Show that $x^{3}-3$ irreducible over $mathbb{Q}(sqrt{-3})$
Is there a slick way to show that $x^{3}-3$ is irreducible over $F= mathbb{Q}(sqrt{-3})$? What I did seems kind of convoluted (showing directly that there is no root in F).
Thanks
abstract-algebra irreducible-polynomials
|
show 5 more comments
Is there a slick way to show that $x^{3}-3$ is irreducible over $F= mathbb{Q}(sqrt{-3})$? What I did seems kind of convoluted (showing directly that there is no root in F).
Thanks
abstract-algebra irreducible-polynomials
Leaving this as a comment because I'm not 100% sure whether this is the right way... I guess you could start by saying that, if $x^3-3$ were reducible, it would be expressed as $(x+A)(x^2+Bx+C)$, where $A$ is a rational number. Then that would mean that there exists a fraction $frac{p}{q}$ with $p$ and $q$ being coprime such that $frac{p^3}{q^3}-3 = 0$. Then that would mean that $p^3 = 3q^3$, which would mean that $p$ is divisible by 3. Let $p = 3r$, and we get $9r^3 = q^3$, which again means that $q$ is divisible by 3. Contradiction.
– 2012ssohn
May 29 '15 at 2:45
1
@2012ssohn How do you know $A$ is a rational number?
– Gregory Grant
May 29 '15 at 2:47
1
this answer would land you a downvote....since its "cumbersome"....and not giving a good insight into the problem.
– DeepSea
May 29 '15 at 2:47
1
Sorry guys, it appears that I have a clear misunderstanding in this field... what exactly does $mathbb Q(sqrt{-3})$ mean?
– 2012ssohn
May 29 '15 at 2:51
1
@2012ssohn It all things of the form $a+bsqrt{-3}$ where $a,binBbb Q$.
– Gregory Grant
May 29 '15 at 2:53
|
show 5 more comments
Is there a slick way to show that $x^{3}-3$ is irreducible over $F= mathbb{Q}(sqrt{-3})$? What I did seems kind of convoluted (showing directly that there is no root in F).
Thanks
abstract-algebra irreducible-polynomials
Is there a slick way to show that $x^{3}-3$ is irreducible over $F= mathbb{Q}(sqrt{-3})$? What I did seems kind of convoluted (showing directly that there is no root in F).
Thanks
abstract-algebra irreducible-polynomials
abstract-algebra irreducible-polynomials
edited Jun 3 '15 at 16:41
Martin Sleziak
44.6k8115271
44.6k8115271
asked May 29 '15 at 2:40
TuoTuoTuoTuo
1,748516
1,748516
Leaving this as a comment because I'm not 100% sure whether this is the right way... I guess you could start by saying that, if $x^3-3$ were reducible, it would be expressed as $(x+A)(x^2+Bx+C)$, where $A$ is a rational number. Then that would mean that there exists a fraction $frac{p}{q}$ with $p$ and $q$ being coprime such that $frac{p^3}{q^3}-3 = 0$. Then that would mean that $p^3 = 3q^3$, which would mean that $p$ is divisible by 3. Let $p = 3r$, and we get $9r^3 = q^3$, which again means that $q$ is divisible by 3. Contradiction.
– 2012ssohn
May 29 '15 at 2:45
1
@2012ssohn How do you know $A$ is a rational number?
– Gregory Grant
May 29 '15 at 2:47
1
this answer would land you a downvote....since its "cumbersome"....and not giving a good insight into the problem.
– DeepSea
May 29 '15 at 2:47
1
Sorry guys, it appears that I have a clear misunderstanding in this field... what exactly does $mathbb Q(sqrt{-3})$ mean?
– 2012ssohn
May 29 '15 at 2:51
1
@2012ssohn It all things of the form $a+bsqrt{-3}$ where $a,binBbb Q$.
– Gregory Grant
May 29 '15 at 2:53
|
show 5 more comments
Leaving this as a comment because I'm not 100% sure whether this is the right way... I guess you could start by saying that, if $x^3-3$ were reducible, it would be expressed as $(x+A)(x^2+Bx+C)$, where $A$ is a rational number. Then that would mean that there exists a fraction $frac{p}{q}$ with $p$ and $q$ being coprime such that $frac{p^3}{q^3}-3 = 0$. Then that would mean that $p^3 = 3q^3$, which would mean that $p$ is divisible by 3. Let $p = 3r$, and we get $9r^3 = q^3$, which again means that $q$ is divisible by 3. Contradiction.
