Mixing
Show that $ a，b，c, sqrt{a}+ sqrt{b}+sqrt{c} inmathbb Q implies sqrt{a},sqrt{b},sqrt{c} inmathbb Q $
$begingroup$
Assume that $a，b，c, sqrt{a}+ sqrt{b}+sqrt{c} inmathbb Q$ are rational，prove $sqrt{a},sqrt{b},sqrt{c} inmathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier way
$sqrt a + sqrt b + sqrt c = p in mathbb Q$,
$sqrt a + sqrt b = p- sqrt c$,
$a+b+2sqrt a sqrt b = p^2+c-2psqrt c$,
$2sqrt asqrt b=p^2+c-a-b-2psqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)sqrt c$,
$sqrt c=frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}inmathbb Q$.
number-theory
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
$endgroup$
add a comment |
$begingroup$
Assume that $a，b，c, sqrt{a}+ sqrt{b}+sqrt{c} inmathbb Q$ are rational，prove $sqrt{a},sqrt{b},sqrt{c} inmathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier way
$sqrt a + sqrt b + sqrt c = p in mathbb Q$,
$sqrt a + sqrt b = p- sqrt c$,
$a+b+2sqrt a sqrt b = p^2+c-2psqrt c$,
$2sqrt asqrt b=p^2+c-a-b-2psqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)sqrt c$,
$sqrt c=frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}inmathbb Q$.
number-theory
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
$endgroup$
$begingroup$
Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$.
$endgroup$
– André Nicolas
Apr 25 '12 at 1:29
1
$begingroup$
Exactly; how do you know $p^2+c-a-bne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it.
$endgroup$
– Gerry Myerson
Apr 25 '12 at 1:42
add a comment |
$begingroup$
Assume that $a，b，c, sqrt{a}+ sqrt{b}+sqrt{c} inmathbb Q$ are rational，prove $sqrt{a},sqrt{b},sqrt{c} inmathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier way
$sqrt a + sqrt b + sqrt c = p in mathbb Q$,
$sqrt a + sqrt b = p- sqrt c$,
$a+b+2sqrt a sqrt b = p^2+c-2psqrt c$,
$2sqrt asqrt b=p^2+c-a-b-2psqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)sqrt c$,
$sqrt c=frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}inmathbb Q$.
number-theory
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
$endgroup$
Assume that $a，b，c, sqrt{a}+ sqrt{b}+sqrt{c} inmathbb Q$ are rational，prove $sqrt{a},sqrt{b},sqrt{c} inmathbb Q$,are rational.
I know that can be proved, would like to know that there is no easier way
$sqrt a + sqrt b + sqrt c = p in mathbb Q$,
$sqrt a + sqrt b = p- sqrt c$,
$a+b+2sqrt a sqrt b = p^2+c-2psqrt c$,
$2sqrt asqrt b=p^2+c-a-b-2psqrt c$,
$4ab=(p^2+c-a-b)+4p^2c-4p(p^2+c-a-b)sqrt c$,
$sqrt c=frac{(p^2+c-a-b)+4p^c-4ab}{4p(p^2+c-a-b)}inmathbb Q$.
number-theory
number-theory
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
edited Apr 25 '12 at 19:15
Martin Sleziak
44.7k10118272
44.7k10118272
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
asked Apr 25 '12 at 1:15
tianzhidaosunyouyutianzhidaosunyouyu
86621118
86621118
$begingroup$
Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$.
$endgroup$
– André Nicolas
Apr 25 '12 at 1:29
1
$begingroup$
Exactly; how do you know $p^2+c-a-bne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it.
$endgroup$
– Gerry Myerson
Apr 25 '12 at 1:42
add a comment |
$begingroup$
Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$.
$endgroup$
– André Nicolas
Apr 25 '12 at 1:29
1
$begingroup$
Exactly; how do you know $p^2+c-a-bne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it.
