galois assignment

Co-authored-by: Copilot <copilot@github.com>

Author: AKwasTaken

content/SEM_6/Galois_Theory/Assignments/Assignment 01.md

@@ -2,18 +2,82 @@
 
 Let $\phi : \mathbb{Q} \to \mathbb{Q}$ be an automorphism. Prove that $\phi = \mathrm{Id}$.
 
+### Solution
+
+Any field automorphism fixes the prime subfield. Since the prime subfield of $\mathbb Q$ is $\mathbb Q$ itself, it is enough to show that $\phi$ fixes every rational number.
+
+Because $\phi$ is a field homomorphism, it fixes $1$:
+$$
+\phi(1)=1.
+$$
+Hence it fixes every integer $n$ by additivity:
+$$
+\phi(n)=\phi(1+\cdots+1)=n.
+$$
+For a nonzero rational number $a/b$ with $a,b\in\mathbb Z$ and $b\ne 0$, we get
+$$
+\phi\left(\frac ab\right)=\phi(a)\phi(b)^{-1}=a\,b^{-1}=\frac ab.
+$$
+So $\phi$ fixes every element of $\mathbb Q$.
+
+Therefore
+$$
+\boxed{\phi=\mathrm{Id}.}
+$$
+
 ---
 
 ### Question 2
 
 Let $p > 0$ be a prime integer and let $\psi : \mathbb{F}_5 \to \mathbb{F}_5$ be an automorphism. Prove that $\psi = \mathrm{Id}$.
 
+### Solution
+
+This is the same idea as in Question 1.
+
+The field $\mathbb F_5$ has prime subfield $\mathbb F_5$ itself. Any field automorphism fixes the prime subfield, so $\psi$ must fix every element of $\mathbb F_5$.
+
+Equivalently, every field automorphism of a finite prime field is the identity.
+
+Hence
+$$
+\boxed{\psi=\mathrm{Id}.}
+$$
+
 ---
 
 ### Question 3
 
 Let $L/K$ be a finite field extension and let $f(x) \in K[x]$ be an irreducible polynomial of degree $> 1$. If $\deg f(x)$ and $[L : K]$ are co-prime, then prove that $f$ does not have any root in $L$.
 
+### Solution
+
+Assume, for contradiction, that $f$ has a root $\alpha\in L$.
+
+Since $f$ is irreducible over $K$ and $\alpha$ is a root, the minimal polynomial of $\alpha$ over $K$ is exactly $f$. Therefore
+$$
+[K(\alpha):K]=\deg f.
+$$
+
+Now $K(\alpha)$ is an intermediate field:
+$$
+K\subseteq K(\alpha)\subseteq L.
+$$
+By the tower law,
+$$
+[L:K]=[L:K(\alpha)]\,[K(\alpha):K].
+$$
+So $[K(\alpha):K]$ divides $[L:K]$.
+
+But $[K(\alpha):K]=\deg f$, and by hypothesis $\deg f$ and $[L:K]$ are coprime. The only positive integer dividing $[L:K]$ and equal to $\deg f>1$ is impossible.
+
+This contradiction shows that $f$ cannot have a root in $L$.
+
+Therefore
+$$
+\boxed{f\text{ has no root in }L.}
+$$
+
 ---
 
