Show that $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)=mathbbQ(sqrt3)$. Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Let $a$ be a complex zero of $x^2+x+1$ over $mathbbQ$. Prove that $mathbbQ(sqrta)=mathbbQ(a)$.Show that $mathbbQ(sqrt-14,sqrt-2sqrt2-1)$ has degree 8 over $mathbbQ$Degree of Field Extension $mathbbQ(sqrt[4]2):mathbbQ(sqrt2)$Prove that there exists no algebraically closed field $F$ such that $barmathbbQ subsetneq F subsetneq overlinemathbbQ(pi)$Alternative way to show $sqrt2+ sqrt3$ is algebraic over $mathbbQ$ of degree 4.Proof that $[overlinemathbb Q:mathbb Rcapoverlinemathbb Q]=2$Show that the algebraic closure of $F$ in $K$ is an algebraic closure of $F$.$mathbb Q(zeta_m)capmathbb Q(zeta_n)=mathbb Q(zeta_d)$Degree of the field of rational numbers extended by a complex numberShow that $mathbbQ(sqrt3,sqrt[4]3, sqrt[8]3,…)$ is algebraic over $mathbbQ$ but not a finite extension.
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Show that $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)=mathbbQ(sqrt3)$.
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Let $a$ be a complex zero of $x^2+x+1$ over $mathbbQ$. Prove that $mathbbQ(sqrta)=mathbbQ(a)$.Show that $mathbbQ(sqrt-14,sqrt-2sqrt2-1)$ has degree 8 over $mathbbQ$Degree of Field Extension $mathbbQ(sqrt[4]2):mathbbQ(sqrt2)$Prove that there exists no algebraically closed field $F$ such that $barmathbbQ subsetneq F subsetneq overlinemathbbQ(pi)$Alternative way to show $sqrt2+ sqrt3$ is algebraic over $mathbbQ$ of degree 4.Proof that $[overlinemathbb Q:mathbb Rcapoverlinemathbb Q]=2$Show that the algebraic closure of $F$ in $K$ is an algebraic closure of $F$.$mathbb Q(zeta_m)capmathbb Q(zeta_n)=mathbb Q(zeta_d)$Degree of the field of rational numbers extended by a complex numberShow that $mathbbQ(sqrt3,sqrt[4]3, sqrt[8]3,…)$ is algebraic over $mathbbQ$ but not a finite extension.
$begingroup$
Show that $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)=mathbbQ(sqrt3)$.
I know that by closure, $sqrt3in mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$, but what about the other containment? I want to use the order of the extensions. I know the degrees of each one are $4$ over the field $mathbbQ$, and I know that $mathbbQ(sqrt1+sqrt3)$ and $ mathbbQ(sqrt1-sqrt3)$ are not equal to each other. But where can I go from here? How do I show that $[mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3]=1$?
abstract-algebra extension-field
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
$endgroup$
add a comment |
$begingroup$
Show that $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)=mathbbQ(sqrt3)$.
I know that by closure, $sqrt3in mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$, but what about the other containment? I want to use the order of the extensions. I know the degrees of each one are $4$ over the field $mathbbQ$, and I know that $mathbbQ(sqrt1+sqrt3)$ and $ mathbbQ(sqrt1-sqrt3)$ are not equal to each other. But where can I go from here? How do I show that $[mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3]=1$?
abstract-algebra extension-field
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
$endgroup$
add a comment |
$begingroup$
Show that $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)=mathbbQ(sqrt3)$.
I know that by closure, $sqrt3in mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$, but what about the other containment? I want to use the order of the extensions. I know the degrees of each one are $4$ over the field $mathbbQ$, and I know that $mathbbQ(sqrt1+sqrt3)$ and $ mathbbQ(sqrt1-sqrt3)$ are not equal to each other. But where can I go from here? How do I show that $[mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3]=1$?
abstract-algebra extension-field
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
$endgroup$
Show that $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)=mathbbQ(sqrt3)$.
I know that by closure, $sqrt3in mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$, but what about the other containment? I want to use the order of the extensions. I know the degrees of each one are $4$ over the field $mathbbQ$, and I know that $mathbbQ(sqrt1+sqrt3)$ and $ mathbbQ(sqrt1-sqrt3)$ are not equal to each other. But where can I go from here? How do I show that $[mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3]=1$?
abstract-algebra extension-field
abstract-algebra extension-field
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
asked Mar 31 at 23:50
numericalorangenumericalorange
1,949314
1,949314
add a comment |
add a comment |
2 Answers
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$begingroup$
$ [mathbbQ(sqrt1+sqrt3):(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)][ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=4$.
You deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]$ divides $4$, it is not $4$ since $mathbbQ(sqrt1+sqrt3)$ is distinct of $ mathbbQ(sqrt1-sqrt3)$, so it is $1$ or $2$, it is not $1$ since $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$ contains $mathbbQ(sqrt3)$, so it is $2$, you deduce that:
$[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)][mathbbQ(sqrt3):mathbbQ]$. Since $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[mathbbQ(sqrt3):mathbbQ]=2$, you deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)]=1$.
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
add a comment |
$begingroup$
Let $alpha:=sqrt1+sqrt3$ and $beta:=sqrt1-sqrt3$. You know that
$$[BbbQ(alpha):BbbQ]=[BbbQ(beta):BbbQ]=4,$$
and that $BbbQ(sqrt3)subsetBbbQ(alpha)capBbbQ(beta)$. Because the degree is multiplicative over towers of field extensions
$$[BbbQ(alpha):BbbQ]=[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]cdot[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]cdot[BbbQ(sqrt3):BbbQ],$$
and because you know the two fields $BbbQ(alpha)$ and $BbbQ(beta)$ are not the same, you have
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]>1,$$
and clearly $[BbbQ(sqrt3):BbbQ]=2$. Because $[BbbQ(alpha):BbbQ]=4$ it follows that
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]=2
qquadtext and qquad
[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]=1,$$
which means that $BbbQ(alpha)capBbbQ(beta)=BbbQ(sqrt3)$.
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$ [mathbbQ(sqrt1+sqrt3):(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)][ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=4$.
You deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]$ divides $4$, it is not $4$ since $mathbbQ(sqrt1+sqrt3)$ is distinct of $ mathbbQ(sqrt1-sqrt3)$, so it is $1$ or $2$, it is not $1$ since $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$ contains $mathbbQ(sqrt3)$, so it is $2$, you deduce that:
$[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)][mathbbQ(sqrt3):mathbbQ]$. Since $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[mathbbQ(sqrt3):mathbbQ]=2$, you deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)]=1$.
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
add a comment |
$begingroup$
$ [mathbbQ(sqrt1+sqrt3):(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)][ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=4$.
You deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]$ divides $4$, it is not $4$ since $mathbbQ(sqrt1+sqrt3)$ is distinct of $ mathbbQ(sqrt1-sqrt3)$, so it is $1$ or $2$, it is not $1$ since $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$ contains $mathbbQ(sqrt3)$, so it is $2$, you deduce that:
$[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)][mathbbQ(sqrt3):mathbbQ]$. Since $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[mathbbQ(sqrt3):mathbbQ]=2$, you deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)]=1$.
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
add a comment |
$begingroup$
$ [mathbbQ(sqrt1+sqrt3):(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)][ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=4$.
You deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]$ divides $4$, it is not $4$ since $mathbbQ(sqrt1+sqrt3)$ is distinct of $ mathbbQ(sqrt1-sqrt3)$, so it is $1$ or $2$, it is not $1$ since $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$ contains $mathbbQ(sqrt3)$, so it is $2$, you deduce that:
$[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)][mathbbQ(sqrt3):mathbbQ]$. Since $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[mathbbQ(sqrt3):mathbbQ]=2$, you deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)]=1$.
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
$endgroup$
$ [mathbbQ(sqrt1+sqrt3):(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)][ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=4$.
You deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]$ divides $4$, it is not $4$ since $mathbbQ(sqrt1+sqrt3)$ is distinct of $ mathbbQ(sqrt1-sqrt3)$, so it is $1$ or $2$, it is not $1$ since $mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3)$ contains $mathbbQ(sqrt3)$, so it is $2$, you deduce that:
$[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)][mathbbQ(sqrt3):mathbbQ]$. Since $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ]=[mathbbQ(sqrt3):mathbbQ]=2$, you deduce that $[ mathbbQ(sqrt1+sqrt3)cap mathbbQ(sqrt1-sqrt3):mathbbQ(sqrt3)]=1$.