– 2012ssohn
May 29 '15 at 2:45
1
@2012ssohn How do you know $A$ is a rational number?
– Gregory Grant
May 29 '15 at 2:47
1
this answer would land you a downvote....since its "cumbersome"....and not giving a good insight into the problem.
– DeepSea
May 29 '15 at 2:47
1
Sorry guys, it appears that I have a clear misunderstanding in this field... what exactly does $mathbb Q(sqrt{-3})$ mean?
– 2012ssohn
May 29 '15 at 2:51
1
@2012ssohn It all things of the form $a+bsqrt{-3}$ where $a,binBbb Q$.
– Gregory Grant
May 29 '15 at 2:53
Leaving this as a comment because I'm not 100% sure whether this is the right way... I guess you could start by saying that, if $x^3-3$ were reducible, it would be expressed as $(x+A)(x^2+Bx+C)$, where $A$ is a rational number. Then that would mean that there exists a fraction $frac{p}{q}$ with $p$ and $q$ being coprime such that $frac{p^3}{q^3}-3 = 0$. Then that would mean that $p^3 = 3q^3$, which would mean that $p$ is divisible by 3. Let $p = 3r$, and we get $9r^3 = q^3$, which again means that $q$ is divisible by 3. Contradiction.
– 2012ssohn
May 29 '15 at 2:45
Leaving this as a comment because I'm not 100% sure whether this is the right way... I guess you could start by saying that, if $x^3-3$ were reducible, it would be expressed as $(x+A)(x^2+Bx+C)$, where $A$ is a rational number. Then that would mean that there exists a fraction $frac{p}{q}$ with $p$ and $q$ being coprime such that $frac{p^3}{q^3}-3 = 0$. Then that would mean that $p^3 = 3q^3$, which would mean that $p$ is divisible by 3. Let $p = 3r$, and we get $9r^3 = q^3$, which again means that $q$ is divisible by 3. Contradiction.
– 2012ssohn
May 29 '15 at 2:45
1
1
@2012ssohn How do you know $A$ is a rational number?
– Gregory Grant
May 29 '15 at 2:47
@2012ssohn How do you know $A$ is a rational number?
– Gregory Grant
May 29 '15 at 2:47
1
1
this answer would land you a downvote....since its "cumbersome"....and not giving a good insight into the problem.
– DeepSea
May 29 '15 at 2:47
this answer would land you a downvote....since its "cumbersome"....and not giving a good insight into the problem.
– DeepSea
May 29 '15 at 2:47
1
1
Sorry guys, it appears that I have a clear misunderstanding in this field... what exactly does $mathbb Q(sqrt{-3})$ mean?
– 2012ssohn
May 29 '15 at 2:51
Sorry guys, it appears that I have a clear misunderstanding in this field... what exactly does $mathbb Q(sqrt{-3})$ mean?
– 2012ssohn
May 29 '15 at 2:51
1
1
@2012ssohn It all things of the form $a+bsqrt{-3}$ where $a,binBbb Q$.
– Gregory Grant
May 29 '15 at 2:53
@2012ssohn It all things of the form $a+bsqrt{-3}$ where $a,binBbb Q$.
– Gregory Grant
May 29 '15 at 2:53
|
show 5 more comments
6 Answers
6
active
oldest
votes
There's a number of ways of attacking this problem. Here's but one way of going about this that requires a little less calculation:
Since $deg(f) leq 3$, we know $f(x) = x^3 - 3$ is reducible in $mathbb{Q}(sqrt{-3}) iff$ a root of $f$ is contained in $mathbb{Q}(sqrt{-3})$.
Let $alpha$ be a root of $f$. If $alpha in mathbb{Q}(sqrt{-3})$, then we must have $mathbb{Q}(alpha) subset mathbb{Q}(sqrt{-3})$.
Given that multiplicativity of degrees gives $Big[ mathbb{Q}(sqrt{-3}): mathbb{Q} Big] = Big[ mathbb{Q}(sqrt{-3}): mathbb{Q}(alpha) Big] cdot Big[ mathbb{Q}(alpha): mathbb{Q} Big]$, think about the degrees of these extensions over $mathbb{Q}$ to arrive at a contradiction.
