Dimension of the field
$begingroup$
let $F$ be a field with $7^6$ elements and let $K$ be a subfield of $F$ with $49$ elements then the dimension of $F$ as a vector space over $K$ is...I don't know how to proceed anyone please help me!!!
linear-algebra
asked Jan 5 at 0:12
Gopi SGopi S
285
$endgroup$
add a comment |
$begingroup$
let $F$ be a field with $7^6$ elements and let $K$ be a subfield of $F$ with $49$ elements then the dimension of $F$ as a vector space over $K$ is...I don't know how to proceed anyone please help me!!!
linear-algebra
asked Jan 5 at 0:12
Gopi SGopi S
285
$endgroup$
add a comment |
$begingroup$
let $F$ be a field with $7^6$ elements and let $K$ be a subfield of $F$ with $49$ elements then the dimension of $F$ as a vector space over $K$ is...I don't know how to proceed anyone please help me!!!
linear-algebra
asked Jan 5 at 0:12
Gopi SGopi S
285
$endgroup$
let $F$ be a field with $7^6$ elements and let $K$ be a subfield of $F$ with $49$ elements then the dimension of $F$ as a vector space over $K$ is...I don't know how to proceed anyone please help me!!!
linear-algebra
linear-algebra
asked Jan 5 at 0:12
Gopi SGopi S
285
asked Jan 5 at 0:12
Gopi SGopi S
285
asked Jan 5 at 0:12
Gopi SGopi S
285
asked Jan 5 at 0:12
Gopi SGopi S
285
asked Jan 5 at 0:12
Gopi SGopi S
285
285
add a comment |
add a comment |
2 Answers
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$begingroup$
Let's say the dimension of $F$ as a vector space over $K$ is $n$. This means there is a basis of $F$ with a cardinality of $n$ elements. We will call this basis $f_1, f_2, ..., f_n$. By the definition of a basis, every element of $F$ can be expressed in the form of $k_1f_1+k_2f_2+...+k_nf_n$ for some $k_1,k_2,...,k_n in K$.
Now, since $K$ has $49$ elements, there are $49$ possible values of $k_i$ for any $1leq i leq n$. Thus, there are $49$ choices for each $k_i$ and $n$ different $k_i$ variables, meaning there are $49^n$ possible values of $k_1f_1+k_2f_2+...+k_nf_n$. Thus, there are $49^n$ elements in the field $F$.
Now, there are $7^6$ elements in the field $F$. Thus, we have the following equation:
$$49^n=7^6rightarrow (7^2)^n=7^6rightarrow 7^{2n}=7^6rightarrow 2n=6rightarrow n=3$$
Therefore, $F$ is a vector space of dimension $3$ over $K$.
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
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$begingroup$
Very simply, we have this formula, considering the prime subfield of $F$:
$$underbrace{[F:mathbf F_7]}_{=,6}=[F:L]cdotunderbrace{[L:mathbf F_7]}_{=,2}$$
answered Jan 5 at 0:31
BernardBernard
124k741116
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2 Answers
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2 Answers
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$begingroup$
Let's say the dimension of $F$ as a vector space over $K$ is $n$. This means there is a basis of $F$ with a cardinality of $n$ elements. We will call this basis $f_1, f_2, ..., f_n$. By the definition of a basis, every element of $F$ can be expressed in the form of $k_1f_1+k_2f_2+...+k_nf_n$ for some $k_1,k_2,...,k_n in K$.
Now, since $K$ has $49$ elements, there are $49$ possible values of $k_i$ for any $1leq i leq n$. Thus, there are $49$ choices for each $k_i$ and $n$ different $k_i$ variables, meaning there are $49^n$ possible values of $k_1f_1+k_2f_2+...+k_nf_n$. Thus, there are $49^n$ elements in the field $F$.
Now, there are $7^6$ elements in the field $F$. Thus, we have the following equation:
$$49^n=7^6rightarrow (7^2)^n=7^6rightarrow 7^{2n}=7^6rightarrow 2n=6rightarrow n=3$$
Therefore, $F$ is a vector space of dimension $3$ over $K$.
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Let's say the dimension of $F$ as a vector space over $K$ is $n$. This means there is a basis of $F$ with a cardinality of $n$ elements. We will call this basis $f_1, f_2, ..., f_n$. By the definition of a basis, every element of $F$ can be expressed in the form of $k_1f_1+k_2f_2+...+k_nf_n$ for some $k_1,k_2,...,k_n in K$.