$endgroup$
– Gerry Myerson
Apr 25 '12 at 1:42
$begingroup$
Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$.
$endgroup$
– André Nicolas
Apr 25 '12 at 1:29
$begingroup$
Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$.
$endgroup$
– André Nicolas
Apr 25 '12 at 1:29
1
1
$begingroup$
Exactly; how do you know $p^2+c-a-bne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it.
$endgroup$
– Gerry Myerson
Apr 25 '12 at 1:42
$begingroup$
Exactly; how do you know $p^2+c-a-bne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it.
$endgroup$
– Gerry Myerson
Apr 25 '12 at 1:42
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As you surmised, induction works, employing our prior Lemma (case $rm:n = 2:!).:$ Put $rm:K = mathbb Q:$ in
Theorem $rm sqrt{c_1}+cdots+!sqrt{c_{n}} = kin K Rightarrow sqrt{c_i}in K:$ for all $rm i,:$ if $rm: 0 < c_iin K:$ an ordered field.
Proof $: $ By induction on $rm n.$ Clear if $rm:n=1.$ It is true for $rm:n=2:$ by said Lemma. Suppose that $rm: n>2.$ It suffices to show one of the square-roots is in $rm K,:$ since then the sum of all of the others is in $rm K,:$ so, by induction, all of the others are in $rm K$.
Note that $rm:sqrt{c_1}+cdots+sqrt{c_{n-1}}: =: k! -! sqrt{c_n}in K(sqrt{c_n}):$ so all $,rmsqrt{c_i}in K(sqrt{c_n}):$ by induction.
Therefore $rm sqrt{c_i} =: a_i + b_isqrt{c_n}:$ for some $rm:a_i,:!b_iin K,:$ for $rm:i=1,ldots,n!-!1$.
Some $rm: b_i < 0:$ $Rightarrow$ $rm: a_i = sqrt{c_i}-b_isqrt{c_n} = sqrt{c_i}+!sqrt{b_i^2 c_n}in K:Rightarrow sqrt{c_i}in K:$ by Lemma $rm(n=2).$
Else all $rm b_i ge 0.:$ Let $rm: b = b_1!+cdots+b_{n-1} ge 0,:$ and let $rm: a = a_1!+cdots+a_{n-1}.:$ Then
$$rm sqrt{c_1}+cdots+!sqrt{c_{n}}: =: a+(b!+!1):sqrt{c_n} = kin K:Rightarrow:!sqrt{c_n}= (k!-!a)/(b!+!1)in K$$
Note $rm:bge0:Rightarrow b!+!1ne 0.:$ Hence, in either case, one of the square-roots is in $rm K. $ QED
Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $:sqrt{2} + (-sqrt{2})in mathbb Q:$ but $rm:sqrt{2}notinmathbb Q.:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $,0$.
See also this post on linear independence of square-roots (Besicovic's theorem).
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
add a comment |
$begingroup$
Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=mathbb{Q}(sqrt{a},sqrt{b},sqrt{c})$ and $n = [K: mathbb{Q}]$. Then since $Tr_{K/mathbb{Q}}(sqrt{a}) = Tr_{mathbb{Q}(sqrt{a})/mathbb{Q}} circ Tr_{K/mathbb{Q}(sqrt{a})} (sqrt{a})$, we have
$$ Tr_{K/mathbb{Q}}(sqrt{a}) = begin{cases} 0,& text{if } sqrt{a} notin mathbb{Q} \ nsqrt{a}, &text{if } sqrt{a} in mathbb{Q}, end{cases}$$
and same for $sqrt{b}$ and $sqrt{c}$.
By hypothesis $sqrt{a} + sqrt{b} +sqrt{c} in mathbb{Q}$, so
$$ Tr_{K/mathbb{Q}}(sqrt{a}) + Tr_{K/mathbb{Q}}(sqrt{b}) + Tr_{K/mathbb{Q}}(sqrt{c}) = nsqrt{a} + n sqrt{b} + n sqrt{c}.$$
It is easy to conclude that $sqrt{a},sqrt{b},sqrt{c} in mathbb{Q}$.