1
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
1
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
2
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
2
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
1
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
add a comment |
Suppose it was reducible. Then it would have a root in $mathbb Q(sqrt{-3})$. But it is irreducible over $Bbb Q$, so that would mean $mathbb Q(sqrt{-3})$ would contain an element of degree three over $Bbb Q$. But it is an extension of degree two so that is impossible.
add a comment |
There are so many ways of proving this!
Consider $f(X)=X^3-3$,over $Bbb Q(sqrt{-3},)$. Now, $f$ is irreducible if and only if $f(sqrt{-3}X)=g(X)=-3sqrt{-3}X^3-3$ is irreducible, if and only if $-g(X)/3=sqrt{-3}X^3+1$ is irreducible, if and only if the reverse of the last, namely $X^3+sqrt{-3}$, is irreducible, and it is so by Eisenstein.
add a comment |
One way may be to using the counting theorem?
The degree of $sqrt{-3}$ over $mathbb{Q}$ is 2.
Since $x^3-3$ is of degree 3, if it was reducible in $mathbb{Q}(sqrt{-3})$, it would have a root $alpha$ in $mathbb{Q}(sqrt{-3})$.
But by Eisenstein we also know that $x^3-3$ is irreducible over $mathbb{Q}$. So assume for contradiction that it was reducible over $mathbb{Q}(sqrt{-3})$, and that $alpha$ existed. Then we would have:
$[mathbb{Q}(sqrt{-3}):mathbb{Q}]=2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)][mathbb{Q}(a):mathbb{Q}]=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] cdot 3$.
But $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]$ must be an integer, so we have our contradiction from $2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]cdot3$.
In order to use this we also must have that $Q(sqrt{-3})$ is a finite extension of $Q(alpha)$. But since $Q(sqrt{-3})$ is a finite extension of $mathbb{Q}$, we have that $1, sqrt{-3}$ is a basis for $Q(sqrt{-3})$ over $mathbb{Q}$. If $sqrt{-3}$ should happen to be in $Q(alpha)$ they must be the same, if not, we then have that $1, sqrt{-3}$ over $Q(alpha)$ span $Q(sqrt{-3})$, but we also must have that $1, sqrt{-3}$ must be linearly independent when having coefficients in $mathbb{Q}(alpha)$, if not $g_1+q_2sqrt{-3}=0$, where not both coefficients in $Q(alpha)$ is zero. But then it is easy to see that $sqrt{-3}$ must be in $mathbb{Q}(alpha)$.
This last fact that $mathbb{Q}(sqrt{-3})$ is a finite extension of $mathbb{Q}(alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] $ is the degree of $(mathbb{Q}(alpha))(sqrt{-3})$ over $mathbb{Q}(alpha)$. And we know that since $sqrt{-3}$ is algebraic over $mathbb{Q}$, it must also be algebraic over $mathbb{Q}(alpha)$.
add a comment |
Hint: The norm squared any root of $X^3 - 3$ is $3^{2/3}$, irrational, while the norm squared of every element in $mathbb{Q}(sqrt{-3})$ is rational, so $X^3 - 3$ has no root in $mathbb{Q}(sqrt{-3})$.
add a comment |
Hint: $f$ in $K[x]$ irreducible of degree $m$, $K subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$.
$bf{Added:}$
Let $alpha$ a root of $f$. We have
$$[L(alpha) colon K]= [L(alpha) colon L]cdot [Lcolon K] le m cdot n$$
But $[L(alpha) colon K]$ is divisible by both $[K(alpha) colon K]=m$ and $[Lcolon K]=n$ and so by $mcdot n$. Therefore, we must have equality, $[L(alpha) colon L] = m$, and thus $f$ is irreducible in $L[x]$.
add a comment |
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6 Answers
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6 Answers
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There's a number of ways of attacking this problem. Here's but one way of going about this that requires a little less calculation:
Since $deg(f) leq 3$, we know $f(x) = x^3 - 3$ is reducible in $mathbb{Q}(sqrt{-3}) iff$ a root of $f$ is contained in $mathbb{Q}(sqrt{-3})$.
