$sqrt{x} notin F(x)$
$begingroup$
I wanted to prove that if $F$ is a field and we consider the fraction field $F(x)$, then $sqrt{x} notin F(x)$.
I said that if $sqrt{x} in F(x)$ then there are $f,g in F[x]$ such that $bigl(frac{f(x)}{g(x)}bigr)^2=x$ so $f(x)^2=xg(x)^2$ so $2(deg(f)-deg(g))=1$ which is not possible. Is this true? Is there a quicker way to see this?
rational-functions
edited Jan 16 at 23:30
egreg
186k1486208
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
$endgroup$
add a comment |
$begingroup$
I wanted to prove that if $F$ is a field and we consider the fraction field $F(x)$, then $sqrt{x} notin F(x)$.
I said that if $sqrt{x} in F(x)$ then there are $f,g in F[x]$ such that $bigl(frac{f(x)}{g(x)}bigr)^2=x$ so $f(x)^2=xg(x)^2$ so $2(deg(f)-deg(g))=1$ which is not possible. Is this true? Is there a quicker way to see this?
rational-functions
edited Jan 16 at 23:30
egreg
186k1486208
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
$endgroup$
$begingroup$
You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $deg 0$ is either undefined, set to $-infty$, or set to 0 but with the proviso that then $deg(fg) = deg(f) + deg(g)$ is only valid when $f, g ne 0$. Otherwise, it looks good.
$endgroup$
– Daniel Schepler
Jan 16 at 22:58
$begingroup$
Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root.
$endgroup$
– Daniel Schepler
Jan 16 at 23:00
$begingroup$
@DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)ne0$ and $g(x)ne0$.
$endgroup$
– egreg
Jan 16 at 23:28
$begingroup$
@egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof.
$endgroup$
– Daniel Schepler
Jan 17 at 0:00
$begingroup$
A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x mid g(x)$, contradiction.
$endgroup$
– Daniel Schepler
Jan 17 at 0:03
add a comment |
$begingroup$
I wanted to prove that if $F$ is a field and we consider the fraction field $F(x)$, then $sqrt{x} notin F(x)$.
I said that if $sqrt{x} in F(x)$ then there are $f,g in F[x]$ such that $bigl(frac{f(x)}{g(x)}bigr)^2=x$ so $f(x)^2=xg(x)^2$ so $2(deg(f)-deg(g))=1$ which is not possible. Is this true? Is there a quicker way to see this?
rational-functions
edited Jan 16 at 23:30
egreg
186k1486208
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
$endgroup$
I wanted to prove that if $F$ is a field and we consider the fraction field $F(x)$, then $sqrt{x} notin F(x)$.
I said that if $sqrt{x} in F(x)$ then there are $f,g in F[x]$ such that $bigl(frac{f(x)}{g(x)}bigr)^2=x$ so $f(x)^2=xg(x)^2$ so $2(deg(f)-deg(g))=1$ which is not possible. Is this true? Is there a quicker way to see this?
rational-functions
rational-functions
edited Jan 16 at 23:30
egreg
186k1486208
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
edited Jan 16 at 23:30
egreg
186k1486208
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
edited Jan 16 at 23:30
egreg
186k1486208
edited Jan 16 at 23:30
egreg
186k1486208
edited Jan 16 at 23:30
egreg
186k1486208
186k1486208
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
asked Jan 16 at 22:50
roi_saumonroi_saumon
63938
63938
$begingroup$
You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $deg 0$ is either undefined, set to $-infty$, or set to 0 but with the proviso that then $deg(fg) = deg(f) + deg(g)$ is only valid when $f, g ne 0$. Otherwise, it looks good.
$endgroup$
– Daniel Schepler
Jan 16 at 22:58
$begingroup$
Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root.
$endgroup$
– Daniel Schepler
Jan 16 at 23:00
$begingroup$
@DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)ne0$ and $g(x)ne0$.
$endgroup$
– egreg
Jan 16 at 23:28
$begingroup$
@egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof.
$endgroup$
– Daniel Schepler
Jan 17 at 0:00
$begingroup$
A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x mid g(x)$, contradiction.
$endgroup$
– Daniel Schepler
Jan 17 at 0:03
add a comment |
$begingroup$
You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $deg 0$ is either undefined, set to $-infty$, or set to 0 but with the proviso that then $deg(fg) = deg(f) + deg(g)$ is only valid when $f, g ne 0$. Otherwise, it looks good.
$endgroup$
– Daniel Schepler
Jan 16 at 22:58
$begingroup$
Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root.
$endgroup$
– Daniel Schepler
Jan 16 at 23:00
$begingroup$
@DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)ne0$ and $g(x)ne0$.
$endgroup$
– egreg
Jan 16 at 23:28
$begingroup$
@egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof.
$endgroup$
– Daniel Schepler
Jan 17 at 0:00
$begingroup$
A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x mid g(x)$, contradiction.
$endgroup$
– Daniel Schepler
Jan 17 at 0:03
$begingroup$
You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $deg 0$ is either undefined, set to $-infty$, or set to 0 but with the proviso that then $deg(fg) = deg(f) + deg(g)$ is only valid when $f, g ne 0$. Otherwise, it looks good.
$endgroup$
– Daniel Schepler
Jan 16 at 22:58
$begingroup$
You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $deg 0$ is either undefined, set to $-infty$, or set to 0 but with the proviso that then $deg(fg) = deg(f) + deg(g)$ is only valid when $f, g ne 0$. Otherwise, it looks good.
