Why can't the Polynomial Ring be a Field?If $F$ is a field show that $F[x]$ is not a field.Let $F$ be a...
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Why can't the Polynomial Ring be a Field?
If $F$ is a field show that $F[x]$ is not a field.Let $F$ be a field. Could the ring $F[x]$ be a field?examples such that (i)$F[x]$ is not a field, (ii) $F[x]$ is also a field.Disproving that $mathbb R[x]$ is a fieldProve that if $R$ is a commutative ring, then $R[x]$ is never a fieldWhy is $Q[pi]$ not a field?The ideal $I= langle x,y ranglesubset k[x,y]$ is not principalFinding inverse of polynomial in a fieldUnderstanding the kernel of the evaluation mapgcd of $x$ and $2$ in $mathbb Z[x]$Definition of RingCommutative property of ring additionProve that $B$ is a subfield of $F$I understand what a division ring is, but cannot find any examples. Can anyone give me one?Is that Ring a field?Where does the proof for commutative rings break down in the non-commutative ring when showing only two ideals implies the ring is a field?Why can't $0_R$ have an inverse under multiplication?Why is $mathbb{Z}_2,+,dot{}$ a commutative Ring?Center of a ring is a subring that contains identity, but what happens in the case of ring of all Even integers?Arbitrary Ring and Field Contruction
$begingroup$
I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.
The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?
abstract-algebra ring-theory commutative-algebra
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
asked Aug 15 '10 at 13:23
IAEIAE
509179
$endgroup$
|
show 4 more comments
$begingroup$
I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.
The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?
abstract-algebra ring-theory commutative-algebra
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
asked Aug 15 '10 at 13:23
IAEIAE
509179
$endgroup$
10
$begingroup$
Take polynomial $f(x) = x$ and check that it can't have inverse.
$endgroup$
– falagar
Aug 15 '10 at 13:50
39
$begingroup$
The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)
$endgroup$
– d.t.
Aug 15 '10 at 14:01
12
$begingroup$
@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't.
$endgroup$
– J. M. is not a mathematician
Aug 15 '10 at 14:55
3
$begingroup$
You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :)
$endgroup$
– BBischof
Aug 15 '10 at 22:17
2
$begingroup$
@Ｊ.Ｍ., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $mathbb{R}[x]$ this makes sense, but don't get boxed in.
$endgroup$
– vonbrand
Mar 12 '14 at 0:05
|
show 4 more comments
$begingroup$
I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.
The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?
abstract-algebra ring-theory commutative-algebra
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
asked Aug 15 '10 at 13:23
IAEIAE
509179
$endgroup$
I'm currently studying Polynomial Rings, but I can't figure out why they are Rings, not Fields. In the definition of a Field, a Set builds a Commutative Group with Addition and Multiplication. This implies an inverse multiple for every Element in the Set.
The book doesn't elaborate on this, however. I don't understand why a Polynomial Ring couldn't have an inverse multiplicative for every element (at least in the Whole numbers, and it's already given that it has a neutral element). Could somebody please explain why this can't be so?
abstract-algebra ring-theory commutative-algebra
abstract-algebra ring-theory commutative-algebra
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
asked Aug 15 '10 at 13:23
IAEIAE
509179
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
asked Aug 15 '10 at 13:23
IAEIAE
509179
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
edited Jul 22 '17 at 19:33
Bill Dubuque
212k29195653
212k29195653
asked Aug 15 '10 at 13:23
IAEIAE
509179
asked Aug 15 '10 at 13:23
IAEIAE
509179
asked Aug 15 '10 at 13:23
IAEIAE
509179
509179
10
$begingroup$
Take polynomial $f(x) = x$ and check that it can't have inverse.
$endgroup$
– falagar
Aug 15 '10 at 13:50
39
$begingroup$
The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)
$endgroup$
– d.t.
Aug 15 '10 at 14:01
12
$begingroup$
@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't.
$endgroup$
– J. M. is not a mathematician
Aug 15 '10 at 14:55
3
$begingroup$
You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :)
$endgroup$
– BBischof
Aug 15 '10 at 22:17
2
$begingroup$
@Ｊ.Ｍ., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $mathbb{R}[x]$ this makes sense, but don't get boxed in.