Let $alpha$ be a root of $f$. If $alpha in mathbb{Q}(sqrt{-3})$, then we must have $mathbb{Q}(alpha) subset mathbb{Q}(sqrt{-3})$.
Given that multiplicativity of degrees gives $Big[ mathbb{Q}(sqrt{-3}): mathbb{Q} Big] = Big[ mathbb{Q}(sqrt{-3}): mathbb{Q}(alpha) Big] cdot Big[ mathbb{Q}(alpha): mathbb{Q} Big]$, think about the degrees of these extensions over $mathbb{Q}$ to arrive at a contradiction.
1
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
1
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
2
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
2
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
1
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
add a comment |
There's a number of ways of attacking this problem. Here's but one way of going about this that requires a little less calculation:
Since $deg(f) leq 3$, we know $f(x) = x^3 - 3$ is reducible in $mathbb{Q}(sqrt{-3}) iff$ a root of $f$ is contained in $mathbb{Q}(sqrt{-3})$.
Let $alpha$ be a root of $f$. If $alpha in mathbb{Q}(sqrt{-3})$, then we must have $mathbb{Q}(alpha) subset mathbb{Q}(sqrt{-3})$.
Given that multiplicativity of degrees gives $Big[ mathbb{Q}(sqrt{-3}): mathbb{Q} Big] = Big[ mathbb{Q}(sqrt{-3}): mathbb{Q}(alpha) Big] cdot Big[ mathbb{Q}(alpha): mathbb{Q} Big]$, think about the degrees of these extensions over $mathbb{Q}$ to arrive at a contradiction.
1
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
1
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
2
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
2
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
1
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
add a comment |
There's a number of ways of attacking this problem. Here's but one way of going about this that requires a little less calculation:
Since $deg(f) leq 3$, we know $f(x) = x^3 - 3$ is reducible in $mathbb{Q}(sqrt{-3}) iff$ a root of $f$ is contained in $mathbb{Q}(sqrt{-3})$.
Let $alpha$ be a root of $f$. If $alpha in mathbb{Q}(sqrt{-3})$, then we must have $mathbb{Q}(alpha) subset mathbb{Q}(sqrt{-3})$.
Given that multiplicativity of degrees gives $Big[ mathbb{Q}(sqrt{-3}): mathbb{Q} Big] = Big[ mathbb{Q}(sqrt{-3}): mathbb{Q}(alpha) Big] cdot Big[ mathbb{Q}(alpha): mathbb{Q} Big]$, think about the degrees of these extensions over $mathbb{Q}$ to arrive at a contradiction.
There's a number of ways of attacking this problem. Here's but one way of going about this that requires a little less calculation:
Since $deg(f) leq 3$, we know $f(x) = x^3 - 3$ is reducible in $mathbb{Q}(sqrt{-3}) iff$ a root of $f$ is contained in $mathbb{Q}(sqrt{-3})$.
Let $alpha$ be a root of $f$. If $alpha in mathbb{Q}(sqrt{-3})$, then we must have $mathbb{Q}(alpha) subset mathbb{Q}(sqrt{-3})$.
Given that multiplicativity of degrees gives $Big[ mathbb{Q}(sqrt{-3}): mathbb{Q} Big] = Big[ mathbb{Q}(sqrt{-3}): mathbb{Q}(alpha) Big] cdot Big[ mathbb{Q}(alpha): mathbb{Q} Big]$, think about the degrees of these extensions over $mathbb{Q}$ to arrive at a contradiction.
edited Dec 8 '16 at 9:53
answered May 29 '15 at 2:43
Kaj HansenKaj Hansen
27.2k43779
27.2k43779
1
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
1
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
2
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
2
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
1
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
add a comment |
1
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
1
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
2
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
2
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
1
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
1
1
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
...Why was I downvoted?
– Kaj Hansen
May 29 '15 at 2:45
1
1
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
your answer seems to be pretty much the same as mine, so not sure why somebody down-voted it.
– Gregory Grant
May 29 '15 at 2:45
2
2
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
There are some trigger happy downvoters, happens to me too all the time. I think if somebody is going to downvote you they should at least give a reason.
– Gregory Grant
May 29 '15 at 2:46
2
2
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
I usually make a few typos that I catch and edit out over the first few minutes after I post. Perhaps someone viewed one of them as particularly egregious.
