Proof that $[BbbQ(sqrtq_1,dots,sqrtq_r):BbbQ]=2^r$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to show that $sqrt2$ is not in $mathbb Q(sqrt3,sqrt5)$?Prove that this splitting field has degree 4 over $BbbQ$.Finding a basis for $BbbQ(sqrt2+sqrt3)$ over $BbbQ$.Determine the minimal polynomial$N|K$ normal $Rightarrow F in K[X]$ factors into $F=Q_1cdots Q_r$ with $textdeg(Q_1)=dots=textdeg(Q_r)$Degree of the extension $Bbb Q(sqrt2, sqrt3)/Bbb Q(sqrt2)$ is at most 2How to prove that the polynomial $x^3-5$ is irreducible over $Bbb Q(sqrt2)$?Find the minimal polynomial of $isqrt-1+2sqrt3 in Bbb C$ over $Bbb R$Question of finding the degree of extension $ Bbb Q( sqrt[3]2,ζ_3sqrt[ 3] 2) $Degree of extension $Bbb Q(sqrt[4]5,sqrt[6]7)$.Degree of $mathbbQ(sqrt2 + sqrt7)$ and splitting field
"... to apply for a visa" or "... and applied for a visa"?
Create an outline of font
Is above average number of years spent on PhD considered a red flag in future academia or industry positions?
A pet rabbit called Belle
Semisimplicity of the category of coherent sheaves?
Why can't wing-mounted spoilers be used to steepen approaches?
Does Parliament hold absolute power in the UK?
Relations between two reciprocal partial derivatives?
how can a perfect fourth interval be considered either consonant or dissonant?
What information about me do stores get via my credit card?
Typeface like Times New Roman but with "tied" percent sign
What do you call a plan that's an alternative plan in case your initial plan fails?
When did F become S? Why?
Make it rain characters
system call string length limit
In horse breeding, what is the female equivalent of putting a horse out "to stud"?
Road tyres vs "Street" tyres for charity ride on MTB Tandem
First use of “packing” as in carrying a gun
Finding the path in a graph from A to B then back to A with a minimum of shared edges
Didn't get enough time to take a Coding Test - what to do now?
Change bounding box of math glyphs in LuaTeX
Why did all the guest students take carriages to the Yule Ball?
University's motivation for having tenure-track positions
ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?
Proof that $[BbbQ(sqrtq_1,dots,sqrtq_r):BbbQ]=2^r$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)How to show that $sqrt2$ is not in $mathbb Q(sqrt3,sqrt5)$?Prove that this splitting field has degree 4 over $BbbQ$.Finding a basis for $BbbQ(sqrt2+sqrt3)$ over $BbbQ$.Determine the minimal polynomial$N|K$ normal $Rightarrow F in K[X]$ factors into $F=Q_1cdots Q_r$ with $textdeg(Q_1)=dots=textdeg(Q_r)$Degree of the extension $Bbb Q(sqrt2, sqrt3)/Bbb Q(sqrt2)$ is at most 2How to prove that the polynomial $x^3-5$ is irreducible over $Bbb Q(sqrt2)$?Find the minimal polynomial of $isqrt-1+2sqrt3 in Bbb C$ over $Bbb R$Question of finding the degree of extension $ Bbb Q( sqrt[3]2,ζ_3sqrt[ 3] 2) $Degree of extension $Bbb Q(sqrt[4]5,sqrt[6]7)$.Degree of $mathbbQ(sqrt2 + sqrt7)$ and splitting field
$begingroup$
Let $2leq q_1<q_2< dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $BbbQ(sqrtq_1,dots,sqrtq_r)|BbbQ$ has degree $$[BbbQ(sqrtq_1, dots, sqrtq_r):BbbQ]=2^r$$
I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $BbbQ(sqrtq)|BbbQ$ has degree $2$, but this is clear since the minimal polynomial of $sqrtq$ over $Q$ is
$$P(t)=t^2-q$$
which has degree $2$.
Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider
$$BbbQ(sqrtq_1, dots, sqrtq_k+1)=BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1)$$
Then we have the following chain of inclusions
$$BbbQ subset BbbQ(sqrtq_1, dots, sqrtq_k) subset BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1) $$
Thus
$$[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ]=[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)]cdot [BbbQ(sqrtq_1, dots, sqrtq_k):BbbQ]=2cdot 2^k$$
since the minimal polynomial of $sqrtq_k+1$ over $ BbbQ(sqrtq_1, dots, sqrtq_r)$ is
$$P(t)=t^2-q_k+1$$
which again has degree $2$.
That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $sqrtq_k+1$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!
abstract-algebra polynomials field-theory extension-field
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
$endgroup$
add a comment |
$begingroup$
Let $2leq q_1<q_2< dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $BbbQ(sqrtq_1,dots,sqrtq_r)|BbbQ$ has degree $$[BbbQ(sqrtq_1, dots, sqrtq_r):BbbQ]=2^r$$
I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $BbbQ(sqrtq)|BbbQ$ has degree $2$, but this is clear since the minimal polynomial of $sqrtq$ over $Q$ is
$$P(t)=t^2-q$$
which has degree $2$.
Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider
$$BbbQ(sqrtq_1, dots, sqrtq_k+1)=BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1)$$
Then we have the following chain of inclusions
$$BbbQ subset BbbQ(sqrtq_1, dots, sqrtq_k) subset BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1) $$
Thus
$$[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ]=[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)]cdot [BbbQ(sqrtq_1, dots, sqrtq_k):BbbQ]=2cdot 2^k$$
since the minimal polynomial of $sqrtq_k+1$ over $ BbbQ(sqrtq_1, dots, sqrtq_r)$ is
$$P(t)=t^2-q_k+1$$
which again has degree $2$.
That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $sqrtq_k+1$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!
abstract-algebra polynomials field-theory extension-field
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
$endgroup$
$begingroup$
You have used the square-free part by having the squareroots not being in rational numbers.
$endgroup$
– Zelos Malum
Mar 8 '16 at 14:58
$begingroup$
For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'geq 2$, then $mathbbQ(sqrtq)=mathbbQ(sqrtq')$. So this hypothesis is really just a way to keep only the interesting part.
$endgroup$
– Arnaud D.
Mar 8 '16 at 15:04
add a comment |
$begingroup$
Let $2leq q_1<q_2< dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $BbbQ(sqrtq_1,dots,sqrtq_r)|BbbQ$ has degree $$[BbbQ(sqrtq_1, dots, sqrtq_r):BbbQ]=2^r$$
I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $BbbQ(sqrtq)|BbbQ$ has degree $2$, but this is clear since the minimal polynomial of $sqrtq$ over $Q$ is
$$P(t)=t^2-q$$
which has degree $2$.
Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider
$$BbbQ(sqrtq_1, dots, sqrtq_k+1)=BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1)$$
Then we have the following chain of inclusions
$$BbbQ subset BbbQ(sqrtq_1, dots, sqrtq_k) subset BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1) $$
Thus
$$[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ]=[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)]cdot [BbbQ(sqrtq_1, dots, sqrtq_k):BbbQ]=2cdot 2^k$$
since the minimal polynomial of $sqrtq_k+1$ over $ BbbQ(sqrtq_1, dots, sqrtq_r)$ is
$$P(t)=t^2-q_k+1$$
which again has degree $2$.
That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $sqrtq_k+1$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!
abstract-algebra polynomials field-theory extension-field
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
$endgroup$
Let $2leq q_1<q_2< dots<q_r$ be square-free natural numbers such that $mcd(q_i,q_j)=1$. Prove that the field extension $BbbQ(sqrtq_1,dots,sqrtq_r)|BbbQ$ has degree $$[BbbQ(sqrtq_1, dots, sqrtq_r):BbbQ]=2^r$$
I tried to prove the assertion by induction over $r$. For $r=1$, we need to prove that $BbbQ(sqrtq)|BbbQ$ has degree $2$, but this is clear since the minimal polynomial of $sqrtq$ over $Q$ is
$$P(t)=t^2-q$$
which has degree $2$.
