By Ghorpade

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**Additional info for A Quick Review Of Commutative Algebra [Short Lecture]**

**Example text**

11) in Q(α), where mα,Q = X 3 − 2. 2 Maple, in fact, already contains routines that handle the calculation of inverses and reduced forms in generated fields: these can be invoked with the commands evala(Normal( )) and reduce. Similarly, Mathematica, with its NumberTheory‘AlgebraicNumberFields‘ package, provides commands for the simplification of algebraic expressions in generated fields. In place of these, the package AlgFields provides functions with a simpler interface that help collect information about the field extensions.

Since {γ1 , γ2 , . . , γm} is a basis for K (α) over K (β), δ may be written as a linear combination of γ1 , γ2 , . . , γm with coefficients in K (β), say m δ= ci γi , ci ∈ K (β). i=1 Then, since {ν1 , ν2 , . . , νn} is a basis for K (β) over K , each of these coefficients ci may be written as a linear combination of ν1 , ν2 , . . , νn with coefficients in K , say as n ci = ci, j ν j , ci, j ∈ K , i = 1, . . , m. j=1 Multiplying out and collecting terms, we see that δ may then be written as a linear combination of the elements of {γi ν j } with coefficients in K : m δ= m n i=1 m n ci, j ν j γi = ci γi = i=1 j=1 ci, j (γi ν j ).

2), p is irreducible. If q = q1 q2 for polynomials q1 , q2 ∈ Q[X ], then p = p1 p2 where pi (X ) = qi (aX + b) for i = 1, 2. Hence q is irreducible also. Now p and q are irreducible polynomials of the same degree, and so the minimal polynomials of α and β over Q have the same degree. 1). Therefore α may be represented by an arithmetic combination of β, and β may be represented by an arithmetic combination of α, each with coefficients in Q. But then K (α) ⊂ K (β) and K (β) ⊂ K (α), from which we conclude that K (α) = K (β).

### A Quick Review Of Commutative Algebra [Short Lecture] by Ghorpade

by Robert

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