– Kaj Hansen
May 29 '15 at 2:47
1
1
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
I gave you an up-vote
– Gregory Grant
May 29 '15 at 2:50
add a comment |
Suppose it was reducible. Then it would have a root in $mathbb Q(sqrt{-3})$. But it is irreducible over $Bbb Q$, so that would mean $mathbb Q(sqrt{-3})$ would contain an element of degree three over $Bbb Q$. But it is an extension of degree two so that is impossible.
add a comment |
Suppose it was reducible. Then it would have a root in $mathbb Q(sqrt{-3})$. But it is irreducible over $Bbb Q$, so that would mean $mathbb Q(sqrt{-3})$ would contain an element of degree three over $Bbb Q$. But it is an extension of degree two so that is impossible.
add a comment |
Suppose it was reducible. Then it would have a root in $mathbb Q(sqrt{-3})$. But it is irreducible over $Bbb Q$, so that would mean $mathbb Q(sqrt{-3})$ would contain an element of degree three over $Bbb Q$. But it is an extension of degree two so that is impossible.
Suppose it was reducible. Then it would have a root in $mathbb Q(sqrt{-3})$. But it is irreducible over $Bbb Q$, so that would mean $mathbb Q(sqrt{-3})$ would contain an element of degree three over $Bbb Q$. But it is an extension of degree two so that is impossible.
answered May 29 '15 at 2:42
Gregory GrantGregory Grant
12.3k42449
12.3k42449
add a comment |
add a comment |
There are so many ways of proving this!
Consider $f(X)=X^3-3$,over $Bbb Q(sqrt{-3},)$. Now, $f$ is irreducible if and only if $f(sqrt{-3}X)=g(X)=-3sqrt{-3}X^3-3$ is irreducible, if and only if $-g(X)/3=sqrt{-3}X^3+1$ is irreducible, if and only if the reverse of the last, namely $X^3+sqrt{-3}$, is irreducible, and it is so by Eisenstein.
add a comment |
There are so many ways of proving this!
Consider $f(X)=X^3-3$,over $Bbb Q(sqrt{-3},)$. Now, $f$ is irreducible if and only if $f(sqrt{-3}X)=g(X)=-3sqrt{-3}X^3-3$ is irreducible, if and only if $-g(X)/3=sqrt{-3}X^3+1$ is irreducible, if and only if the reverse of the last, namely $X^3+sqrt{-3}$, is irreducible, and it is so by Eisenstein.
add a comment |
There are so many ways of proving this!
Consider $f(X)=X^3-3$,over $Bbb Q(sqrt{-3},)$. Now, $f$ is irreducible if and only if $f(sqrt{-3}X)=g(X)=-3sqrt{-3}X^3-3$ is irreducible, if and only if $-g(X)/3=sqrt{-3}X^3+1$ is irreducible, if and only if the reverse of the last, namely $X^3+sqrt{-3}$, is irreducible, and it is so by Eisenstein.
There are so many ways of proving this!
Consider $f(X)=X^3-3$,over $Bbb Q(sqrt{-3},)$. Now, $f$ is irreducible if and only if $f(sqrt{-3}X)=g(X)=-3sqrt{-3}X^3-3$ is irreducible, if and only if $-g(X)/3=sqrt{-3}X^3+1$ is irreducible, if and only if the reverse of the last, namely $X^3+sqrt{-3}$, is irreducible, and it is so by Eisenstein.
answered May 29 '15 at 4:17
LubinLubin
43.8k44585
43.8k44585
add a comment |
add a comment |
One way may be to using the counting theorem?
The degree of $sqrt{-3}$ over $mathbb{Q}$ is 2.
Since $x^3-3$ is of degree 3, if it was reducible in $mathbb{Q}(sqrt{-3})$, it would have a root $alpha$ in $mathbb{Q}(sqrt{-3})$.
But by Eisenstein we also know that $x^3-3$ is irreducible over $mathbb{Q}$. So assume for contradiction that it was reducible over $mathbb{Q}(sqrt{-3})$, and that $alpha$ existed. Then we would have:
$[mathbb{Q}(sqrt{-3}):mathbb{Q}]=2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)][mathbb{Q}(a):mathbb{Q}]=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] cdot 3$.