Now let's assume the assertion is true for $r=k$ and prove it for $k+1$. Consider
$$BbbQ(sqrtq_1, dots, sqrtq_k+1)=BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1)$$
Then we have the following chain of inclusions
$$BbbQ subset BbbQ(sqrtq_1, dots, sqrtq_k) subset BbbQ(sqrtq_1, dots, sqrtq_k)( sqrtq_k+1) $$
Thus
$$[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ]=[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)]cdot [BbbQ(sqrtq_1, dots, sqrtq_k):BbbQ]=2cdot 2^k$$
since the minimal polynomial of $sqrtq_k+1$ over $ BbbQ(sqrtq_1, dots, sqrtq_r)$ is
$$P(t)=t^2-q_k+1$$
which again has degree $2$.
That should conclude the proof. However, something seems to be wrong, since I didn't use the hypothesis that $mcd(q_i,q_j)=1$ and that each $q_i$ is square-free at all. So I'm afraid that I left something unproved, that requires the use of the hypothesis. I think that might be the part where I stated that the minimal polynomial of $sqrtq_k+1$ over the "small field" was the one I said it was, but I really don't know how to prove it. Any help would be appreciated. Thanks in advance!
abstract-algebra polynomials field-theory extension-field
abstract-algebra polynomials field-theory extension-field
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
edited Mar 31 at 12:33
J. W. Tanner
4,7721420
4,7721420
asked Mar 8 '16 at 14:15
user281593
$begingroup$
You have used the square-free part by having the squareroots not being in rational numbers.
$endgroup$
– Zelos Malum
Mar 8 '16 at 14:58
$begingroup$
For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'geq 2$, then $mathbbQ(sqrtq)=mathbbQ(sqrtq')$. So this hypothesis is really just a way to keep only the interesting part.
$endgroup$
– Arnaud D.
Mar 8 '16 at 15:04
add a comment |
$begingroup$
You have used the square-free part by having the squareroots not being in rational numbers.
$endgroup$
– Zelos Malum
Mar 8 '16 at 14:58
$begingroup$
For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'geq 2$, then $mathbbQ(sqrtq)=mathbbQ(sqrtq')$. So this hypothesis is really just a way to keep only the interesting part.
$endgroup$
– Arnaud D.
Mar 8 '16 at 15:04
$begingroup$
You have used the square-free part by having the squareroots not being in rational numbers.
$endgroup$
– Zelos Malum
Mar 8 '16 at 14:58
$begingroup$
You have used the square-free part by having the squareroots not being in rational numbers.
$endgroup$
– Zelos Malum
Mar 8 '16 at 14:58
$begingroup$
For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'geq 2$, then $mathbbQ(sqrtq)=mathbbQ(sqrtq')$. So this hypothesis is really just a way to keep only the interesting part.
$endgroup$
– Arnaud D.
Mar 8 '16 at 15:04
$begingroup$
For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'geq 2$, then $mathbbQ(sqrtq)=mathbbQ(sqrtq')$. So this hypothesis is really just a way to keep only the interesting part.
$endgroup$
– Arnaud D.
Mar 8 '16 at 15:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I would like to give a detailed proof of why $[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$. Then you can see where you have implicitly used what.
Using the inductive hypothesis, $sqrtq_k,sqrtq_k+1,sqrtq_k+1q_k notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$ (and because $gcd(q_i, q_j) = 1$).
So suppose $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$.
Then there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ so that $$sqrtq_k+1 = x + y cdot sqrtq_k$$ $$text(inductive hyp. provides [BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2) text)$$
(note: $x neq 0$, because otherwise $q_k mid q_k+1$. And also $y neq 0$, because $sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$)
So we get $q_k+1 = x^2 + y^2q_k + 2xysqrtq_k$ and this implies that
$$underbrace2xysqrtq_k_notin L:= BbbQ(sqrtq_1,dots, sqrtq_k-1) = -underbracex^2_in L - underbracey^2q_k_in L + underbraceq_k+1_in BbbN subset L $$
which is a contradiction. So
$$sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k) implies [BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$$
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
$endgroup$
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
add a comment |
$begingroup$
You have used it, albeit unstated. As $q_i$ is square-free, we know that $sqrtq_inotinmathbbQ,$ which is necessary, as otherwise it's redundant. You have also used that $gcd(q_i,q_j)=1$ by having that $q_jnotinmathbbQ(sqrtq_i),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1688441%2fproof-that-bbbq-sqrtq-1-dots-sqrtq-r-bbbq-2r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I would like to give a detailed proof of why $[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$. Then you can see where you have implicitly used what.
