$mathbb{F}_q(t)$ vs $mathbb{F}_q[t]$?
up vote
1
down vote
favorite
I am having certain troubles understanding certain manuscript. What is the difference between
$$
mathbb{F}_q(t) quad text{ and } quad mathbb{F}_q[t]?
$$
algebraic-geometry field-theory notation finite-fields function-fields
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
asked Nov 26 at 18:26
mindrilla
163
add a comment |
up vote
1
down vote
favorite
I am having certain troubles understanding certain manuscript. What is the difference between
$$
mathbb{F}_q(t) quad text{ and } quad mathbb{F}_q[t]?
$$
algebraic-geometry field-theory notation finite-fields function-fields
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
asked Nov 26 at 18:26
mindrilla
163
2
Typically the difference between $k[t]$ and $k(t) is that the first denotes the polynomial ring in t with coefficients in k, and the latter denotes the field of fraction in t (k is a field).
– user458276
Nov 26 at 18:29
By the field of fraction you mean $A/B$, where $A, B$ are polynomials with coefficients in $mathbb{F}_q$?
– mindrilla
Nov 26 at 18:34
1
Yes - with the condition that $Bneq 0$, and $frac{A}{B}$ is in lowest terms.
– user458276
Nov 26 at 18:36
Thank you a lot!
– mindrilla
Nov 26 at 18:38
1
Also about the notation $mathbb{Q}[i]$ (polynomials in $i$) is also $ {a+ib, (a,b) in mathbb{Q}}$ and $mathbb{Q}(i)$ is the smallest field containing $mathbb{Q}$ and $i$ which is $ {a+ib, (a,b) in mathbb{Q}}$ again.
– reuns
Nov 26 at 18:46
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am having certain troubles understanding certain manuscript. What is the difference between
$$
mathbb{F}_q(t) quad text{ and } quad mathbb{F}_q[t]?
$$
algebraic-geometry field-theory notation finite-fields function-fields
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
asked Nov 26 at 18:26
mindrilla
163
I am having certain troubles understanding certain manuscript. What is the difference between
$$
mathbb{F}_q(t) quad text{ and } quad mathbb{F}_q[t]?
$$
algebraic-geometry field-theory notation finite-fields function-fields
algebraic-geometry field-theory notation finite-fields function-fields
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
asked Nov 26 at 18:26
mindrilla
163
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
asked Nov 26 at 18:26
mindrilla
163
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
edited Nov 26 at 18:38
Robert Lewis
42.8k22863
42.8k22863
asked Nov 26 at 18:26
mindrilla
163
asked Nov 26 at 18:26
mindrilla
163
asked Nov 26 at 18:26
mindrilla
163
163
2
Typically the difference between $k[t]$ and $k(t) is that the first denotes the polynomial ring in t with coefficients in k, and the latter denotes the field of fraction in t (k is a field).
– user458276
Nov 26 at 18:29
By the field of fraction you mean $A/B$, where $A, B$ are polynomials with coefficients in $mathbb{F}_q$?
– mindrilla
Nov 26 at 18:34
1
Yes - with the condition that $Bneq 0$, and $frac{A}{B}$ is in lowest terms.
– user458276
Nov 26 at 18:36
Thank you a lot!
– mindrilla
Nov 26 at 18:38
1
Also about the notation $mathbb{Q}[i]$ (polynomials in $i$) is also $ {a+ib, (a,b) in mathbb{Q}}$ and $mathbb{Q}(i)$ is the smallest field containing $mathbb{Q}$ and $i$ which is $ {a+ib, (a,b) in mathbb{Q}}$ again.
– reuns
Nov 26 at 18:46
add a comment |
2
Typically the difference between $k[t]$ and $k(t) is that the first denotes the polynomial ring in t with coefficients in k, and the latter denotes the field of fraction in t (k is a field).
– user458276
Nov 26 at 18:29
By the field of fraction you mean $A/B$, where $A, B$ are polynomials with coefficients in $mathbb{F}_q$?
– mindrilla
Nov 26 at 18:34
1
Yes - with the condition that $Bneq 0$, and $frac{A}{B}$ is in lowest terms.
– user458276
Nov 26 at 18:36
Thank you a lot!