But $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]$ must be an integer, so we have our contradiction from $2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]cdot3$.
In order to use this we also must have that $Q(sqrt{-3})$ is a finite extension of $Q(alpha)$. But since $Q(sqrt{-3})$ is a finite extension of $mathbb{Q}$, we have that $1, sqrt{-3}$ is a basis for $Q(sqrt{-3})$ over $mathbb{Q}$. If $sqrt{-3}$ should happen to be in $Q(alpha)$ they must be the same, if not, we then have that $1, sqrt{-3}$ over $Q(alpha)$ span $Q(sqrt{-3})$, but we also must have that $1, sqrt{-3}$ must be linearly independent when having coefficients in $mathbb{Q}(alpha)$, if not $g_1+q_2sqrt{-3}=0$, where not both coefficients in $Q(alpha)$ is zero. But then it is easy to see that $sqrt{-3}$ must be in $mathbb{Q}(alpha)$.
This last fact that $mathbb{Q}(sqrt{-3})$ is a finite extension of $mathbb{Q}(alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] $ is the degree of $(mathbb{Q}(alpha))(sqrt{-3})$ over $mathbb{Q}(alpha)$. And we know that since $sqrt{-3}$ is algebraic over $mathbb{Q}$, it must also be algebraic over $mathbb{Q}(alpha)$.
add a comment |
One way may be to using the counting theorem?
The degree of $sqrt{-3}$ over $mathbb{Q}$ is 2.
Since $x^3-3$ is of degree 3, if it was reducible in $mathbb{Q}(sqrt{-3})$, it would have a root $alpha$ in $mathbb{Q}(sqrt{-3})$.
But by Eisenstein we also know that $x^3-3$ is irreducible over $mathbb{Q}$. So assume for contradiction that it was reducible over $mathbb{Q}(sqrt{-3})$, and that $alpha$ existed. Then we would have:
$[mathbb{Q}(sqrt{-3}):mathbb{Q}]=2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)][mathbb{Q}(a):mathbb{Q}]=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] cdot 3$.
But $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]$ must be an integer, so we have our contradiction from $2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]cdot3$.
In order to use this we also must have that $Q(sqrt{-3})$ is a finite extension of $Q(alpha)$. But since $Q(sqrt{-3})$ is a finite extension of $mathbb{Q}$, we have that $1, sqrt{-3}$ is a basis for $Q(sqrt{-3})$ over $mathbb{Q}$. If $sqrt{-3}$ should happen to be in $Q(alpha)$ they must be the same, if not, we then have that $1, sqrt{-3}$ over $Q(alpha)$ span $Q(sqrt{-3})$, but we also must have that $1, sqrt{-3}$ must be linearly independent when having coefficients in $mathbb{Q}(alpha)$, if not $g_1+q_2sqrt{-3}=0$, where not both coefficients in $Q(alpha)$ is zero. But then it is easy to see that $sqrt{-3}$ must be in $mathbb{Q}(alpha)$.
This last fact that $mathbb{Q}(sqrt{-3})$ is a finite extension of $mathbb{Q}(alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] $ is the degree of $(mathbb{Q}(alpha))(sqrt{-3})$ over $mathbb{Q}(alpha)$. And we know that since $sqrt{-3}$ is algebraic over $mathbb{Q}$, it must also be algebraic over $mathbb{Q}(alpha)$.
add a comment |
One way may be to using the counting theorem?
The degree of $sqrt{-3}$ over $mathbb{Q}$ is 2.
Since $x^3-3$ is of degree 3, if it was reducible in $mathbb{Q}(sqrt{-3})$, it would have a root $alpha$ in $mathbb{Q}(sqrt{-3})$.
But by Eisenstein we also know that $x^3-3$ is irreducible over $mathbb{Q}$. So assume for contradiction that it was reducible over $mathbb{Q}(sqrt{-3})$, and that $alpha$ existed. Then we would have:
$[mathbb{Q}(sqrt{-3}):mathbb{Q}]=2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)][mathbb{Q}(a):mathbb{Q}]=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] cdot 3$.
But $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]$ must be an integer, so we have our contradiction from $2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]cdot3$.