$endgroup$
– vonbrand
Mar 12 '14 at 0:05
|
show 4 more comments
10
$begingroup$
Take polynomial $f(x) = x$ and check that it can't have inverse.
$endgroup$
– falagar
Aug 15 '10 at 13:50
39
$begingroup$
The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)
$endgroup$
– d.t.
Aug 15 '10 at 14:01
12
$begingroup$
@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't.
$endgroup$
– J. M. is not a mathematician
Aug 15 '10 at 14:55
3
$begingroup$
You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :)
$endgroup$
– BBischof
Aug 15 '10 at 22:17
2
$begingroup$
@Ｊ.Ｍ., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $mathbb{R}[x]$ this makes sense, but don't get boxed in.
$endgroup$
– vonbrand
Mar 12 '14 at 0:05
10
10
$begingroup$
Take polynomial $f(x) = x$ and check that it can't have inverse.
$endgroup$
– falagar
Aug 15 '10 at 13:50
$begingroup$
Take polynomial $f(x) = x$ and check that it can't have inverse.
$endgroup$
– falagar
Aug 15 '10 at 13:50
39
39
$begingroup$
The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)
$endgroup$
– d.t.
Aug 15 '10 at 14:01
$begingroup$
The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)
$endgroup$
– d.t.
Aug 15 '10 at 14:01
12
12
$begingroup$
@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't.
$endgroup$
– J. M. is not a mathematician
Aug 15 '10 at 14:55
$begingroup$
@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't.
$endgroup$
– J. M. is not a mathematician
Aug 15 '10 at 14:55
3
3
$begingroup$
You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :)
$endgroup$
– BBischof
Aug 15 '10 at 22:17
$begingroup$
You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :)
$endgroup$
– BBischof
Aug 15 '10 at 22:17
2
2
$begingroup$
@Ｊ.Ｍ., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $mathbb{R}[x]$ this makes sense, but don't get boxed in.
$endgroup$
– vonbrand
Mar 12 '14 at 0:05
$begingroup$
@Ｊ.Ｍ., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $mathbb{R}[x]$ this makes sense, but don't get boxed in.
$endgroup$
– vonbrand
Mar 12 '14 at 0:05
|
show 4 more comments
5 Answers
5
active
oldest
votes
$begingroup$
Hint $rmquadrm x , f(x) = 1 ,$ in $,rm R[x] Rightarrow 0 = 1 , $ in $,rm R, , $ by evaluating at $rm x = 0 $
Remark $ $ This has a very instructive universal interpretation: if $rm, x,$ is a unit in $rm, R[x],$ then so too is every $rm, R$-algebra element $rm, r,,$ as follows by evaluating $ rm x f(x) = 1 $ at $rm x = r,.,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $rm R$-algebra. $ $ A natural choice is the nonunit $,rm 0in R,,$ which yields the above proof.
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
$endgroup$
9
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
3
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
add a comment |
$begingroup$
Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-infty$. Non-zero constants have degree $0$. You then have the degree equation: $deg (fg) = deg (f) + deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
$endgroup$
4
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
add a comment |
$begingroup$
For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x in F[x]$, and clearly $x ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
$endgroup$
add a comment |
$begingroup$
The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
$endgroup$
3
$begingroup$
While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
$endgroup$
– Robin Chapman
Aug 15 '10 at 15:55
$begingroup$
@Robin: edited, thanks.
$endgroup$
– lhf
Aug 15 '10 at 20:28
add a comment |
$begingroup$
Consider $mathbb{C}[x]$ the ring of polynomials with coefficients from $mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint $rmquadrm x , f(x) = 1 ,$ in $,rm R[x] Rightarrow 0 = 1 , $ in $,rm R, , $ by evaluating at $rm x = 0 $
Remark $ $ This has a very instructive universal interpretation: if $rm, x,$ is a unit in $rm, R[x],$ then so too is every $rm, R$-algebra element $rm, r,,$ as follows by evaluating $ rm x f(x) = 1 $ at $rm x = r,.,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $rm R$-algebra. $ $ A natural choice is the nonunit $,rm 0in R,,$ which yields the above proof.