– mindrilla
Nov 26 at 18:38
1
Also about the notation $mathbb{Q}[i]$ (polynomials in $i$) is also $ {a+ib, (a,b) in mathbb{Q}}$ and $mathbb{Q}(i)$ is the smallest field containing $mathbb{Q}$ and $i$ which is $ {a+ib, (a,b) in mathbb{Q}}$ again.
– reuns
Nov 26 at 18:46
2
2
Typically the difference between $k[t]$ and $k(t) is that the first denotes the polynomial ring in t with coefficients in k, and the latter denotes the field of fraction in t (k is a field).
– user458276
Nov 26 at 18:29
Typically the difference between $k[t]$ and $k(t) is that the first denotes the polynomial ring in t with coefficients in k, and the latter denotes the field of fraction in t (k is a field).
– user458276
Nov 26 at 18:29
By the field of fraction you mean $A/B$, where $A, B$ are polynomials with coefficients in $mathbb{F}_q$?
– mindrilla
Nov 26 at 18:34
By the field of fraction you mean $A/B$, where $A, B$ are polynomials with coefficients in $mathbb{F}_q$?
– mindrilla
Nov 26 at 18:34
1
1
Yes - with the condition that $Bneq 0$, and $frac{A}{B}$ is in lowest terms.
– user458276
Nov 26 at 18:36
Yes - with the condition that $Bneq 0$, and $frac{A}{B}$ is in lowest terms.
– user458276
Nov 26 at 18:36
Thank you a lot!
– mindrilla
Nov 26 at 18:38
Thank you a lot!
– mindrilla
Nov 26 at 18:38
1
1
Also about the notation $mathbb{Q}[i]$ (polynomials in $i$) is also $ {a+ib, (a,b) in mathbb{Q}}$ and $mathbb{Q}(i)$ is the smallest field containing $mathbb{Q}$ and $i$ which is $ {a+ib, (a,b) in mathbb{Q}}$ again.
– reuns
Nov 26 at 18:46
Also about the notation $mathbb{Q}[i]$ (polynomials in $i$) is also $ {a+ib, (a,b) in mathbb{Q}}$ and $mathbb{Q}(i)$ is the smallest field containing $mathbb{Q}$ and $i$ which is $ {a+ib, (a,b) in mathbb{Q}}$ again.
– reuns
Nov 26 at 18:46
add a comment |
1 Answer
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For any field $Bbb K$, $Bbb K[t]$ is the algebra of polynomials in the indeterminate $t$, having coefficients in the field $Bbb K$:
$Bbb K[t] = left { displaystyle sum_0^n a_i t^i, ; a_i in Bbb K, ; 0 le n in Bbb Z right }; tag 1$
we note, as is well-known, that $Bbb K[t]$ is an integral domain; that is, it has no zero-divisors:
$forall a(t), b(t) in Bbb K[t], ; a(t) ne 0 ne b(t) Longrightarrow a(t)b(t) ne 0; tag 2$
this follows easily from the fact that the product of the leading terms of
$a(t) = displaystyle sum_0^{deg a} a_i t^i, ; b(t) = displaystyle sum_0^{deg b} b_i t^i, tag 3$
which is
$a_{deg a} b_{deg b} t^{deg a + deg b} ne 0, tag 4$
since
$a_{deg a} ne 0 ne b_{deg b}, tag 5$
by the definition of "leading term". In fact, it is also well-known that $Bbb K[t]$ is a principal ideal domain, but this need not concern us further here.
Since $Bbb K[t]$ is an integral domain, it is possessed of well-defined field of fractions $mathcal{Frac}(Bbb K[t])$, which prima facie may be taken to be the set of quotients of elements of $Bbb K[t]$:
$mathcal{Frac}(Bbb K[t]) = left { dfrac{a(t)}{b(t)}, ; a(t), 0 ne b(t) in
Bbb K[t] right }; tag 6$
$mathcal{Frac}(Bbb K[t])$ is of course easily recognized as the field of rational functions in the indeterminate $t$, with coefficients in the field $Bbb K$.