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
As you surmised, induction works, employing our prior Lemma (case $rm:n = 2:!).:$ Put $rm:K = mathbb Q:$ in
Theorem $rm sqrt{c_1}+cdots+!sqrt{c_{n}} = kin K Rightarrow sqrt{c_i}in K:$ for all $rm i,:$ if $rm: 0 < c_iin K:$ an ordered field.
Proof $: $ By induction on $rm n.$ Clear if $rm:n=1.$ It is true for $rm:n=2:$ by said Lemma. Suppose that $rm: n>2.$ It suffices to show one of the square-roots is in $rm K,:$ since then the sum of all of the others is in $rm K,:$ so, by induction, all of the others are in $rm K$.
Note that $rm:sqrt{c_1}+cdots+sqrt{c_{n-1}}: =: k! -! sqrt{c_n}in K(sqrt{c_n}):$ so all $,rmsqrt{c_i}in K(sqrt{c_n}):$ by induction.
Therefore $rm sqrt{c_i} =: a_i + b_isqrt{c_n}:$ for some $rm:a_i,:!b_iin K,:$ for $rm:i=1,ldots,n!-!1$.
Some $rm: b_i < 0:$ $Rightarrow$ $rm: a_i = sqrt{c_i}-b_isqrt{c_n} = sqrt{c_i}+!sqrt{b_i^2 c_n}in K:Rightarrow sqrt{c_i}in K:$ by Lemma $rm(n=2).$
Else all $rm b_i ge 0.:$ Let $rm: b = b_1!+cdots+b_{n-1} ge 0,:$ and let $rm: a = a_1!+cdots+a_{n-1}.:$ Then
$$rm sqrt{c_1}+cdots+!sqrt{c_{n}}: =: a+(b!+!1):sqrt{c_n} = kin K:Rightarrow:!sqrt{c_n}= (k!-!a)/(b!+!1)in K$$
Note $rm:bge0:Rightarrow b!+!1ne 0.:$ Hence, in either case, one of the square-roots is in $rm K. $ QED
Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $:sqrt{2} + (-sqrt{2})in mathbb Q:$ but $rm:sqrt{2}notinmathbb Q.:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $,0$.
See also this post on linear independence of square-roots (Besicovic's theorem).
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
add a comment |
$begingroup$
As you surmised, induction works, employing our prior Lemma (case $rm:n = 2:!).:$ Put $rm:K = mathbb Q:$ in
Theorem $rm sqrt{c_1}+cdots+!sqrt{c_{n}} = kin K Rightarrow sqrt{c_i}in K:$ for all $rm i,:$ if $rm: 0 < c_iin K:$ an ordered field.
Proof $: $ By induction on $rm n.$ Clear if $rm:n=1.$ It is true for $rm:n=2:$ by said Lemma. Suppose that $rm: n>2.$ It suffices to show one of the square-roots is in $rm K,:$ since then the sum of all of the others is in $rm K,:$ so, by induction, all of the others are in $rm K$.
Note that $rm:sqrt{c_1}+cdots+sqrt{c_{n-1}}: =: k! -! sqrt{c_n}in K(sqrt{c_n}):$ so all $,rmsqrt{c_i}in K(sqrt{c_n}):$ by induction.
Therefore $rm sqrt{c_i} =: a_i + b_isqrt{c_n}:$ for some $rm:a_i,:!b_iin K,:$ for $rm:i=1,ldots,n!-!1$.
