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Field Extensions
Def. A field 𝐸 is an extension field of a field 𝐹 if 𝐹 is a subfield of 𝐸 (𝐹 ≤ 𝐸).
Ex. ℝ is an extension field of ℚ and ℂ is an extension field of ℝ and ℚ.
Kronecker’s Theorem: Let 𝐹 be a field and let 𝑔(𝑥) be a nonconstant
polynomial in 𝐹 [𝑥 ]. Then there exists an extension field 𝐸 of 𝐹 and an
𝛼 ∈ 𝐸 such that 𝑔(𝛼 ) = 0.
Ex. Let 𝐹 = ℝ and let 𝑔(𝑥 ) = 𝑥 2 + 1. 𝑔(𝑥) has no zeros in ℝ and thus
is irreducible over ℝ. < 𝑥 2 + 1 > is a maximal ideal in ℝ[𝑥 ] so
ℝ[𝑥 ]/< 𝑥 2 + 1 > is a field.
We can view ℝ as a subfield of ℝ[𝑥 ]/< 𝑥 2 + 1 > through the mapping:
𝜑: ℝ → ℝ[𝑥 ]/< 𝑥 2 + 1 > by 𝜑(𝑡) = 𝑡+< 𝑥 2 + 1 >, 𝑡 ∈ ℝ.
Let 𝛼 = 𝑥+< 𝑥 2 + 1 > ∈ ℝ[𝑥 ]/< 𝑥 2 + 1 >,
then 𝛼 2 + 1 = (𝑥+< 𝑥 2 + 1 >)2 + (1+< 𝑥 2 + 1 >)
= ( 𝑥 2 + 1) + < 𝑥 2 + 1 >
= 0.
Thus 𝛼 is a zero of 𝑥 2 + 1. So we can think of ℝ[𝑥 ]/< 𝑥 2 + 1 > as
an extension field of ℝ, which has an element 𝛼 where 𝛼 2 + 1 = 0.
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Ex. Let 𝐹 = ℚ and consider 𝑓 (𝑥 ) = 𝑥 4 − 7𝑥 2 + 10.
In ℚ[𝑥 ], 𝑓 (𝑥 ) = (𝑥 2 − 2)(𝑥 2 − 5), where 𝑥 2 − 2 and 𝑥 2 − 5 are
irreducible over ℚ.
We can construct a field ℚ[𝑥 ]/ < 𝑥 2 − 2 >, which can be thought
of as an extension field of ℚ, which has an element 𝛼 such that
𝛼 2 − 2 = 0 (just let 𝛼 = 𝑥+< 𝑥 2 − 2 >).
We can also construct an extension field of ℚ, ℚ[𝑥 ]/ < 𝑥 2 − 5 >,
which has an element 𝛼 such that 𝛼 2 − 5 = 0.
Def. An element 𝛼 of an extension field 𝐸 of a field 𝐹 is algebraic over 𝐹 if
𝑓(𝛼) = 0 for some 𝑓(𝑥 ) = 𝐹[𝑥]. If 𝛼 is not algebraic over 𝐹, then 𝛼 is
transcendental over 𝐹.
Ex. ℂ is an extension field of ℚ. Since √3 is a zero of 𝑥 2 − 3, √3 is an
algebraic element over ℚ. Since 𝑖 is a zero of 𝑥 2 + 1, 𝑖 is also
algebraic over ℚ.
Ex. Although it’s not that easy to prove, 𝜋 and 𝑒 are transcendental
numbers over ℚ.
Field Extensions
Def. A field 𝐸 is an extension field of a field 𝐹 if 𝐹 is a subfield of 𝐸 (𝐹 ≤ 𝐸).
Ex. ℝ is an extension field of ℚ and ℂ is an extension field of ℝ and ℚ.
Kronecker’s Theorem: Let 𝐹 be a field and let 𝑔(𝑥) be a nonconstant
polynomial in 𝐹 [𝑥 ]. Then there exists an extension field 𝐸 of 𝐹 and an
𝛼 ∈ 𝐸 such that 𝑔(𝛼 ) = 0.
Ex. Let 𝐹 = ℝ and let 𝑔(𝑥 ) = 𝑥 2 + 1. 𝑔(𝑥) has no zeros in ℝ and thus
is irreducible over ℝ. < 𝑥 2 + 1 > is a maximal ideal in ℝ[𝑥 ] so
ℝ[𝑥 ]/< 𝑥 2 + 1 > is a field.
We can view ℝ as a subfield of ℝ[𝑥 ]/< 𝑥 2 + 1 > through the mapping:
𝜑: ℝ → ℝ[𝑥 ]/< 𝑥 2 + 1 > by 𝜑(𝑡) = 𝑡+< 𝑥 2 + 1 >, 𝑡 ∈ ℝ.
Let 𝛼 = 𝑥+< 𝑥 2 + 1 > ∈ ℝ[𝑥 ]/< 𝑥 2 + 1 >,
then 𝛼 2 + 1 = (𝑥+< 𝑥 2 + 1 >)2 + (1+< 𝑥 2 + 1 >)
= ( 𝑥 2 + 1) + < 𝑥 2 + 1 >
= 0.
Thus 𝛼 is a zero of 𝑥 2 + 1. So we can think of ℝ[𝑥 ]/< 𝑥 2 + 1 > as
an extension field of ℝ, which has an element 𝛼 where 𝛼 2 + 1 = 0.
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Ex. Let 𝐹 = ℚ and consider 𝑓 (𝑥 ) = 𝑥 4 − 7𝑥 2 + 10.
In ℚ[𝑥 ], 𝑓 (𝑥 ) = (𝑥 2 − 2)(𝑥 2 − 5), where 𝑥 2 − 2 and 𝑥 2 − 5 are
irreducible over ℚ.
We can construct a field ℚ[𝑥 ]/ < 𝑥 2 − 2 >, which can be thought
of as an extension field of ℚ, which has an element 𝛼 such that
𝛼 2 − 2 = 0 (just let 𝛼 = 𝑥+< 𝑥 2 − 2 >).
We can also construct an extension field of ℚ, ℚ[𝑥 ]/ < 𝑥 2 − 5 >,
which has an element 𝛼 such that 𝛼 2 − 5 = 0.
Def. An element 𝛼 of an extension field 𝐸 of a field 𝐹 is algebraic over 𝐹 if
𝑓(𝛼) = 0 for some 𝑓(𝑥 ) = 𝐹[𝑥]. If 𝛼 is not algebraic over 𝐹, then 𝛼 is
transcendental over 𝐹.
Ex. ℂ is an extension field of ℚ. Since √3 is a zero of 𝑥 2 − 3, √3 is an
algebraic element over ℚ. Since 𝑖 is a zero of 𝑥 2 + 1, 𝑖 is also
algebraic over ℚ.
Ex. Although it’s not that easy to prove, 𝜋 and 𝑒 are transcendental
numbers over ℚ.
