Interpreting statements in Lang's Undergraduate Algebra
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So, I've been reading this book and I've come across two sentences that I find a little confusing.
On pg. 109:
The polynomial ring $R[t]$ is generated by the variable $t$ over $R$, and $t$ is transcendental over $R$.
Context: $R[t]$ is the polynomial ring, and, for a fixed $x in R$, $R[x] = { f(x) : f in R[t] }$. $x$ is transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
Problem: $t$ doesn't seem to be an element of $R$.
On pg. 117:
Let $F$ be a field and $sigma : F[t] rightarrow F[t]$ is an automorphism of the polynomial ring such that $sigma$ restricts to the identity on $F$.
Problem: This seems to act as if $F subset F[t]$?
Thanks the help.
abstract-algebra
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
$endgroup$
|
show 8 more comments
$begingroup$
So, I've been reading this book and I've come across two sentences that I find a little confusing.
On pg. 109:
The polynomial ring $R[t]$ is generated by the variable $t$ over $R$, and $t$ is transcendental over $R$.
Context: $R[t]$ is the polynomial ring, and, for a fixed $x in R$, $R[x] = { f(x) : f in R[t] }$. $x$ is transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
Problem: $t$ doesn't seem to be an element of $R$.
On pg. 117:
Let $F$ be a field and $sigma : F[t] rightarrow F[t]$ is an automorphism of the polynomial ring such that $sigma$ restricts to the identity on $F$.
Problem: This seems to act as if $F subset F[t]$?
Thanks the help.
abstract-algebra
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
$endgroup$
3
$begingroup$
There's a natural embedding $F hookrightarrow F[t]$ by identifying field elements with constant polynomials.
$endgroup$
– user61527
Jul 1 '14 at 5:13
$begingroup$
That would've been my guess, but just wanted to make sure. Thanks!
$endgroup$
– Roy D.
Jul 1 '14 at 5:14
$begingroup$
Is there something similar for the first question about transcendentals?
$endgroup$
– Roy D.
Jul 1 '14 at 5:15
$begingroup$
I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem?
$endgroup$
– Gerry Myerson
Jul 1 '14 at 5:31
$begingroup$
The property of something being transcendental applies to elements of $R$, right?
$endgroup$
– Roy D.
Jul 1 '14 at 5:32
|
show 8 more comments
$begingroup$
So, I've been reading this book and I've come across two sentences that I find a little confusing.
On pg. 109:
The polynomial ring $R[t]$ is generated by the variable $t$ over $R$, and $t$ is transcendental over $R$.
Context: $R[t]$ is the polynomial ring, and, for a fixed $x in R$, $R[x] = { f(x) : f in R[t] }$. $x$ is transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
Problem: $t$ doesn't seem to be an element of $R$.
On pg. 117:
Let $F$ be a field and $sigma : F[t] rightarrow F[t]$ is an automorphism of the polynomial ring such that $sigma$ restricts to the identity on $F$.
Problem: This seems to act as if $F subset F[t]$?
Thanks the help.
abstract-algebra
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
$endgroup$
So, I've been reading this book and I've come across two sentences that I find a little confusing.
On pg. 109:
The polynomial ring $R[t]$ is generated by the variable $t$ over $R$, and $t$ is transcendental over $R$.
Context: $R[t]$ is the polynomial ring, and, for a fixed $x in R$, $R[x] = { f(x) : f in R[t] }$. $x$ is transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
Problem: $t$ doesn't seem to be an element of $R$.
On pg. 117:
Let $F$ be a field and $sigma : F[t] rightarrow F[t]$ is an automorphism of the polynomial ring such that $sigma$ restricts to the identity on $F$.
Problem: This seems to act as if $F subset F[t]$?
Thanks the help.
abstract-algebra
abstract-algebra
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
asked Jul 1 '14 at 5:12
Roy D.Roy D.
404211
404211
3
$begingroup$
There's a natural embedding $F hookrightarrow F[t]$ by identifying field elements with constant polynomials.
$endgroup$
– user61527
Jul 1 '14 at 5:13
$begingroup$
That would've been my guess, but just wanted to make sure. Thanks!