Some $rm: b_i < 0:$ $Rightarrow$ $rm: a_i = sqrt{c_i}-b_isqrt{c_n} = sqrt{c_i}+!sqrt{b_i^2 c_n}in K:Rightarrow sqrt{c_i}in K:$ by Lemma $rm(n=2).$
Else all $rm b_i ge 0.:$ Let $rm: b = b_1!+cdots+b_{n-1} ge 0,:$ and let $rm: a = a_1!+cdots+a_{n-1}.:$ Then
$$rm sqrt{c_1}+cdots+!sqrt{c_{n}}: =: a+(b!+!1):sqrt{c_n} = kin K:Rightarrow:!sqrt{c_n}= (k!-!a)/(b!+!1)in K$$
Note $rm:bge0:Rightarrow b!+!1ne 0.:$ Hence, in either case, one of the square-roots is in $rm K. $ QED
Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $:sqrt{2} + (-sqrt{2})in mathbb Q:$ but $rm:sqrt{2}notinmathbb Q.:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $,0$.
See also this post on linear independence of square-roots (Besicovic's theorem).
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
add a comment |
$begingroup$
As you surmised, induction works, employing our prior Lemma (case $rm:n = 2:!).:$ Put $rm:K = mathbb Q:$ in
Theorem $rm sqrt{c_1}+cdots+!sqrt{c_{n}} = kin K Rightarrow sqrt{c_i}in K:$ for all $rm i,:$ if $rm: 0 < c_iin K:$ an ordered field.
Proof $: $ By induction on $rm n.$ Clear if $rm:n=1.$ It is true for $rm:n=2:$ by said Lemma. Suppose that $rm: n>2.$ It suffices to show one of the square-roots is in $rm K,:$ since then the sum of all of the others is in $rm K,:$ so, by induction, all of the others are in $rm K$.
Note that $rm:sqrt{c_1}+cdots+sqrt{c_{n-1}}: =: k! -! sqrt{c_n}in K(sqrt{c_n}):$ so all $,rmsqrt{c_i}in K(sqrt{c_n}):$ by induction.
Therefore $rm sqrt{c_i} =: a_i + b_isqrt{c_n}:$ for some $rm:a_i,:!b_iin K,:$ for $rm:i=1,ldots,n!-!1$.
Some $rm: b_i < 0:$ $Rightarrow$ $rm: a_i = sqrt{c_i}-b_isqrt{c_n} = sqrt{c_i}+!sqrt{b_i^2 c_n}in K:Rightarrow sqrt{c_i}in K:$ by Lemma $rm(n=2).$
Else all $rm b_i ge 0.:$ Let $rm: b = b_1!+cdots+b_{n-1} ge 0,:$ and let $rm: a = a_1!+cdots+a_{n-1}.:$ Then
$$rm sqrt{c_1}+cdots+!sqrt{c_{n}}: =: a+(b!+!1):sqrt{c_n} = kin K:Rightarrow:!sqrt{c_n}= (k!-!a)/(b!+!1)in K$$
Note $rm:bge0:Rightarrow b!+!1ne 0.:$ Hence, in either case, one of the square-roots is in $rm K. $ QED
Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $:sqrt{2} + (-sqrt{2})in mathbb Q:$ but $rm:sqrt{2}notinmathbb Q.:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $,0$.
See also this post on linear independence of square-roots (Besicovic's theorem).
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
$endgroup$
As you surmised, induction works, employing our prior Lemma (case $rm:n = 2:!).:$ Put $rm:K = mathbb Q:$ in
Theorem $rm sqrt{c_1}+cdots+!sqrt{c_{n}} = kin K Rightarrow sqrt{c_i}in K:$ for all $rm i,:$ if $rm: 0 < c_iin K:$ an ordered field.
Proof $: $ By induction on $rm n.$ Clear if $rm:n=1.$ It is true for $rm:n=2:$ by said Lemma. Suppose that $rm: n>2.$ It suffices to show one of the square-roots is in $rm K,:$ since then the sum of all of the others is in $rm K,:$ so, by induction, all of the others are in $rm K$.
