Groups, rings, fields, vector spaces
... Proof xq − x has at most q roots in F. It now suffices to show that for all α ∈ F, x − α divides xq − x, or equivalently that α is a root. If α = 0, then it is clear. Otherwise, note that non-zero α is contained in F∗ , which has order q − 1. By Lagrange’s theorem αq = α, and we are done. We now are ...
... Proof xq − x has at most q roots in F. It now suffices to show that for all α ∈ F, x − α divides xq − x, or equivalently that α is a root. If α = 0, then it is clear. Otherwise, note that non-zero α is contained in F∗ , which has order q − 1. By Lagrange’s theorem αq = α, and we are done. We now are ...
Complex Numbers
... claim that ({a,b},#,*) is a field, because all algebraic properties are preserved by an isomorphism. Example 1.3. (R×R,+,·) where + and · are defined “componentwise”, i.e. (a,b)+(c,d) = (a+c, b+d) and (a,b) · (c,d) = (a·c, b·d) is NOT a field, since no element of the form (0,b) or (a,0) is invertibl ...
... claim that ({a,b},#,*) is a field, because all algebraic properties are preserved by an isomorphism. Example 1.3. (R×R,+,·) where + and · are defined “componentwise”, i.e. (a,b)+(c,d) = (a+c, b+d) and (a,b) · (c,d) = (a·c, b·d) is NOT a field, since no element of the form (0,b) or (a,0) is invertibl ...
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WHAT IS A POLYNOMIAL? 1. A Construction of the Complex
... meaning an associative, commutative ring A having scalar multiplication by R. (From now on in this writeup, algebras are understood to be commutative.) • The algebraic structure is not described by internal details of what its elements are, but rather by how it interacts with other R-algebras. Speci ...
... meaning an associative, commutative ring A having scalar multiplication by R. (From now on in this writeup, algebras are understood to be commutative.) • The algebraic structure is not described by internal details of what its elements are, but rather by how it interacts with other R-algebras. Speci ...
Sol 2 - D-MATH
... 3. (a) Let F [x] be a polynomial ring over a field F . Prove that the maximal ideals of F [x] are the principal ideals generated by monic irreducible polynomials. Proof : We already know that ideals in a polynomial ring over a field are principal ideals, and that any non-zero ideal is generated by t ...
... 3. (a) Let F [x] be a polynomial ring over a field F . Prove that the maximal ideals of F [x] are the principal ideals generated by monic irreducible polynomials. Proof : We already know that ideals in a polynomial ring over a field are principal ideals, and that any non-zero ideal is generated by t ...
Algebraic Expressions and Terms
... You are familiar with the following type of numerical expressions: ...
... You are familiar with the following type of numerical expressions: ...
Problem Set 1 - University of Oxford
... Thus deg : k[t]\{0} → Z≥0 . It is easy to check that deg(f.g) = deg(f ) + deg(g) (in fact this holds in any polynomial ring R[t] where R is an integral domain). Thus if f ∈ k[t] is a unit, so that there is some g ∈ k[t] with f.g = 1, taking degrees we see that deg(f ) + deg(g) = 0 and since both deg ...
... Thus deg : k[t]\{0} → Z≥0 . It is easy to check that deg(f.g) = deg(f ) + deg(g) (in fact this holds in any polynomial ring R[t] where R is an integral domain). Thus if f ∈ k[t] is a unit, so that there is some g ∈ k[t] with f.g = 1, taking degrees we see that deg(f ) + deg(g) = 0 and since both deg ...