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
$endgroup$
9
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
3
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
add a comment |
$begingroup$
Hint $rmquadrm x , f(x) = 1 ,$ in $,rm R[x] Rightarrow 0 = 1 , $ in $,rm R, , $ by evaluating at $rm x = 0 $
Remark $ $ This has a very instructive universal interpretation: if $rm, x,$ is a unit in $rm, R[x],$ then so too is every $rm, R$-algebra element $rm, r,,$ as follows by evaluating $ rm x f(x) = 1 $ at $rm x = r,.,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $rm R$-algebra. $ $ A natural choice is the nonunit $,rm 0in R,,$ which yields the above proof.
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
$endgroup$
9
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
3
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
add a comment |
$begingroup$
Hint $rmquadrm x , f(x) = 1 ,$ in $,rm R[x] Rightarrow 0 = 1 , $ in $,rm R, , $ by evaluating at $rm x = 0 $
Remark $ $ This has a very instructive universal interpretation: if $rm, x,$ is a unit in $rm, R[x],$ then so too is every $rm, R$-algebra element $rm, r,,$ as follows by evaluating $ rm x f(x) = 1 $ at $rm x = r,.,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $rm R$-algebra. $ $ A natural choice is the nonunit $,rm 0in R,,$ which yields the above proof.
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
$endgroup$
Hint $rmquadrm x , f(x) = 1 ,$ in $,rm R[x] Rightarrow 0 = 1 , $ in $,rm R, , $ by evaluating at $rm x = 0 $
Remark $ $ This has a very instructive universal interpretation: if $rm, x,$ is a unit in $rm, R[x],$ then so too is every $rm, R$-algebra element $rm, r,,$ as follows by evaluating $ rm x f(x) = 1 $ at $rm x = r,.,$ Therefore to present a counterexample it suffices to exhibit any nonunit in any $rm R$-algebra. $ $ A natural choice is the nonunit $,rm 0in R,,$ which yields the above proof.
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
edited Dec 22 '16 at 20:03
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
answered Aug 15 '10 at 16:10
Bill DubuqueBill Dubuque
212k29195653
212k29195653
9
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
3
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
add a comment |
9
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
3
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
9
9
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
$begingroup$
This is a good answer! The type of answer I really want to remember when my students ask this question! So much better than the degree argument! In short, I love it!!!
$endgroup$
– BBischof
Aug 15 '10 at 22:15
3
3
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
$begingroup$
I agree--this is extremely elegant, exactly what you want in a proof. I upvoted it!
$endgroup$
– user452
Aug 19 '10 at 12:49
add a comment |
$begingroup$
Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-infty$. Non-zero constants have degree $0$. You then have the degree equation: $deg (fg) = deg (f) + deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
$endgroup$
4
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
add a comment |
$begingroup$
Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-infty$. Non-zero constants have degree $0$. You then have the degree equation: $deg (fg) = deg (f) + deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
$endgroup$
4
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
add a comment |
$begingroup$
Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-infty$. Non-zero constants have degree $0$. You then have the degree equation: $deg (fg) = deg (f) + deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
$endgroup$
Because by definition, the only polynomial that can have a negative degree is $0$, which is defined to have a degree of $-infty$. Non-zero constants have degree $0$. You then have the degree equation: $deg (fg) = deg (f) + deg (g)$ for any polynomials $f,g$. By inspection, any polynomial of degree $n geq 1$ would need as an inverse a polynomial of degree $-n$, which does not exist (i.e. what Agusti Roig said!) The set you want does exist, however: it is called the field of rational functions, and is precisely the set of ratios of polynomials. It is constructed the same way that the field of rational numbers is from the ring of integers.
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
edited Jan 13 '13 at 6:12
Parth Kohli
6,04512961
6,04512961
answered Aug 15 '10 at 14:34
user452
4
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
add a comment |
4
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
4
4
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
$begingroup$
The degree equation only works for polynomial rings over domains.
