# Galois Theory

References: This video course in Galois Theory is based on Abstract Algebra by  Dummit and Foote, and online lecture notes on Galois Theory of Milne [course notes]. The latter is my favourite and I have borrowed proof style and examples from Milne’s notes.

The videos lectures are linked via square brackets

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We begin the course by introducing [irreducible polynomials] and four criteria (Rational Roots, Gauss theorem. mod p test, Eisentein’s Criterion) to determine irreducibility of the polynomial. Why do we even bother with irreducible polynomials ? This is because an irreducible polynomial $f(x)\in F[x]$ for $F$ field is also maximal and prime.

In the [second lecture] we introduce field extensions which can be considered as a vector space over the base field. For example $[\mathbf{C}:\mathbf{R}]$. Kronecker’s theorem ensures that we can always find an extension field in which $f(x)\in F[x]$ will have a root. This extension field is constructed by simply adjoining the root of the polynomial to the base field. More formally we have $E=F[x]/$. This is the key to Galois theory. If the polynomial has m roots. Then $E\simeq F[\alpha_1]\simeq F[\alpha_2]\simeq\ldots\simeq F[\alpha_m]$, the field cannot tell the roots apart.

Let us now introduce [ Algebraic Extensions], the key idea is to associate a minimal polynomial with each algebraic element. If $\alpha$ is algebraic over $F$ then $[F(\alpha):F]=$ degree minimal polynomial and hence finite. This automatically leads to result that $[E:F]$ finite implies that the extension is algebraic.

In this lecture we prove that if [E:F] is algebraic and [L:E] is algebraic then [L:F] is algebraic. This lecture collects the techniques we have developed till now and uses them in a non-trivial way to prove an important result  [Algebraic over Algebraic is Algebraic]

In this lecture we give the algebraic counterpart of [Geometric Constructions] using ruler and compass. A point (x,y) is constructible then $[Q(x,y):Q]=2^s$ where $s=\{0,1,2,3,\ldots\}$. This helps us prove the impossibility of duplicating a cube and squaring a circle

In the following post we introduce splitting fields. Splitting Fields