Algabraic field extensions (true false questions)

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In summary, every field has nontrivial extensions. Every extension is simple. Every minimal polynomial is monic. Monic polynomails are always irreducible. x^2+4x+2 is not a solution to Q(pi).
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cap.r
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Homework Statement


Can you please check my answers and help me develop explanation for the ones I can't explain.

a) every field has nontrivial extensions.
b) every field has nontrivial algebraic extensions.
c) every simple extension is algebraic.
d) every extension is simple.
e) all simple algebraic extensions of a given subfield of C are isomorphic.
f) all simple transcendental extensions of a given subfield of C are isomorphic.
g) every minimal polynomial is monic.
h) monic polynomails are always irreducible.
i) every polynomial is a constant multiple of an irreducible polynomial.

2. The attempt at a solution
a) true I am thinking of C and adding something that's not in it. like a cube rotation or something call it J. a in C. a+J is a*90degree rotation of the cube. a*J is number of flips. and J+J is a 180 rotation. J*J is again a flip.
b) false. a flip wouldn't be a solution to anything in C. so it has to be a transcendental extension.
c) false Q(pi) is simple and transcendetal.
d) false. only true for when the field has characteristic 0.
e) false? need explanation
f) true? need explanation
g) true. definition of minimal polynomial was that it's monic.
h) false. x^2+4x+2
i) false. need explanation
 
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  • #2
a) Any field can be the base field for a polynomial ring, so true.

b) An algebraically closed field has no nontrivial algebraic extensions.

c) Like you said.

d) Like you said, except only finite extensions in characteristic zero are necessarily simple.

e) Q is a subfield of C, and [tex]Q(\sqrt{2})[/tex] is not isomorphic to [tex]Q(\sqrt{3})[/tex].

f) Transcendentals don't solve any polynomial, so one is as good as another.

g) Right.

h) Right.

i) It doesn't say nonzero constant, and zero is an irreducible polynomial...but that's pretty cheesy.
 
  • #3
Tinyboss said:
zero is an irreducible polynomial...
Au contraire, I bet I can write zero as a product of two non-units...
 

1. Is every field extension algebraic?

False. Not every field extension is algebraic. Counterexamples include transcendental extensions such as the extension of the rational numbers by π.

2. Do algebraic field extensions always have finite degree?

False. Some algebraic field extensions have infinite degree, such as the algebraic closure of the rational numbers.

3. Can every element in an algebraic field extension be expressed as a polynomial in a single element?

True. This is known as the primitive element theorem and states that every finite separable extension is generated by a single element.

4. Are all algebraic field extensions Galois extensions?

False. Not all algebraic field extensions are Galois extensions. A Galois extension must also be normal and separable.

5. Is the Galois group of an algebraic field extension always a finite group?

True. The Galois group of an algebraic field extension is always a finite group, as it is a subgroup of the symmetric group on the roots of the minimal polynomial.

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