In order to use this we also must have that $Q(sqrt{-3})$ is a finite extension of $Q(alpha)$. But since $Q(sqrt{-3})$ is a finite extension of $mathbb{Q}$, we have that $1, sqrt{-3}$ is a basis for $Q(sqrt{-3})$ over $mathbb{Q}$. If $sqrt{-3}$ should happen to be in $Q(alpha)$ they must be the same, if not, we then have that $1, sqrt{-3}$ over $Q(alpha)$ span $Q(sqrt{-3})$, but we also must have that $1, sqrt{-3}$ must be linearly independent when having coefficients in $mathbb{Q}(alpha)$, if not $g_1+q_2sqrt{-3}=0$, where not both coefficients in $Q(alpha)$ is zero. But then it is easy to see that $sqrt{-3}$ must be in $mathbb{Q}(alpha)$.
This last fact that $mathbb{Q}(sqrt{-3})$ is a finite extension of $mathbb{Q}(alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] $ is the degree of $(mathbb{Q}(alpha))(sqrt{-3})$ over $mathbb{Q}(alpha)$. And we know that since $sqrt{-3}$ is algebraic over $mathbb{Q}$, it must also be algebraic over $mathbb{Q}(alpha)$.
One way may be to using the counting theorem?
The degree of $sqrt{-3}$ over $mathbb{Q}$ is 2.
Since $x^3-3$ is of degree 3, if it was reducible in $mathbb{Q}(sqrt{-3})$, it would have a root $alpha$ in $mathbb{Q}(sqrt{-3})$.
But by Eisenstein we also know that $x^3-3$ is irreducible over $mathbb{Q}$. So assume for contradiction that it was reducible over $mathbb{Q}(sqrt{-3})$, and that $alpha$ existed. Then we would have:
$[mathbb{Q}(sqrt{-3}):mathbb{Q}]=2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)][mathbb{Q}(a):mathbb{Q}]=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] cdot 3$.
But $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]$ must be an integer, so we have our contradiction from $2=[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)]cdot3$.
In order to use this we also must have that $Q(sqrt{-3})$ is a finite extension of $Q(alpha)$. But since $Q(sqrt{-3})$ is a finite extension of $mathbb{Q}$, we have that $1, sqrt{-3}$ is a basis for $Q(sqrt{-3})$ over $mathbb{Q}$. If $sqrt{-3}$ should happen to be in $Q(alpha)$ they must be the same, if not, we then have that $1, sqrt{-3}$ over $Q(alpha)$ span $Q(sqrt{-3})$, but we also must have that $1, sqrt{-3}$ must be linearly independent when having coefficients in $mathbb{Q}(alpha)$, if not $g_1+q_2sqrt{-3}=0$, where not both coefficients in $Q(alpha)$ is zero. But then it is easy to see that $sqrt{-3}$ must be in $mathbb{Q}(alpha)$.
This last fact that $mathbb{Q}(sqrt{-3})$ is a finite extension of $mathbb{Q}(alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[mathbb{Q}(sqrt{-3}):mathbb{Q}(alpha)] $ is the degree of $(mathbb{Q}(alpha))(sqrt{-3})$ over $mathbb{Q}(alpha)$. And we know that since $sqrt{-3}$ is algebraic over $mathbb{Q}$, it must also be algebraic over $mathbb{Q}(alpha)$.
edited Nov 27 '18 at 20:27
rae306
5,51831134
5,51831134
answered May 29 '15 at 3:03
user119615user119615
3,86531745
3,86531745
add a comment |
add a comment |
Hint: The norm squared any root of $X^3 - 3$ is $3^{2/3}$, irrational, while the norm squared of every element in $mathbb{Q}(sqrt{-3})$ is rational, so $X^3 - 3$ has no root in $mathbb{Q}(sqrt{-3})$.
add a comment |
Hint: The norm squared any root of $X^3 - 3$ is $3^{2/3}$, irrational, while the norm squared of every element in $mathbb{Q}(sqrt{-3})$ is rational, so $X^3 - 3$ has no root in $mathbb{Q}(sqrt{-3})$.
add a comment |
Hint: The norm squared any root of $X^3 - 3$ is $3^{2/3}$, irrational, while the norm squared of every element in $mathbb{Q}(sqrt{-3})$ is rational, so $X^3 - 3$ has no root in $mathbb{Q}(sqrt{-3})$.