$endgroup$
– Roy D.
Jul 1 '14 at 5:14
$begingroup$
Is there something similar for the first question about transcendentals?
$endgroup$
– Roy D.
Jul 1 '14 at 5:15
$begingroup$
I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem?
$endgroup$
– Gerry Myerson
Jul 1 '14 at 5:31
$begingroup$
The property of something being transcendental applies to elements of $R$, right?
$endgroup$
– Roy D.
Jul 1 '14 at 5:32
|
show 8 more comments
3
$begingroup$
There's a natural embedding $F hookrightarrow F[t]$ by identifying field elements with constant polynomials.
$endgroup$
– user61527
Jul 1 '14 at 5:13
$begingroup$
That would've been my guess, but just wanted to make sure. Thanks!
$endgroup$
– Roy D.
Jul 1 '14 at 5:14
$begingroup$
Is there something similar for the first question about transcendentals?
$endgroup$
– Roy D.
Jul 1 '14 at 5:15
$begingroup$
I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem?
$endgroup$
– Gerry Myerson
Jul 1 '14 at 5:31
$begingroup$
The property of something being transcendental applies to elements of $R$, right?
$endgroup$
– Roy D.
Jul 1 '14 at 5:32
3
3
$begingroup$
There's a natural embedding $F hookrightarrow F[t]$ by identifying field elements with constant polynomials.
$endgroup$
– user61527
Jul 1 '14 at 5:13
$begingroup$
There's a natural embedding $F hookrightarrow F[t]$ by identifying field elements with constant polynomials.
$endgroup$
– user61527
Jul 1 '14 at 5:13
$begingroup$
That would've been my guess, but just wanted to make sure. Thanks!
$endgroup$
– Roy D.
Jul 1 '14 at 5:14
$begingroup$
That would've been my guess, but just wanted to make sure. Thanks!
$endgroup$
– Roy D.
Jul 1 '14 at 5:14
$begingroup$
Is there something similar for the first question about transcendentals?
$endgroup$
– Roy D.
Jul 1 '14 at 5:15
$begingroup$
Is there something similar for the first question about transcendentals?
$endgroup$
– Roy D.
Jul 1 '14 at 5:15
$begingroup$
I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem?
$endgroup$
– Gerry Myerson
Jul 1 '14 at 5:31
$begingroup$
I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem?
$endgroup$
– Gerry Myerson
Jul 1 '14 at 5:31
$begingroup$
The property of something being transcendental applies to elements of $R$, right?
$endgroup$
– Roy D.
Jul 1 '14 at 5:32
$begingroup$
The property of something being transcendental applies to elements of $R$, right?
$endgroup$
– Roy D.
Jul 1 '14 at 5:32
|
show 8 more comments
1 Answer
1
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$begingroup$
From the comments above.
For the first question:
There is a problem with the definition you have of a transcendental element. You say that
[F]or a fixed $xin R$, $R[x]={f(x):fin R[t]}$ [and] $x$ is transcendental if $fmapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
I take it that $R[x]={f(x):fin R[t]}$ is meant to be a definition of $R[x]$, in which case it is quite restrictive. What should be said is that if $R$ and $S$ are rings with $R subset S$, then for any fixed $x in S$ we define $R[x] = { f(x) : f in R[t] }$. Now, $x in S$ is said to be transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
The earlier definition defined transcendence (over $R$) only on elements of $R$, which is inappropriate because no element of $R$ is transcendental over $R$.
For the second question:
There is a natural inclusion of $F$ inside $F[t]$, given by identifying $a in F$ with the constant polynomial $at^0$. In this way we can view $F$ as a subset of $F[t]$.
$endgroup$
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$begingroup$
From the comments above.
For the first question:
There is a problem with the definition you have of a transcendental element. You say that
[F]or a fixed $xin R$, $R[x]={f(x):fin R[t]}$ [and] $x$ is transcendental if $fmapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
I take it that $R[x]={f(x):fin R[t]}$ is meant to be a definition of $R[x]$, in which case it is quite restrictive. What should be said is that if $R$ and $S$ are rings with $R subset S$, then for any fixed $x in S$ we define $R[x] = { f(x) : f in R[t] }$. Now, $x in S$ is said to be transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
The earlier definition defined transcendence (over $R$) only on elements of $R$, which is inappropriate because no element of $R$ is transcendental over $R$.