Now, since $K$ has $49$ elements, there are $49$ possible values of $k_i$ for any $1leq i leq n$. Thus, there are $49$ choices for each $k_i$ and $n$ different $k_i$ variables, meaning there are $49^n$ possible values of $k_1f_1+k_2f_2+...+k_nf_n$. Thus, there are $49^n$ elements in the field $F$.
Now, there are $7^6$ elements in the field $F$. Thus, we have the following equation:
$$49^n=7^6rightarrow (7^2)^n=7^6rightarrow 7^{2n}=7^6rightarrow 2n=6rightarrow n=3$$
Therefore, $F$ is a vector space of dimension $3$ over $K$.
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
Let's say the dimension of $F$ as a vector space over $K$ is $n$. This means there is a basis of $F$ with a cardinality of $n$ elements. We will call this basis $f_1, f_2, ..., f_n$. By the definition of a basis, every element of $F$ can be expressed in the form of $k_1f_1+k_2f_2+...+k_nf_n$ for some $k_1,k_2,...,k_n in K$.
Now, since $K$ has $49$ elements, there are $49$ possible values of $k_i$ for any $1leq i leq n$. Thus, there are $49$ choices for each $k_i$ and $n$ different $k_i$ variables, meaning there are $49^n$ possible values of $k_1f_1+k_2f_2+...+k_nf_n$. Thus, there are $49^n$ elements in the field $F$.
Now, there are $7^6$ elements in the field $F$. Thus, we have the following equation:
$$49^n=7^6rightarrow (7^2)^n=7^6rightarrow 7^{2n}=7^6rightarrow 2n=6rightarrow n=3$$
Therefore, $F$ is a vector space of dimension $3$ over $K$.
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
Let's say the dimension of $F$ as a vector space over $K$ is $n$. This means there is a basis of $F$ with a cardinality of $n$ elements. We will call this basis $f_1, f_2, ..., f_n$. By the definition of a basis, every element of $F$ can be expressed in the form of $k_1f_1+k_2f_2+...+k_nf_n$ for some $k_1,k_2,...,k_n in K$.
Now, since $K$ has $49$ elements, there are $49$ possible values of $k_i$ for any $1leq i leq n$. Thus, there are $49$ choices for each $k_i$ and $n$ different $k_i$ variables, meaning there are $49^n$ possible values of $k_1f_1+k_2f_2+...+k_nf_n$. Thus, there are $49^n$ elements in the field $F$.
Now, there are $7^6$ elements in the field $F$. Thus, we have the following equation:
$$49^n=7^6rightarrow (7^2)^n=7^6rightarrow 7^{2n}=7^6rightarrow 2n=6rightarrow n=3$$
Therefore, $F$ is a vector space of dimension $3$ over $K$.
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 5 at 0:19
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
$begingroup$
Very simply, we have this formula, considering the prime subfield of $F$:
$$underbrace{[F:mathbf F_7]}_{=,6}=[F:L]cdotunderbrace{[L:mathbf F_7]}_{=,2}$$
answered Jan 5 at 0:31
BernardBernard
124k741116
$endgroup$
add a comment |
$begingroup$
Very simply, we have this formula, considering the prime subfield of $F$:
$$underbrace{[F:mathbf F_7]}_{=,6}=[F:L]cdotunderbrace{[L:mathbf F_7]}_{=,2}$$
answered Jan 5 at 0:31
BernardBernard
124k741116
$endgroup$
add a comment |
$begingroup$
Very simply, we have this formula, considering the prime subfield of $F$:
$$underbrace{[F:mathbf F_7]}_{=,6}=[F:L]cdotunderbrace{[L:mathbf F_7]}_{=,2}$$
answered Jan 5 at 0:31
BernardBernard
124k741116
$endgroup$
Very simply, we have this formula, considering the prime subfield of $F$:
$$underbrace{[F:mathbf F_7]}_{=,6}=[F:L]cdotunderbrace{[L:mathbf F_7]}_{=,2}$$
answered Jan 5 at 0:31
BernardBernard
124k741116
answered Jan 5 at 0:31
BernardBernard
124k741116
answered Jan 5 at 0:31
BernardBernard
124k741116
answered Jan 5 at 0:31
BernardBernard
124k741116
124k741116
add a comment |
add a comment |
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