In the light of these remarks, we consider the field $Bbb K(t)$, the smallest field, in the sense of set inclusion, containing both $Bbb K$ and $t$, that is,
$Bbb K subset Bbb K(t), ; t in Bbb K(t), tag 7$
and therefore, since $Bbb K(t)$ is closed under the algebraic operations addition and multiplication,
$Bbb K[t] subset Bbb K(t); tag 8$
furthermore, since $Bbb K(t)$ is a field, it must also contain the quotients of elements of $K[t]$, that is,
$mathcal{Frac}(Bbb K[t]) subset K(t) tag 9$
as well. Now since $K(t)$ is by definition the smallest field $K[t]$, it follows that in fact
$mathcal{Frac}(Bbb K[t]) = K(t), tag{10}$
and thus $K(t)$ is in fact the field of rational functions in $t$ over $Bbb K$. We now see that the difference 'twixt $Bbb K[t]$ and $Bbb K(t)$ is concisely expressed by (10).
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
add a comment |
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For any field $Bbb K$, $Bbb K[t]$ is the algebra of polynomials in the indeterminate $t$, having coefficients in the field $Bbb K$:
$Bbb K[t] = left { displaystyle sum_0^n a_i t^i, ; a_i in Bbb K, ; 0 le n in Bbb Z right }; tag 1$
we note, as is well-known, that $Bbb K[t]$ is an integral domain; that is, it has no zero-divisors:
$forall a(t), b(t) in Bbb K[t], ; a(t) ne 0 ne b(t) Longrightarrow a(t)b(t) ne 0; tag 2$
this follows easily from the fact that the product of the leading terms of
$a(t) = displaystyle sum_0^{deg a} a_i t^i, ; b(t) = displaystyle sum_0^{deg b} b_i t^i, tag 3$
which is
$a_{deg a} b_{deg b} t^{deg a + deg b} ne 0, tag 4$
since
$a_{deg a} ne 0 ne b_{deg b}, tag 5$
by the definition of "leading term". In fact, it is also well-known that $Bbb K[t]$ is a principal ideal domain, but this need not concern us further here.
Since $Bbb K[t]$ is an integral domain, it is possessed of well-defined field of fractions $mathcal{Frac}(Bbb K[t])$, which prima facie may be taken to be the set of quotients of elements of $Bbb K[t]$:
$mathcal{Frac}(Bbb K[t]) = left { dfrac{a(t)}{b(t)}, ; a(t), 0 ne b(t) in
Bbb K[t] right }; tag 6$
$mathcal{Frac}(Bbb K[t])$ is of course easily recognized as the field of rational functions in the indeterminate $t$, with coefficients in the field $Bbb K$.
In the light of these remarks, we consider the field $Bbb K(t)$, the smallest field, in the sense of set inclusion, containing both $Bbb K$ and $t$, that is,
$Bbb K subset Bbb K(t), ; t in Bbb K(t), tag 7$
and therefore, since $Bbb K(t)$ is closed under the algebraic operations addition and multiplication,
$Bbb K[t] subset Bbb K(t); tag 8$
furthermore, since $Bbb K(t)$ is a field, it must also contain the quotients of elements of $K[t]$, that is,
$mathcal{Frac}(Bbb K[t]) subset K(t) tag 9$
as well. Now since $K(t)$ is by definition the smallest field $K[t]$, it follows that in fact
$mathcal{Frac}(Bbb K[t]) = K(t), tag{10}$
and thus $K(t)$ is in fact the field of rational functions in $t$ over $Bbb K$. We now see that the difference 'twixt $Bbb K[t]$ and $Bbb K(t)$ is concisely expressed by (10).
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
add a comment |
up vote
2
down vote
For any field $Bbb K$, $Bbb K[t]$ is the algebra of polynomials in the indeterminate $t$, having coefficients in the field $Bbb K$:
$Bbb K[t] = left { displaystyle sum_0^n a_i t^i, ; a_i in Bbb K, ; 0 le n in Bbb Z right }; tag 1$
we note, as is well-known, that $Bbb K[t]$ is an integral domain; that is, it has no zero-divisors:
$forall a(t), b(t) in Bbb K[t], ; a(t) ne 0 ne b(t) Longrightarrow a(t)b(t) ne 0; tag 2$
this follows easily from the fact that the product of the leading terms of
$a(t) = displaystyle sum_0^{deg a} a_i t^i, ; b(t) = displaystyle sum_0^{deg b} b_i t^i, tag 3$
which is
$a_{deg a} b_{deg b} t^{deg a + deg b} ne 0, tag 4$
since
$a_{deg a} ne 0 ne b_{deg b}, tag 5$
by the definition of "leading term". In fact, it is also well-known that $Bbb K[t]$ is a principal ideal domain, but this need not concern us further here.