Note that $rm:sqrt{c_1}+cdots+sqrt{c_{n-1}}: =: k! -! sqrt{c_n}in K(sqrt{c_n}):$ so all $,rmsqrt{c_i}in K(sqrt{c_n}):$ by induction.
Therefore $rm sqrt{c_i} =: a_i + b_isqrt{c_n}:$ for some $rm:a_i,:!b_iin K,:$ for $rm:i=1,ldots,n!-!1$.
Some $rm: b_i < 0:$ $Rightarrow$ $rm: a_i = sqrt{c_i}-b_isqrt{c_n} = sqrt{c_i}+!sqrt{b_i^2 c_n}in K:Rightarrow sqrt{c_i}in K:$ by Lemma $rm(n=2).$
Else all $rm b_i ge 0.:$ Let $rm: b = b_1!+cdots+b_{n-1} ge 0,:$ and let $rm: a = a_1!+cdots+a_{n-1}.:$ Then
$$rm sqrt{c_1}+cdots+!sqrt{c_{n}}: =: a+(b!+!1):sqrt{c_n} = kin K:Rightarrow:!sqrt{c_n}= (k!-!a)/(b!+!1)in K$$
Note $rm:bge0:Rightarrow b!+!1ne 0.:$ Hence, in either case, one of the square-roots is in $rm K. $ QED
Remark $ $ Note that the proof depends crucially on the positivity of the square-root summands. Without such the proof fails, e.g. $:sqrt{2} + (-sqrt{2})in mathbb Q:$ but $rm:sqrt{2}notinmathbb Q.:$ It is instructive to examine all of the spots where positivity is used in the proof (above and Lemma), e.g. to avoid dividing by $,0$.
See also this post on linear independence of square-roots (Besicovic's theorem).
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
edited Jan 14 at 22:40
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
answered Apr 25 '12 at 6:00
Bill DubuqueBill Dubuque
211k29192645
211k29192645
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
add a comment |
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
The current solution, I do not quite understand. I would like to see your simple proof of, thank you
$endgroup$
– tianzhidaosunyouyu
Apr 26 '12 at 11:52
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
$begingroup$
@pxc417 I added further details - see above.
$endgroup$
– Bill Dubuque
Apr 26 '12 at 23:31
add a comment |
$begingroup$
Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=mathbb{Q}(sqrt{a},sqrt{b},sqrt{c})$ and $n = [K: mathbb{Q}]$. Then since $Tr_{K/mathbb{Q}}(sqrt{a}) = Tr_{mathbb{Q}(sqrt{a})/mathbb{Q}} circ Tr_{K/mathbb{Q}(sqrt{a})} (sqrt{a})$, we have
$$ Tr_{K/mathbb{Q}}(sqrt{a}) = begin{cases} 0,& text{if } sqrt{a} notin mathbb{Q} \ nsqrt{a}, &text{if } sqrt{a} in mathbb{Q}, end{cases}$$
and same for $sqrt{b}$ and $sqrt{c}$.
By hypothesis $sqrt{a} + sqrt{b} +sqrt{c} in mathbb{Q}$, so
$$ Tr_{K/mathbb{Q}}(sqrt{a}) + Tr_{K/mathbb{Q}}(sqrt{b}) + Tr_{K/mathbb{Q}}(sqrt{c}) = nsqrt{a} + n sqrt{b} + n sqrt{c}.$$
It is easy to conclude that $sqrt{a},sqrt{b},sqrt{c} in mathbb{Q}$.
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
$endgroup$
add a comment |
$begingroup$
Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=mathbb{Q}(sqrt{a},sqrt{b},sqrt{c})$ and $n = [K: mathbb{Q}]$. Then since $Tr_{K/mathbb{Q}}(sqrt{a}) = Tr_{mathbb{Q}(sqrt{a})/mathbb{Q}} circ Tr_{K/mathbb{Q}(sqrt{a})} (sqrt{a})$, we have
$$ Tr_{K/mathbb{Q}}(sqrt{a}) = begin{cases} 0,& text{if } sqrt{a} notin mathbb{Q} \ nsqrt{a}, &text{if } sqrt{a} in mathbb{Q}, end{cases}$$
and same for $sqrt{b}$ and $sqrt{c}$.
