(x²)/g(x²)} — sirf even powers of x allowed. Yeh k(x) se chhota hai.
For any field \(k\), the field \(k(x)\) is an intermediate extension of \(k((x))/k\).
🇮🇳 Hinglish Explanation
k ⊆ k(x) ⊆ k((x)). Rational functions naturally Laurent series mein fit hote hain (no negative powers needed for rationals, but Laurent allows them).
Let \(k(t)\) be the field of rational functions over \(k\). Show that any element of \(k(t)\) is a root of some quadratic polynomial with coefficients in \(k(t^2)\). Determine all field homomorphisms from \(k(t)\) to itself that fix \(k(t^2)\) pointwise.
🇮🇳 Hinglish Solution
Quadratic root: Koi f(t)/g(t) ∈ k(t) lo. Likho P(X) = g(t)X−f(t) = 0... wait, actually t satisfies X²−t²=0 over k(t²) (since t²∈k(t²)). Toh har element of k(t) = rational in t, aur t degree-2 over k(t²).
Homomorphisms fixing k(t²): σ(t²)=t² zaruri hai. Toh σ(t)²=t². Matlab σ(t)=t ya σ(t)=−t.
Dono valid hain → exactly do homomorphisms: id aur t↦−t.
English Solution
Let \(F=k(t^2)\subseteq k(t)\).
Part 1. Every element \(\alpha\in k(t)\) can be written in the form \[ \alpha=A(t^2)+tB(t^2),\qquad A,B\in k(t^2). \] Indeed, this is just separating even and odd powers of \(t\) in a rational function.
Define its conjugate by \[ \alpha' = A(t^2)-tB(t^2). \] Then \[ \alpha+\alpha' = 2A(t^2)\in k(t^2), \] and \[ \alpha\alpha' = A(t^2)^2 - t^2 B(t^2)^2 \in k(t^2). \] Hence the polynomial \[ P(X)=(X-\alpha)(X-\alpha') = X^2-(\alpha+\alpha')X+\alpha\alpha' \in k(t^2)[X] \] has \(\alpha\) as a root. Therefore every element of \(k(t)\) is a root of a quadratic polynomial over \(k(t^2)\).
Part 2. Let \(\sigma:k(t)\to k(t)\) be a field homomorphism fixing \(k(t^2)\) pointwise. Since \(t\) satisfies \[ X^2-t^2=0, \] its image satisfies \[ \sigma(t)^2 - t^2 =0, \] so \[ \sigma(t)^2=t^2. \] Thus \[ \sigma(t)=t\quad\text{or}\quad \sigma(t)=-t. \] A field homomorphism is determined by the image of the generator \(t\), so only two possibilities remain:
- the identity map \(t\mapsto t\),
- the sign-change map \(t\mapsto -t\).
Therefore, \[ \boxed{\operatorname{Aut}_{k(t^2)}(k(t))=\{\mathrm{id},\,t\mapsto -t\}\cong C_2.} \] In characteristic \(2\), the two values coincide, so the only automorphism is the identity. \(\blacksquare\)
🇮🇳 Hinglish Solution
Step 1: Let \[ F=k(t^2)\subseteq k(t). \] Hum jaante hain ki \[ [k(t):k(t^2)]=2. \] Isliye \(k(t)\) ka har element \(k(t^2)\) ke upar quadratic hota hai.
Step 2: Har \(\alpha\in k(t)\) ko is prakar likha ja sakta hai: \[ \alpha=A(t^2)+tB(t^2),\qquad A,B\in k(t^2). \] Uska conjugate define karte hain: \[ \alpha' = A(t^2)-tB(t^2). \] Ab \[ \alpha+\alpha' = 2A(t^2)\in k(t^2), \] \[ \alpha\alpha' = A(t^2)^2 - t^2B(t^2)^2 \in k(t^2). \] Isliye polynomial \[ P(X)=(X-\alpha)(X-\alpha')=X^2-(\alpha+\alpha')X+\alpha\alpha' \in k(t^2)[X] \] mein \(\alpha\) ek root hai. Iska matlab har element quadratic over \(k(t^2)\) hai.
Step 3: Agar \(\sigma\) field homomorphism hai jo \(k(t^2)\) ko pointwise fix karta hai, to \[ \sigma(t)^2 = \sigma(t^2)=t^2. \] Therefore \[ \sigma(t)=t\quad\text{or}\quad \sigma(t)=-t. \] Sirf do possibilities bachein:
- identity: \(t\mapsto t\),
- nontrivial automorphism: \(t\mapsto -t\).
Isliye \[ \boxed{\operatorname{Aut}_{k(t^2)}(k(t))=\{\mathrm{id},\,t\mapsto -t\}\cong C_2.} \] Aur characteristic 2 mein dono coincide ho jaate hain, isliye sirf identity bachega. \(\blacksquare\)