Since $Bbb K[t]$ is an integral domain, it is possessed of well-defined field of fractions $mathcal{Frac}(Bbb K[t])$, which prima facie may be taken to be the set of quotients of elements of $Bbb K[t]$:
$mathcal{Frac}(Bbb K[t]) = left { dfrac{a(t)}{b(t)}, ; a(t), 0 ne b(t) in
Bbb K[t] right }; tag 6$
$mathcal{Frac}(Bbb K[t])$ is of course easily recognized as the field of rational functions in the indeterminate $t$, with coefficients in the field $Bbb K$.
In the light of these remarks, we consider the field $Bbb K(t)$, the smallest field, in the sense of set inclusion, containing both $Bbb K$ and $t$, that is,
$Bbb K subset Bbb K(t), ; t in Bbb K(t), tag 7$
and therefore, since $Bbb K(t)$ is closed under the algebraic operations addition and multiplication,
$Bbb K[t] subset Bbb K(t); tag 8$
furthermore, since $Bbb K(t)$ is a field, it must also contain the quotients of elements of $K[t]$, that is,
$mathcal{Frac}(Bbb K[t]) subset K(t) tag 9$
as well. Now since $K(t)$ is by definition the smallest field $K[t]$, it follows that in fact
$mathcal{Frac}(Bbb K[t]) = K(t), tag{10}$
and thus $K(t)$ is in fact the field of rational functions in $t$ over $Bbb K$. We now see that the difference 'twixt $Bbb K[t]$ and $Bbb K(t)$ is concisely expressed by (10).
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
add a comment |
up vote
2
down vote
up vote
2
down vote
For any field $Bbb K$, $Bbb K[t]$ is the algebra of polynomials in the indeterminate $t$, having coefficients in the field $Bbb K$:
$Bbb K[t] = left { displaystyle sum_0^n a_i t^i, ; a_i in Bbb K, ; 0 le n in Bbb Z right }; tag 1$
we note, as is well-known, that $Bbb K[t]$ is an integral domain; that is, it has no zero-divisors:
$forall a(t), b(t) in Bbb K[t], ; a(t) ne 0 ne b(t) Longrightarrow a(t)b(t) ne 0; tag 2$
this follows easily from the fact that the product of the leading terms of
$a(t) = displaystyle sum_0^{deg a} a_i t^i, ; b(t) = displaystyle sum_0^{deg b} b_i t^i, tag 3$
which is
$a_{deg a} b_{deg b} t^{deg a + deg b} ne 0, tag 4$
since
$a_{deg a} ne 0 ne b_{deg b}, tag 5$
by the definition of "leading term". In fact, it is also well-known that $Bbb K[t]$ is a principal ideal domain, but this need not concern us further here.
Since $Bbb K[t]$ is an integral domain, it is possessed of well-defined field of fractions $mathcal{Frac}(Bbb K[t])$, which prima facie may be taken to be the set of quotients of elements of $Bbb K[t]$:
$mathcal{Frac}(Bbb K[t]) = left { dfrac{a(t)}{b(t)}, ; a(t), 0 ne b(t) in
Bbb K[t] right }; tag 6$
$mathcal{Frac}(Bbb K[t])$ is of course easily recognized as the field of rational functions in the indeterminate $t$, with coefficients in the field $Bbb K$.
In the light of these remarks, we consider the field $Bbb K(t)$, the smallest field, in the sense of set inclusion, containing both $Bbb K$ and $t$, that is,
$Bbb K subset Bbb K(t), ; t in Bbb K(t), tag 7$
and therefore, since $Bbb K(t)$ is closed under the algebraic operations addition and multiplication,
$Bbb K[t] subset Bbb K(t); tag 8$
furthermore, since $Bbb K(t)$ is a field, it must also contain the quotients of elements of $K[t]$, that is,
$mathcal{Frac}(Bbb K[t]) subset K(t) tag 9$
as well. Now since $K(t)$ is by definition the smallest field $K[t]$, it follows that in fact
$mathcal{Frac}(Bbb K[t]) = K(t), tag{10}$
and thus $K(t)$ is in fact the field of rational functions in $t$ over $Bbb K$. We now see that the difference 'twixt $Bbb K[t]$ and $Bbb K(t)$ is concisely expressed by (10).
