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Snis Pilbor
science forum addict
Joined: 11 May 2005
Posts: 50
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 Posted: Sat Jun 17, 2006 12:08 am Post subject:
Looking for feedback for a True/False question from a practice qual
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Hello, this is from a practice qual and since it's a T/F I have very
little way of solo figuring out whether I got them all right or if I
tripped up on something. And also whether the explanations I give to
justify my answers are thorough enough (it says "Briefly justify your
answers")
a) "For each positive integer n, there exists an extension field E of
Q such that [E:Q]=n"
My answer: TRUE, just take Q(sqrt(p1),...,sqrt(p_(n-1))), where p_i is
the ith prime. It's well known that these are independent of
eachother.
b) "Every irreducible polynomial of degree n in Q[x] has n distinct
roots in some extension field"
My answer: FALSE. Certainly there are irreducible polynomials in Q[x]
of degree n>1 which have multiple roots over C. This, together with
fundamental theorem of algebra and uniqueness of splitting fields,
implies the answer is false.
c) "The fields Q(cube root of 2) and Q(w * cube root of 2) are
isomorphic, where w is a primitive cube root of unity."
My answer: TRUE. They are both splitting fields of the same
polynomial.
d) "The fields Q(sqrt(2)) and Q(sqrt(3)) are isomorphic."
My answer: FALSE. Suppose f:Q(sqrt(2))->Q(sqrt(3)) were a field
isomorphism. Then
1=f(1)=f(1/2 sqrt(2)^2)=1/2 f(sqrt(2))^2
Thus
2 = f(sqrt(2))^2, absurd since Q(sqrt(3)) contains no square root of 2.
I very much appreciate any feedback you can give!!!
-SP |
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mskirvin@gmail.com
science forum addict
Joined: 24 Nov 2005
Posts: 66
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| Posted: Sat Jun 17, 2006 2:00 am Post subject:
Re: Looking for feedback for a True/False question from a practice qual
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Snis Pilbor wrote:
| Quote: | Hello, this is from a practice qual and since it's a T/F I have very
little way of solo figuring out whether I got them all right or if I
tripped up on something. And also whether the explanations I give to
justify my answers are thorough enough (it says "Briefly justify your
answers")
a) "For each positive integer n, there exists an extension field E of
Q such that [E:Q]=n"
My answer: TRUE, just take Q(sqrt(p1),...,sqrt(p_(n-1))), where p_i is
the ith prime. It's well known that these are independent of
eachother.
b) "Every irreducible polynomial of degree n in Q[x] has n distinct
roots in some extension field"
My answer: FALSE. Certainly there are irreducible polynomials in Q[x]
of degree n>1 which have multiple roots over C. This, together with
fundamental theorem of algebra and uniqueness of splitting fields,
implies the answer is false.
c) "The fields Q(cube root of 2) and Q(w * cube root of 2) are
isomorphic, where w is a primitive cube root of unity."
My answer: TRUE. They are both splitting fields of the same
polynomial.
d) "The fields Q(sqrt(2)) and Q(sqrt(3)) are isomorphic."
My answer: FALSE. Suppose f:Q(sqrt(2))->Q(sqrt(3)) were a field
isomorphism. Then
1=f(1)=f(1/2 sqrt(2)^2)=1/2 f(sqrt(2))^2
Thus
2 = f(sqrt(2))^2, absurd since Q(sqrt(3)) contains no square root of 2.
I very much appreciate any feedback you can give!!!
-SP
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a) This statement is true, but your proof is wrong. Obviously,
Q(sqrt(p)) is a degree 2 extension, but then you seem to be saying that
each sqrt(p_i) adjoined only increases the degree by one. This is far
from true, as even Q(sqrt(2), sqrt(3)) is a degree 4 extension, with
basis given by {1, sqrt(2), sqrt(3), sqrt(6)}.
To get a field extension of degree n for any n, consider the polynomial
x^n - 2. This is irreducible by Eisenstein's Criterion, so just adjoin
one of its roots to Q to get a degree n extension.
c) I think you are assuming that both these are splittling fields for
x^3 - 2, but that is not true. The splitting field of x^3 - 2 is Q(w,
cbrt(2)) with w a third root of unity, and has degree 6. Try proving
this if it is unclear.
To show they're not isomorphic, you can mimic your proof of d) to show
that if they were isomorphic, Q(w*cbrt(2)) would have to contain
cbrt(2), and hence w as well. But then Q(w*cbrt(2)) = Q(w, cbrt(2)),
contradicting the above paragraph.
b) and d) both look fine to me.
Mike |
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Falagar
science forum beginner
Joined: 11 Jun 2006
Posts: 8
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| Posted: Sat Jun 17, 2006 8:39 am Post subject:
Re: Looking for feedback for a True/False question from a practice qual
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mskirvin@gmail.com wrote:
| Quote: | Snis Pilbor wrote:
Hello, this is from a practice qual and since it's a T/F I have very
little way of solo figuring out whether I got them all right or if I
tripped up on something. And also whether the explanations I give to
justify my answers are thorough enough (it says "Briefly justify your
answers")
a) "For each positive integer n, there exists an extension field E of
Q such that [E:Q]=n"
My answer: TRUE, just take Q(sqrt(p1),...,sqrt(p_(n-1))), where p_i is
the ith prime. It's well known that these are independent of
eachother.
b) "Every irreducible polynomial of degree n in Q[x] has n distinct
roots in some extension field"
My answer: FALSE. Certainly there are irreducible polynomials in Q[x]
of degree n>1 which have multiple roots over C. This, together with
fundamental theorem of algebra and uniqueness of splitting fields,
implies the answer is false.
c) "The fields Q(cube root of 2) and Q(w * cube root of 2) are
isomorphic, where w is a primitive cube root of unity."