Hint: The norm squared any root of $X^3 - 3$ is $3^{2/3}$, irrational, while the norm squared of every element in $mathbb{Q}(sqrt{-3})$ is rational, so $X^3 - 3$ has no root in $mathbb{Q}(sqrt{-3})$.
answered May 29 '15 at 4:47
Orest BucicovschiOrest Bucicovschi
28.4k31746
28.4k31746
add a comment |
add a comment |
Hint: $f$ in $K[x]$ irreducible of degree $m$, $K subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$.
$bf{Added:}$
Let $alpha$ a root of $f$. We have
$$[L(alpha) colon K]= [L(alpha) colon L]cdot [Lcolon K] le m cdot n$$
But $[L(alpha) colon K]$ is divisible by both $[K(alpha) colon K]=m$ and $[Lcolon K]=n$ and so by $mcdot n$. Therefore, we must have equality, $[L(alpha) colon L] = m$, and thus $f$ is irreducible in $L[x]$.
add a comment |
Hint: $f$ in $K[x]$ irreducible of degree $m$, $K subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$.
$bf{Added:}$
Let $alpha$ a root of $f$. We have
$$[L(alpha) colon K]= [L(alpha) colon L]cdot [Lcolon K] le m cdot n$$
But $[L(alpha) colon K]$ is divisible by both $[K(alpha) colon K]=m$ and $[Lcolon K]=n$ and so by $mcdot n$. Therefore, we must have equality, $[L(alpha) colon L] = m$, and thus $f$ is irreducible in $L[x]$.
add a comment |
Hint: $f$ in $K[x]$ irreducible of degree $m$, $K subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$.
$bf{Added:}$
Let $alpha$ a root of $f$. We have
$$[L(alpha) colon K]= [L(alpha) colon L]cdot [Lcolon K] le m cdot n$$
But $[L(alpha) colon K]$ is divisible by both $[K(alpha) colon K]=m$ and $[Lcolon K]=n$ and so by $mcdot n$. Therefore, we must have equality, $[L(alpha) colon L] = m$, and thus $f$ is irreducible in $L[x]$.
Hint: $f$ in $K[x]$ irreducible of degree $m$, $K subset L$ of degree $n$, $(m,n)=1$, implies $f$ stays irreducible in $L[x]$.
$bf{Added:}$
Let $alpha$ a root of $f$. We have
$$[L(alpha) colon K]= [L(alpha) colon L]cdot [Lcolon K] le m cdot n$$
But $[L(alpha) colon K]$ is divisible by both $[K(alpha) colon K]=m$ and $[Lcolon K]=n$ and so by $mcdot n$. Therefore, we must have equality, $[L(alpha) colon L] = m$, and thus $f$ is irreducible in $L[x]$.
edited May 29 '15 at 8:06
answered May 29 '15 at 4:51
Orest BucicovschiOrest Bucicovschi
28.4k31746
28.4k31746
add a comment |
add a comment |
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Leaving this as a comment because I'm not 100% sure whether this is the right way... I guess you could start by saying that, if $x^3-3$ were reducible, it would be expressed as $(x+A)(x^2+Bx+C)$, where $A$ is a rational number. Then that would mean that there exists a fraction $frac{p}{q}$ with $p$ and $q$ being coprime such that $frac{p^3}{q^3}-3 = 0$. Then that would mean that $p^3 = 3q^3$, which would mean that $p$ is divisible by 3. Let $p = 3r$, and we get $9r^3 = q^3$, which again means that $q$ is divisible by 3. Contradiction.
– 2012ssohn
May 29 '15 at 2:45
1
@2012ssohn How do you know $A$ is a rational number?
– Gregory Grant
May 29 '15 at 2:47
1
this answer would land you a downvote....since its "cumbersome"....and not giving a good insight into the problem.
– DeepSea
May 29 '15 at 2:47
1
Sorry guys, it appears that I have a clear misunderstanding in this field... what exactly does $mathbb Q(sqrt{-3})$ mean?
– 2012ssohn
May 29 '15 at 2:51
1
@2012ssohn It all things of the form $a+bsqrt{-3}$ where $a,binBbb Q$.
– Gregory Grant
May 29 '15 at 2:53