$endgroup$
– Daniel Schepler
Jan 16 at 22:58
$begingroup$
Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root.
$endgroup$
– Daniel Schepler
Jan 16 at 23:00
$begingroup$
Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root.
$endgroup$
– Daniel Schepler
Jan 16 at 23:00
$begingroup$
@DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)ne0$ and $g(x)ne0$.
$endgroup$
– egreg
Jan 16 at 23:28
$begingroup$
@DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)ne0$ and $g(x)ne0$.
$endgroup$
– egreg
Jan 16 at 23:28
$begingroup$
@egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof.
$endgroup$
– Daniel Schepler
Jan 17 at 0:00
$begingroup$
@egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof.
$endgroup$
– Daniel Schepler
Jan 17 at 0:00
$begingroup$
A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x mid g(x)$, contradiction.
$endgroup$
– Daniel Schepler
Jan 17 at 0:03
$begingroup$
A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x mid g(x)$, contradiction.
$endgroup$
– Daniel Schepler
Jan 17 at 0:03
add a comment |
1 Answer
1
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$begingroup$
That's correct and I can't see a quicker way. You should observe that $f(x)ne0$ and $g(x)ne0$, but otherwise it's fine.
Probably I'd write
$$
2deg f(x)=1+2deg g(x)
$$
and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.
By the way, if one defines a “degree” in $mathbb{Z}$ by declaring that $deg n$ is, for $nne0$, the number of prime factors in $n$, so $deg 1=0$, $deg 2=1$ and $deg12=3$; this function is well defined because of uniqueness of factorization and $deg(mn)=deg m+deg n$. Using this degree, irrationality of $sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $sqrt{d}$ for every square free positive integer $d$.
answered Jan 16 at 23:31
egregegreg
186k1486208
$endgroup$
add a comment |
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1 Answer
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$begingroup$
That's correct and I can't see a quicker way. You should observe that $f(x)ne0$ and $g(x)ne0$, but otherwise it's fine.
Probably I'd write
$$
2deg f(x)=1+2deg g(x)
$$
and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.
By the way, if one defines a “degree” in $mathbb{Z}$ by declaring that $deg n$ is, for $nne0$, the number of prime factors in $n$, so $deg 1=0$, $deg 2=1$ and $deg12=3$; this function is well defined because of uniqueness of factorization and $deg(mn)=deg m+deg n$. Using this degree, irrationality of $sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $sqrt{d}$ for every square free positive integer $d$.
answered Jan 16 at 23:31
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
That's correct and I can't see a quicker way. You should observe that $f(x)ne0$ and $g(x)ne0$, but otherwise it's fine.
Probably I'd write
$$
2deg f(x)=1+2deg g(x)
$$
and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.
By the way, if one defines a “degree” in $mathbb{Z}$ by declaring that $deg n$ is, for $nne0$, the number of prime factors in $n$, so $deg 1=0$, $deg 2=1$ and $deg12=3$; this function is well defined because of uniqueness of factorization and $deg(mn)=deg m+deg n$. Using this degree, irrationality of $sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $sqrt{d}$ for every square free positive integer $d$.
answered Jan 16 at 23:31
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
That's correct and I can't see a quicker way. You should observe that $f(x)ne0$ and $g(x)ne0$, but otherwise it's fine.
Probably I'd write
$$
2deg f(x)=1+2deg g(x)
$$
and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.
By the way, if one defines a “degree” in $mathbb{Z}$ by declaring that $deg n$ is, for $nne0$, the number of prime factors in $n$, so $deg 1=0$, $deg 2=1$ and $deg12=3$; this function is well defined because of uniqueness of factorization and $deg(mn)=deg m+deg n$. Using this degree, irrationality of $sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $sqrt{d}$ for every square free positive integer $d$.
answered Jan 16 at 23:31
egregegreg
186k1486208
$endgroup$
That's correct and I can't see a quicker way. You should observe that $f(x)ne0$ and $g(x)ne0$, but otherwise it's fine.
Probably I'd write
$$
2deg f(x)=1+2deg g(x)
$$
and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.
By the way, if one defines a “degree” in $mathbb{Z}$ by declaring that $deg n$ is, for $nne0$, the number of prime factors in $n$, so $deg 1=0$, $deg 2=1$ and $deg12=3$; this function is well defined because of uniqueness of factorization and $deg(mn)=deg m+deg n$. Using this degree, irrationality of $sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $sqrt{d}$ for every square free positive integer $d$.
answered Jan 16 at 23:31
egregegreg
186k1486208
edited Jan 17 at 14:08
answered Jan 16 at 23:31
egregegreg
186k1486208
answered Jan 16 at 23:31
egregegreg
186k1486208
answered Jan 16 at 23:31
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
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$begingroup$
You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $deg 0$ is either undefined, set to $-infty$, or set to 0 but with the proviso that then $deg(fg) = deg(f) + deg(g)$ is only valid when $f, g ne 0$. Otherwise, it looks good.
$endgroup$
– Daniel Schepler
Jan 16 at 22:58
$begingroup$
Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root.
$endgroup$
– Daniel Schepler
Jan 16 at 23:00
$begingroup$
@DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)ne0$ and $g(x)ne0$.
$endgroup$
– egreg
Jan 16 at 23:28
$begingroup$
@egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof.
$endgroup$
– Daniel Schepler
Jan 17 at 0:00
$begingroup$
A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x mid g(x)$, contradiction.
$endgroup$
– Daniel Schepler
Jan 17 at 0:03