For the second question:
There is a natural inclusion of $F$ inside $F[t]$, given by identifying $a in F$ with the constant polynomial $at^0$. In this way we can view $F$ as a subset of $F[t]$.
$endgroup$
add a comment |
$begingroup$
From the comments above.
For the first question:
There is a problem with the definition you have of a transcendental element. You say that
[F]or a fixed $xin R$, $R[x]={f(x):fin R[t]}$ [and] $x$ is transcendental if $fmapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
I take it that $R[x]={f(x):fin R[t]}$ is meant to be a definition of $R[x]$, in which case it is quite restrictive. What should be said is that if $R$ and $S$ are rings with $R subset S$, then for any fixed $x in S$ we define $R[x] = { f(x) : f in R[t] }$. Now, $x in S$ is said to be transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
The earlier definition defined transcendence (over $R$) only on elements of $R$, which is inappropriate because no element of $R$ is transcendental over $R$.
For the second question:
There is a natural inclusion of $F$ inside $F[t]$, given by identifying $a in F$ with the constant polynomial $at^0$. In this way we can view $F$ as a subset of $F[t]$.
$endgroup$
add a comment |
$begingroup$
From the comments above.
For the first question:
There is a problem with the definition you have of a transcendental element. You say that
[F]or a fixed $xin R$, $R[x]={f(x):fin R[t]}$ [and] $x$ is transcendental if $fmapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
I take it that $R[x]={f(x):fin R[t]}$ is meant to be a definition of $R[x]$, in which case it is quite restrictive. What should be said is that if $R$ and $S$ are rings with $R subset S$, then for any fixed $x in S$ we define $R[x] = { f(x) : f in R[t] }$. Now, $x in S$ is said to be transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
The earlier definition defined transcendence (over $R$) only on elements of $R$, which is inappropriate because no element of $R$ is transcendental over $R$.
For the second question:
There is a natural inclusion of $F$ inside $F[t]$, given by identifying $a in F$ with the constant polynomial $at^0$. In this way we can view $F$ as a subset of $F[t]$.
$endgroup$
From the comments above.
For the first question:
There is a problem with the definition you have of a transcendental element. You say that
[F]or a fixed $xin R$, $R[x]={f(x):fin R[t]}$ [and] $x$ is transcendental if $fmapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
I take it that $R[x]={f(x):fin R[t]}$ is meant to be a definition of $R[x]$, in which case it is quite restrictive. What should be said is that if $R$ and $S$ are rings with $R subset S$, then for any fixed $x in S$ we define $R[x] = { f(x) : f in R[t] }$. Now, $x in S$ is said to be transcendental if $f mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.
The earlier definition defined transcendence (over $R$) only on elements of $R$, which is inappropriate because no element of $R$ is transcendental over $R$.
For the second question:
There is a natural inclusion of $F$ inside $F[t]$, given by identifying $a in F$ with the constant polynomial $at^0$. In this way we can view $F$ as a subset of $F[t]$.
answered Dec 12 '18 at 11:06
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Brahadeesh
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$begingroup$
There's a natural embedding $F hookrightarrow F[t]$ by identifying field elements with constant polynomials.
$endgroup$
– user61527
Jul 1 '14 at 5:13
$begingroup$
That would've been my guess, but just wanted to make sure. Thanks!
$endgroup$
– Roy D.
Jul 1 '14 at 5:14
$begingroup$
Is there something similar for the first question about transcendentals?
$endgroup$
– Roy D.
Jul 1 '14 at 5:15
$begingroup$
I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem?
$endgroup$
– Gerry Myerson
Jul 1 '14 at 5:31
$begingroup$
The property of something being transcendental applies to elements of $R$, right?
$endgroup$
– Roy D.
Jul 1 '14 at 5:32