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
For any field $Bbb K$, $Bbb K[t]$ is the algebra of polynomials in the indeterminate $t$, having coefficients in the field $Bbb K$:
$Bbb K[t] = left { displaystyle sum_0^n a_i t^i, ; a_i in Bbb K, ; 0 le n in Bbb Z right }; tag 1$
we note, as is well-known, that $Bbb K[t]$ is an integral domain; that is, it has no zero-divisors:
$forall a(t), b(t) in Bbb K[t], ; a(t) ne 0 ne b(t) Longrightarrow a(t)b(t) ne 0; tag 2$
this follows easily from the fact that the product of the leading terms of
$a(t) = displaystyle sum_0^{deg a} a_i t^i, ; b(t) = displaystyle sum_0^{deg b} b_i t^i, tag 3$
which is
$a_{deg a} b_{deg b} t^{deg a + deg b} ne 0, tag 4$
since
$a_{deg a} ne 0 ne b_{deg b}, tag 5$
by the definition of "leading term". In fact, it is also well-known that $Bbb K[t]$ is a principal ideal domain, but this need not concern us further here.
Since $Bbb K[t]$ is an integral domain, it is possessed of well-defined field of fractions $mathcal{Frac}(Bbb K[t])$, which prima facie may be taken to be the set of quotients of elements of $Bbb K[t]$:
$mathcal{Frac}(Bbb K[t]) = left { dfrac{a(t)}{b(t)}, ; a(t), 0 ne b(t) in
Bbb K[t] right }; tag 6$
$mathcal{Frac}(Bbb K[t])$ is of course easily recognized as the field of rational functions in the indeterminate $t$, with coefficients in the field $Bbb K$.
In the light of these remarks, we consider the field $Bbb K(t)$, the smallest field, in the sense of set inclusion, containing both $Bbb K$ and $t$, that is,
$Bbb K subset Bbb K(t), ; t in Bbb K(t), tag 7$
and therefore, since $Bbb K(t)$ is closed under the algebraic operations addition and multiplication,
$Bbb K[t] subset Bbb K(t); tag 8$
furthermore, since $Bbb K(t)$ is a field, it must also contain the quotients of elements of $K[t]$, that is,
$mathcal{Frac}(Bbb K[t]) subset K(t) tag 9$
as well. Now since $K(t)$ is by definition the smallest field $K[t]$, it follows that in fact
$mathcal{Frac}(Bbb K[t]) = K(t), tag{10}$
and thus $K(t)$ is in fact the field of rational functions in $t$ over $Bbb K$. We now see that the difference 'twixt $Bbb K[t]$ and $Bbb K(t)$ is concisely expressed by (10).
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
answered Nov 26 at 20:52
Robert Lewis
42.8k22863
42.8k22863
add a comment |
add a comment |
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2
Typically the difference between $k[t]$ and $k(t) is that the first denotes the polynomial ring in t with coefficients in k, and the latter denotes the field of fraction in t (k is a field).
– user458276
Nov 26 at 18:29
By the field of fraction you mean $A/B$, where $A, B$ are polynomials with coefficients in $mathbb{F}_q$?
– mindrilla
Nov 26 at 18:34
1
Yes - with the condition that $Bneq 0$, and $frac{A}{B}$ is in lowest terms.
– user458276
Nov 26 at 18:36
Thank you a lot!
– mindrilla
Nov 26 at 18:38
1
Also about the notation $mathbb{Q}[i]$ (polynomials in $i$) is also $ {a+ib, (a,b) in mathbb{Q}}$ and $mathbb{Q}(i)$ is the smallest field containing $mathbb{Q}$ and $i$ which is $ {a+ib, (a,b) in mathbb{Q}}$ again.
– reuns
Nov 26 at 18:46