My answer: TRUE. They are both splitting fields of the same
polynomial.
d) "The fields Q(sqrt(2)) and Q(sqrt(3)) are isomorphic."
My answer: FALSE. Suppose f:Q(sqrt(2))->Q(sqrt(3)) were a field
isomorphism. Then
1=f(1)=f(1/2 sqrt(2)^2)=1/2 f(sqrt(2))^2
Thus
2 = f(sqrt(2))^2, absurd since Q(sqrt(3)) contains no square root of 2.
I very much appreciate any feedback you can give!!!
-SP
a) This statement is true, but your proof is wrong. Obviously,
Q(sqrt(p)) is a degree 2 extension, but then you seem to be saying that
each sqrt(p_i) adjoined only increases the degree by one. This is far
from true, as even Q(sqrt(2), sqrt(3)) is a degree 4 extension, with
basis given by {1, sqrt(2), sqrt(3), sqrt(6)}.
To get a field extension of degree n for any n, consider the polynomial
x^n - 2. This is irreducible by Eisenstein's Criterion, so just adjoin
one of its roots to Q to get a degree n extension.
c) I think you are assuming that both these are splittling fields for
x^3 - 2, but that is not true. The splitting field of x^3 - 2 is Q(w,
cbrt(2)) with w a third root of unity, and has degree 6. Try proving
this if it is unclear.
To show they're not isomorphic, you can mimic your proof of d) to show
that if they were isomorphic, Q(w*cbrt(2)) would have to contain
cbrt(2), and hence w as well. But then Q(w*cbrt(2)) = Q(w, cbrt(2)),
contradicting the above paragraph.
b) and d) both look fine to me.
Mike
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b) I think you are wrong here.
If f is a irreducible polynomial of degree n in Q[x] which have
multiple roots over C.
Let us consider f' (formal derivative of f).
It is a polynomial over Q[x].
Since f has multiple roots (over C), gcd(f, f') is a polynomial over
Q[x] with degree > 0.
But gcd(f, f') is a divisor of f.
So f is not irreducible polynomial. |
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mskirvin@gmail.com
science forum addict
Joined: 24 Nov 2005
Posts: 66
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| Posted: Sat Jun 17, 2006 8:18 pm Post subject:
Re: Looking for feedback for a True/False question from a practice qual
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Falagar wrote:
| Quote: | mskirvin@gmail.com wrote:
Snis Pilbor wrote:
Hello, this is from a practice qual and since it's a T/F I have very
little way of solo figuring out whether I got them all right or if I
tripped up on something. And also whether the explanations I give to
justify my answers are thorough enough (it says "Briefly justify your
answers")
a) "For each positive integer n, there exists an extension field E of
Q such that [E:Q]=n"
My answer: TRUE, just take Q(sqrt(p1),...,sqrt(p_(n-1))), where p_i is
the ith prime. It's well known that these are independent of
eachother.
b) "Every irreducible polynomial of degree n in Q[x] has n distinct
roots in some extension field"
My answer: FALSE. Certainly there are irreducible polynomials in Q[x]
of degree n>1 which have multiple roots over C. This, together with
fundamental theorem of algebra and uniqueness of splitting fields,
implies the answer is false.
c) "The fields Q(cube root of 2) and Q(w * cube root of 2) are
isomorphic, where w is a primitive cube root of unity."
My answer: TRUE. They are both splitting fields of the same
polynomial.
d) "The fields Q(sqrt(2)) and Q(sqrt(3)) are isomorphic."
My answer: FALSE. Suppose f:Q(sqrt(2))->Q(sqrt(3)) were a field
isomorphism. Then
1=f(1)=f(1/2 sqrt(2)^2)=1/2 f(sqrt(2))^2
Thus
2 = f(sqrt(2))^2, absurd since Q(sqrt(3)) contains no square root of 2.
I very much appreciate any feedback you can give!!!
-SP
a) This statement is true, but your proof is wrong. Obviously,
Q(sqrt(p)) is a degree 2 extension, but then you seem to be saying that
each sqrt(p_i) adjoined only increases the degree by one. This is far
from true, as even Q(sqrt(2), sqrt(3)) is a degree 4 extension, with
basis given by {1, sqrt(2), sqrt(3), sqrt(6)}.
To get a field extension of degree n for any n, consider the polynomial
x^n - 2. This is irreducible by Eisenstein's Criterion, so just adjoin
one of its roots to Q to get a degree n extension.
c) I think you are assuming that both these are splittling fields for
x^3 - 2, but that is not true. The splitting field of x^3 - 2 is Q(w,
cbrt(2)) with w a third root of unity, and has degree 6. Try proving
this if it is unclear.
To show they're not isomorphic, you can mimic your proof of d) to show
that if they were isomorphic, Q(w*cbrt(2)) would have to contain
cbrt(2), and hence w as well. But then Q(w*cbrt(2)) = Q(w, cbrt(2)),
contradicting the above paragraph.
b) and d) both look fine to me.
Mike
b) I think you are wrong here.
If f is a irreducible polynomial of degree n in Q[x] which have
multiple roots over C.
Let us consider f' (formal derivative of f).
It is a polynomial over Q[x].
Since f has multiple roots (over C), gcd(f, f') is a polynomial over
Q[x] with degree > 0.
But gcd(f, f') is a divisor of f.
So f is not irreducible polynomial.
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You're right.
Mike |
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