By hypothesis $sqrt{a} + sqrt{b} +sqrt{c} in mathbb{Q}$, so
$$ Tr_{K/mathbb{Q}}(sqrt{a}) + Tr_{K/mathbb{Q}}(sqrt{b}) + Tr_{K/mathbb{Q}}(sqrt{c}) = nsqrt{a} + n sqrt{b} + n sqrt{c}.$$
It is easy to conclude that $sqrt{a},sqrt{b},sqrt{c} in mathbb{Q}$.
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
$endgroup$
add a comment |
$begingroup$
Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=mathbb{Q}(sqrt{a},sqrt{b},sqrt{c})$ and $n = [K: mathbb{Q}]$. Then since $Tr_{K/mathbb{Q}}(sqrt{a}) = Tr_{mathbb{Q}(sqrt{a})/mathbb{Q}} circ Tr_{K/mathbb{Q}(sqrt{a})} (sqrt{a})$, we have
$$ Tr_{K/mathbb{Q}}(sqrt{a}) = begin{cases} 0,& text{if } sqrt{a} notin mathbb{Q} \ nsqrt{a}, &text{if } sqrt{a} in mathbb{Q}, end{cases}$$
and same for $sqrt{b}$ and $sqrt{c}$.
By hypothesis $sqrt{a} + sqrt{b} +sqrt{c} in mathbb{Q}$, so
$$ Tr_{K/mathbb{Q}}(sqrt{a}) + Tr_{K/mathbb{Q}}(sqrt{b}) + Tr_{K/mathbb{Q}}(sqrt{c}) = nsqrt{a} + n sqrt{b} + n sqrt{c}.$$
It is easy to conclude that $sqrt{a},sqrt{b},sqrt{c} in mathbb{Q}$.
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
$endgroup$
Maybe not easier, but quite elegant :
Suppose that $a,b,c$ are all non zero. Let $K=mathbb{Q}(sqrt{a},sqrt{b},sqrt{c})$ and $n = [K: mathbb{Q}]$. Then since $Tr_{K/mathbb{Q}}(sqrt{a}) = Tr_{mathbb{Q}(sqrt{a})/mathbb{Q}} circ Tr_{K/mathbb{Q}(sqrt{a})} (sqrt{a})$, we have
$$ Tr_{K/mathbb{Q}}(sqrt{a}) = begin{cases} 0,& text{if } sqrt{a} notin mathbb{Q} \ nsqrt{a}, &text{if } sqrt{a} in mathbb{Q}, end{cases}$$
and same for $sqrt{b}$ and $sqrt{c}$.
By hypothesis $sqrt{a} + sqrt{b} +sqrt{c} in mathbb{Q}$, so
$$ Tr_{K/mathbb{Q}}(sqrt{a}) + Tr_{K/mathbb{Q}}(sqrt{b}) + Tr_{K/mathbb{Q}}(sqrt{c}) = nsqrt{a} + n sqrt{b} + n sqrt{c}.$$
It is easy to conclude that $sqrt{a},sqrt{b},sqrt{c} in mathbb{Q}$.
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
answered Apr 26 '12 at 11:20
user10676user10676
6,29021737
6,29021737
add a comment |
add a comment |
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$begingroup$
Looks good, the only thing that needs to be done is to mqke sure we are not dividing by $0$.
$endgroup$
– André Nicolas
Apr 25 '12 at 1:29
1
$begingroup$
Exactly; how do you know $p^2+c-a-bne0$? Also, the line that starts $4ab=$ should have $(p^2+c-a-b)^2$ in it.
$endgroup$
– Gerry Myerson
Apr 25 '12 at 1:42