$endgroup$
– azimut
May 8 '15 at 8:59
add a comment |
$begingroup$
For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x in F[x]$, and clearly $x ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
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add a comment |
$begingroup$
For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x in F[x]$, and clearly $x ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
$endgroup$
add a comment |
$begingroup$
For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x in F[x]$, and clearly $x ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
$endgroup$
For $F[x]$ to be a field, you need to show there is an inverse for each element that isn't 0. Now $x in F[x]$, and clearly $x ne 0$ (considered as a polynomial). But if you multiply $x$ by any non-zero polynomial, the result will always contain $x$ or higher powers, so it has no inverse.
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
answered Mar 12 '14 at 0:11
vonbrandvonbrand
20k63260
20k63260
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add a comment |
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The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
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3
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While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
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– Robin Chapman
Aug 15 '10 at 15:55
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@Robin: edited, thanks.
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– lhf
Aug 15 '10 at 20:28
add a comment |
$begingroup$
The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
$endgroup$
3
$begingroup$
While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
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– Robin Chapman
Aug 15 '10 at 15:55
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@Robin: edited, thanks.
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– lhf
Aug 15 '10 at 20:28
add a comment |
$begingroup$
The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
$endgroup$
The units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
edited Aug 15 '10 at 20:27
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
answered Aug 15 '10 at 14:52
lhflhf
166k10171400
166k10171400
3
$begingroup$
While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
$endgroup$
– Robin Chapman
Aug 15 '10 at 15:55
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@Robin: edited, thanks.
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– lhf
Aug 15 '10 at 20:28
add a comment |
3
$begingroup$
While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
$endgroup$
– Robin Chapman
Aug 15 '10 at 15:55
$begingroup$
@Robin: edited, thanks.
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– lhf
Aug 15 '10 at 20:28
3
3
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While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
$endgroup$
– Robin Chapman
Aug 15 '10 at 15:55
$begingroup$
While this is true when $D$ is an integral domain, it fails for general commutative rings (but still $D[X]$ is never a field).
$endgroup$
– Robin Chapman
Aug 15 '10 at 15:55
$begingroup$
@Robin: edited, thanks.
$endgroup$
– lhf
Aug 15 '10 at 20:28
$begingroup$
@Robin: edited, thanks.
$endgroup$
– lhf
Aug 15 '10 at 20:28
add a comment |
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Consider $mathbb{C}[x]$ the ring of polynomials with coefficients from $mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.
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add a comment |
$begingroup$
Consider $mathbb{C}[x]$ the ring of polynomials with coefficients from $mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.
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add a comment |
$begingroup$
Consider $mathbb{C}[x]$ the ring of polynomials with coefficients from $mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.
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Consider $mathbb{C}[x]$ the ring of polynomials with coefficients from $mathbb{C}$. This is an example of polynomial ring which is not a field, because $x$ has no multiplicativ inverse.
answered Aug 15 '10 at 14:37
anonymous
add a comment |
add a comment |
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Take polynomial $f(x) = x$ and check that it can't have inverse.
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– falagar
Aug 15 '10 at 13:50
39
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The favourite question of my students at this point: "Why isn't $1/x$ a polynomial?" :-)
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– d.t.
Aug 15 '10 at 14:01
12
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@SB: Well, a simpler way of looking at it is that a polynomial has to be continuous everywhere; 1/x clearly isn't.
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– J. M. is not a mathematician
Aug 15 '10 at 14:55
3
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You might find it useful to look at the field of fractions of a polynomial ring. By looking at this and comparing the differences, I think it will be illuminating for you. Plus it will get you thinking in the direction of localization, which is a good direction to be thinking. :)
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– BBischof
Aug 15 '10 at 22:17
2
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@Ｊ.Ｍ., polynomials over arbitrary fields aren't "continuous" anywhere. In fact, you can use $x$ just as a mark, and not ever consider the polynomial as a function. Yes, for $mathbb{R}[x]$ this makes sense, but don't get boxed in.
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– vonbrand
Mar 12 '14 at 0:05