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
answered Mar 31 at 23:58
Tsemo AristideTsemo Aristide
60.7k11446
60.7k11446
add a comment |
add a comment |
$begingroup$
Let $alpha:=sqrt1+sqrt3$ and $beta:=sqrt1-sqrt3$. You know that
$$[BbbQ(alpha):BbbQ]=[BbbQ(beta):BbbQ]=4,$$
and that $BbbQ(sqrt3)subsetBbbQ(alpha)capBbbQ(beta)$. Because the degree is multiplicative over towers of field extensions
$$[BbbQ(alpha):BbbQ]=[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]cdot[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]cdot[BbbQ(sqrt3):BbbQ],$$
and because you know the two fields $BbbQ(alpha)$ and $BbbQ(beta)$ are not the same, you have
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]>1,$$
and clearly $[BbbQ(sqrt3):BbbQ]=2$. Because $[BbbQ(alpha):BbbQ]=4$ it follows that
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]=2
qquadtext and qquad
[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]=1,$$
which means that $BbbQ(alpha)capBbbQ(beta)=BbbQ(sqrt3)$.
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
Let $alpha:=sqrt1+sqrt3$ and $beta:=sqrt1-sqrt3$. You know that
$$[BbbQ(alpha):BbbQ]=[BbbQ(beta):BbbQ]=4,$$
and that $BbbQ(sqrt3)subsetBbbQ(alpha)capBbbQ(beta)$. Because the degree is multiplicative over towers of field extensions
$$[BbbQ(alpha):BbbQ]=[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]cdot[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]cdot[BbbQ(sqrt3):BbbQ],$$
and because you know the two fields $BbbQ(alpha)$ and $BbbQ(beta)$ are not the same, you have
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]>1,$$
and clearly $[BbbQ(sqrt3):BbbQ]=2$. Because $[BbbQ(alpha):BbbQ]=4$ it follows that
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]=2
qquadtext and qquad
[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]=1,$$
which means that $BbbQ(alpha)capBbbQ(beta)=BbbQ(sqrt3)$.
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
Let $alpha:=sqrt1+sqrt3$ and $beta:=sqrt1-sqrt3$. You know that
$$[BbbQ(alpha):BbbQ]=[BbbQ(beta):BbbQ]=4,$$
and that $BbbQ(sqrt3)subsetBbbQ(alpha)capBbbQ(beta)$. Because the degree is multiplicative over towers of field extensions
$$[BbbQ(alpha):BbbQ]=[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]cdot[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]cdot[BbbQ(sqrt3):BbbQ],$$
and because you know the two fields $BbbQ(alpha)$ and $BbbQ(beta)$ are not the same, you have
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]>1,$$
and clearly $[BbbQ(sqrt3):BbbQ]=2$. Because $[BbbQ(alpha):BbbQ]=4$ it follows that
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]=2
qquadtext and qquad
[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]=1,$$
which means that $BbbQ(alpha)capBbbQ(beta)=BbbQ(sqrt3)$.
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
$endgroup$
Let $alpha:=sqrt1+sqrt3$ and $beta:=sqrt1-sqrt3$. You know that
$$[BbbQ(alpha):BbbQ]=[BbbQ(beta):BbbQ]=4,$$
and that $BbbQ(sqrt3)subsetBbbQ(alpha)capBbbQ(beta)$. Because the degree is multiplicative over towers of field extensions
$$[BbbQ(alpha):BbbQ]=[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]cdot[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]cdot[BbbQ(sqrt3):BbbQ],$$
and because you know the two fields $BbbQ(alpha)$ and $BbbQ(beta)$ are not the same, you have
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]>1,$$
and clearly $[BbbQ(sqrt3):BbbQ]=2$. Because $[BbbQ(alpha):BbbQ]=4$ it follows that
$$[BbbQ(alpha):BbbQ(alpha)capBbbQ(beta)]=2
qquadtext and qquad
[BbbQ(alpha)capBbbQ(beta):BbbQ(sqrt3)]=1,$$
which means that $BbbQ(alpha)capBbbQ(beta)=BbbQ(sqrt3)$.
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
edited Apr 1 at 0:16
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
answered Apr 1 at 0:10
ServaesServaes
30.6k342101
30.6k342101
add a comment |
add a comment |
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