Using the inductive hypothesis, $sqrtq_k,sqrtq_k+1,sqrtq_k+1q_k notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$ (and because $gcd(q_i, q_j) = 1$).
So suppose $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$.
Then there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ so that $$sqrtq_k+1 = x + y cdot sqrtq_k$$ $$text(inductive hyp. provides [BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2) text)$$
(note: $x neq 0$, because otherwise $q_k mid q_k+1$. And also $y neq 0$, because $sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$)
So we get $q_k+1 = x^2 + y^2q_k + 2xysqrtq_k$ and this implies that
$$underbrace2xysqrtq_k_notin L:= BbbQ(sqrtq_1,dots, sqrtq_k-1) = -underbracex^2_in L - underbracey^2q_k_in L + underbraceq_k+1_in BbbN subset L $$
which is a contradiction. So
$$sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k) implies [BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$$
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
$endgroup$
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
add a comment |
$begingroup$
I would like to give a detailed proof of why $[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$. Then you can see where you have implicitly used what.
Using the inductive hypothesis, $sqrtq_k,sqrtq_k+1,sqrtq_k+1q_k notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$ (and because $gcd(q_i, q_j) = 1$).
So suppose $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$.
Then there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ so that $$sqrtq_k+1 = x + y cdot sqrtq_k$$ $$text(inductive hyp. provides [BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2) text)$$
(note: $x neq 0$, because otherwise $q_k mid q_k+1$. And also $y neq 0$, because $sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$)
So we get $q_k+1 = x^2 + y^2q_k + 2xysqrtq_k$ and this implies that
$$underbrace2xysqrtq_k_notin L:= BbbQ(sqrtq_1,dots, sqrtq_k-1) = -underbracex^2_in L - underbracey^2q_k_in L + underbraceq_k+1_in BbbN subset L $$
which is a contradiction. So
$$sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k) implies [BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$$
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
$endgroup$
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
add a comment |
$begingroup$
I would like to give a detailed proof of why $[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$. Then you can see where you have implicitly used what.
Using the inductive hypothesis, $sqrtq_k,sqrtq_k+1,sqrtq_k+1q_k notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$ (and because $gcd(q_i, q_j) = 1$).
So suppose $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$.
Then there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ so that $$sqrtq_k+1 = x + y cdot sqrtq_k$$ $$text(inductive hyp. provides [BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2) text)$$
(note: $x neq 0$, because otherwise $q_k mid q_k+1$. And also $y neq 0$, because $sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$)
So we get $q_k+1 = x^2 + y^2q_k + 2xysqrtq_k$ and this implies that
$$underbrace2xysqrtq_k_notin L:= BbbQ(sqrtq_1,dots, sqrtq_k-1) = -underbracex^2_in L - underbracey^2q_k_in L + underbraceq_k+1_in BbbN subset L $$
which is a contradiction. So
$$sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k) implies [BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$$
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
$endgroup$
I would like to give a detailed proof of why $[BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$. Then you can see where you have implicitly used what.
Using the inductive hypothesis, $sqrtq_k,sqrtq_k+1,sqrtq_k+1q_k notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$ (and because $gcd(q_i, q_j) = 1$).
So suppose $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$.