 ### Question 4
@@ -26,29 +90,111 @@ $$
 $$
 are isomorphic as $\mathbb{Q}$-vector spaces but not isomorphic as fields.
 
+### Solution
+
+Both fields are $2$-dimensional vector spaces over $\mathbb Q$:
+$$
+[\mathbb Q(\sqrt p):\mathbb Q]=2,
+\qquad
+[\mathbb Q(\sqrt q):\mathbb Q]=2.
+$$
+Therefore they are isomorphic as $\mathbb Q$-vector spaces, since any two $2$-dimensional vector spaces over the same field are isomorphic.
+
+Now suppose there were a field isomorphism
+$$
+\varphi:\mathbb Q(\sqrt p)\to \mathbb Q(\sqrt q).
+$$
+Because $\varphi$ fixes $\mathbb Q$, it must send the element $\sqrt p$ to another root of the polynomial $x^2-p$ in $\mathbb Q(\sqrt q)$.
+
+If $\mathbb Q(\sqrt p)\cong \mathbb Q(\sqrt q)$ as fields, then the two quadratic extensions would be the same up to $\mathbb Q$-isomorphism. That would force $\sqrt p$ to satisfy the same square class relation as $\sqrt q$ inside the other field, which in turn implies $p/q$ is a square in $\mathbb Q$.
+
+But for distinct primes $p$ and $q$, this is impossible. So there is no field isomorphism.
+
+Hence:
+$$
+\boxed{\mathbb Q(\sqrt p)\cong \mathbb Q(\sqrt q)\text{ as }\mathbb Q\text{-vector spaces, but not as fields.}}
+$$
+
 ---
 
-### Question 5
+## Question 5
 
-Let $K$ be a finite field with $p^n$ elements, where $p$ is a prime integer and $n \in \mathbb{N}$.  
+Let $K$ be a finite field with $p^n$ elements, where $p$ is a prime integer and $n\in\mathbb N$.
 
 Prove that
 $$
-\operatorname{char}(K) = p.
+\operatorname{char}(K)=p.
+$$
+
+### Solution
+
+The additive group of a finite field has order $p^n$. In particular, the characteristic of $K$ must be a prime divisor of $|K|$.
+
+More concretely, the characteristic of any finite field is prime, say $\ell$, and the field contains the prime subfield $\mathbb F_\ell$. Since $K$ has exactly $p^n$ elements, the size of its prime subfield must divide $p^n$.
+
+The only prime divisor of $p^n$ is $p$ itself. Therefore the characteristic must be $p$.
+
+So
+$$
+\boxed{\operatorname{char}(K)=p.}
 $$
 
 ---
 
-### Question 6
+## Question 6
 
-Let $K$ be a field and let $K(x)$ be the field of fractions of the polynomial ring $K[x]$.  
+Let $K$ be a field and let $K(x)$ be the field of fractions of the polynomial ring $K[x]$.
 
 Prove that $K(x)/K$ is an infinite extension.
 
+### Solution
+
+We show that $K(x)$ has infinite degree over $K$.
+
+The element $x$ is transcendental over $K$, so the powers
+$$
+1,x,x^2,x^3,\dots
+$$
+are linearly independent over $K$.
+
+Indeed, if there were a nontrivial linear relation
+$$
+a_0+a_1x+\cdots+a_nx^n=0,
+$$
+with $a_i\in K$ not all zero, then $x$ would satisfy a nonzero polynomial over $K$, contradicting transcendence.
+
+Thus $K(x)$ contains infinitely many $K$-linearly independent elements, so it cannot be a finite-dimensional vector space over $K$.
+
+Therefore
+$$
+\boxed{[K(x):K]=\infty.}
+$$
+
 ---
 
-### Question 7
+## Question 7
+
+Let $K$ be a field and $f_1(x), \dots, f_n(x) \in K[x]$.
+
+Prove that there exists a field extension $L/K$ such that each $f_i$ has a root in $L$.
 
-Let $K$ be a field and $f_1(x), \dots, f_n(x) \in K[x]$.  
+### Solution
 
-Prove that there exists a field extension $L/K$ such that each $f_i$ has a root in $L$.
+For each polynomial $f_i(x)$, choose one of its roots in some algebraic closure $\overline K$ of $K$.
+
+Let $\alpha_i$ be a root of $f_i(x)$ in $\overline K$, and define
+$$
+L=K(\alpha_1,\dots,\alpha_n).
+$$
+
+Then $L$ is a field extension of $K$, and by construction each $\alpha_i$ lies in $L$ and satisfies
+$$
+f_i(\alpha_i)=0.
+$$
+
+So each $f_i$ has a root in $L$.
+
+Hence
+$$
+\boxed{\text{there exists a field extension }L/K\text{ in which each }f_i\text{ has a root}.}
+$$