Then there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ so that $$sqrtq_k+1 = x + y cdot sqrtq_k$$ $$text(inductive hyp. provides [BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2) text)$$
(note: $x neq 0$, because otherwise $q_k mid q_k+1$. And also $y neq 0$, because $sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k-1)$)
So we get $q_k+1 = x^2 + y^2q_k + 2xysqrtq_k$ and this implies that
$$underbrace2xysqrtq_k_notin L:= BbbQ(sqrtq_1,dots, sqrtq_k-1) = -underbracex^2_in L - underbracey^2q_k_in L + underbraceq_k+1_in BbbN subset L $$
which is a contradiction. So
$$sqrtq_k+1 notin BbbQ(sqrtq_1,dots, sqrtq_k) implies [BbbQ(sqrtq_1, dots, sqrtq_k+1):BbbQ(sqrtq_1,dots, sqrtq_k)] = 2$$
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
answered Mar 8 '16 at 16:13
johnnycrabjohnnycrab
1,414614
1,414614
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
add a comment |
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
what is the argument for $sqrtq_k+1 in BbbQ(sqrtq_1,dots, sqrtq_k)$. $implies$ there are $x,y in BbbQ(sqrtq_1,dots, sqrtq_k-1)$ such that $sqrtq_k+1 = x + y cdot sqrtq_k$ ? shouldn't it be $x,y in BbbQ(sqrtq_1,dots, sqrtq_k)$ ?
$endgroup$
– reuns
Mar 8 '16 at 23:05
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
@user1952009 This is because the inductive hyp. provides $[BbbQ(sqrtq_1,dots, sqrtq_k):BbbQ(sqrtq_1,dots, sqrtq_k-1)] = 2$. That means, that $L:= BbbQ(sqrtq_1,dots, sqrtq_k)$ is a 2-dimensional $BbbQ(sqrtq_1,dots, sqrtq_k-1)$-vector space. Thus $x, y in L$, because $(1, sqrtq_k)$ is a $L$-basis of $BbbQ(sqrtq_1,dots, sqrtq_k)$
$endgroup$
– johnnycrab
Mar 8 '16 at 23:55
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
Sorry, there is a typo in the second sentence. Of course $L$ is the smaller field :) everything else is good :)
$endgroup$
– johnnycrab
Mar 9 '16 at 7:42
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
$begingroup$
I've spent hours filling in the 'elementary details' of your detailed proof. Although your inductive/algebraic logic might have some 'jumps or rough spots', it is 'dazzling and delicate'. (+1)
$endgroup$
– CopyPasteIt
Apr 1 at 18:06
add a comment |
$begingroup$
You have used it, albeit unstated. As $q_i$ is square-free, we know that $sqrtq_inotinmathbbQ,$ which is necessary, as otherwise it's redundant. You have also used that $gcd(q_i,q_j)=1$ by having that $q_jnotinmathbbQ(sqrtq_i),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
$endgroup$
add a comment |
$begingroup$
You have used it, albeit unstated. As $q_i$ is square-free, we know that $sqrtq_inotinmathbbQ,$ which is necessary, as otherwise it's redundant. You have also used that $gcd(q_i,q_j)=1$ by having that $q_jnotinmathbbQ(sqrtq_i),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
$endgroup$
add a comment |
$begingroup$
You have used it, albeit unstated. As $q_i$ is square-free, we know that $sqrtq_inotinmathbbQ,$ which is necessary, as otherwise it's redundant. You have also used that $gcd(q_i,q_j)=1$ by having that $q_jnotinmathbbQ(sqrtq_i),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
$endgroup$
You have used it, albeit unstated. As $q_i$ is square-free, we know that $sqrtq_inotinmathbbQ,$ which is necessary, as otherwise it's redundant. You have also used that $gcd(q_i,q_j)=1$ by having that $q_jnotinmathbbQ(sqrtq_i),$ as otherwise that might occur. With them not sharing any prime divisors, we know that the previous statement holds true.
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
edited Mar 31 at 12:36
J. W. Tanner
4,7721420
4,7721420
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
answered Mar 8 '16 at 15:01
Zelos MalumZelos Malum
5,5032823
5,5032823
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1688441%2fproof-that-bbbq-sqrtq-1-dots-sqrtq-r-bbbq-2r%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
You have used the square-free part by having the squareroots not being in rational numbers.
$endgroup$
– Zelos Malum
Mar 8 '16 at 14:58
$begingroup$
For the "square-free" part this is not really important, because if $q=a^2q'$ with $a,q,q'geq 2$, then $mathbbQ(sqrtq)=mathbbQ(sqrtq')$. So this hypothesis is really just a way to keep only the interesting part.
$endgroup$
– Arnaud D.
Mar 8 '16 at 15:04