Mixing
Turning $mathbb R^n$ into field
$begingroup$
I am reading Apostol's fascinating text Mathematical Analysis. In a footnote on P117, he writes:
If it were possible to define multiplication in $mathbb R^3$ so as to make $mathbb R^3$ a field including $mathbb C,$ we could argue as follows: for every $bf x$ in $mathbb R^3$, the vectors $1,bf x,bf x^2,bf x^3$ would be linearly dependent. Hence for each $bf x$ in $mathbb R^3,$ a relation of the form $a_0+a_1{bf x}+a_2{bf x^2}+a_3{bf x^3}=0$ would hold, where $a_0,a_1,a_2,a_3$ are real numbers.But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers.
I have these couple of questions:
Does above argument show that $mathbb R^3$ can't be made a field? Or just that $mathbb R^3$ can't be field such that $mathbb C$ is its subfield?
How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!
real-analysis complex-analysis complex-numbers field-theory extension-field
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
asked Jan 8 at 7:15
SilentSilent
2,78032150
$endgroup$
|
show 1 more comment
$begingroup$
I am reading Apostol's fascinating text Mathematical Analysis. In a footnote on P117, he writes:
If it were possible to define multiplication in $mathbb R^3$ so as to make $mathbb R^3$ a field including $mathbb C,$ we could argue as follows: for every $bf x$ in $mathbb R^3$, the vectors $1,bf x,bf x^2,bf x^3$ would be linearly dependent. Hence for each $bf x$ in $mathbb R^3,$ a relation of the form $a_0+a_1{bf x}+a_2{bf x^2}+a_3{bf x^3}=0$ would hold, where $a_0,a_1,a_2,a_3$ are real numbers.But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers.
I have these couple of questions:
Does above argument show that $mathbb R^3$ can't be made a field? Or just that $mathbb R^3$ can't be field such that $mathbb C$ is its subfield?
How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!
real-analysis complex-analysis complex-numbers field-theory extension-field
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
asked Jan 8 at 7:15
SilentSilent
2,78032150
$endgroup$
1
$begingroup$
math.stackexchange.com/q/216820
$endgroup$
– Jean Marie
Jan 8 at 7:42
1
$begingroup$
In a field, a polynomial of degree $n$ cannot have more than $n$ roots.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:43
1
$begingroup$
Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $Bbb C$.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:49
1
$begingroup$
I added the "field-theory" and "extension-field" tags to your post. Cheers!
$endgroup$
– Robert Lewis
Jan 9 at 0:29
1
$begingroup$
While $mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y varepsilon+z varepsilon^2$ where $varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135
$endgroup$
– James S. Cook
Jan 9 at 0:48
|
show 1 more comment
$begingroup$
I am reading Apostol's fascinating text Mathematical Analysis. In a footnote on P117, he writes:
If it were possible to define multiplication in $mathbb R^3$ so as to make $mathbb R^3$ a field including $mathbb C,$ we could argue as follows: for every $bf x$ in $mathbb R^3$, the vectors $1,bf x,bf x^2,bf x^3$ would be linearly dependent. Hence for each $bf x$ in $mathbb R^3,$ a relation of the form $a_0+a_1{bf x}+a_2{bf x^2}+a_3{bf x^3}=0$ would hold, where $a_0,a_1,a_2,a_3$ are real numbers.But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers.
I have these couple of questions:
Does above argument show that $mathbb R^3$ can't be made a field? Or just that $mathbb R^3$ can't be field such that $mathbb C$ is its subfield?
How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!
real-analysis complex-analysis complex-numbers field-theory extension-field
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
asked Jan 8 at 7:15
SilentSilent
2,78032150
$endgroup$
I am reading Apostol's fascinating text Mathematical Analysis. In a footnote on P117, he writes:
If it were possible to define multiplication in $mathbb R^3$ so as to make $mathbb R^3$ a field including $mathbb C,$ we could argue as follows: for every $bf x$ in $mathbb R^3$, the vectors $1,bf x,bf x^2,bf x^3$ would be linearly dependent. Hence for each $bf x$ in $mathbb R^3,$ a relation of the form $a_0+a_1{bf x}+a_2{bf x^2}+a_3{bf x^3}=0$ would hold, where $a_0,a_1,a_2,a_3$ are real numbers.But every polynomial of degree three with real coefficients is a product of a linear polynomial and a quadratic polynomial with real coefficients. The only roots such polynomials can have are either real numbers or complex numbers.
I have these couple of questions:
Does above argument show that $mathbb R^3$ can't be made a field? Or just that $mathbb R^3$ can't be field such that $mathbb C$ is its subfield?
How are we so sure that there are no other roots than complex numbers? Maybe we have not explored enough!
real-analysis complex-analysis complex-numbers field-theory extension-field
real-analysis complex-analysis complex-numbers field-theory extension-field
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
asked Jan 8 at 7:15
SilentSilent
2,78032150
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
asked Jan 8 at 7:15
SilentSilent
2,78032150
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
edited Jan 9 at 0:29
Robert Lewis
45.5k23065
45.5k23065
asked Jan 8 at 7:15
SilentSilent
2,78032150
asked Jan 8 at 7:15
SilentSilent
2,78032150
asked Jan 8 at 7:15
SilentSilent
2,78032150
2,78032150
1
$begingroup$
math.stackexchange.com/q/216820
$endgroup$
– Jean Marie
Jan 8 at 7:42
1
$begingroup$
In a field, a polynomial of degree $n$ cannot have more than $n$ roots.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:43
1
$begingroup$
Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $Bbb C$.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:49
1
$begingroup$
I added the "field-theory" and "extension-field" tags to your post. Cheers!
$endgroup$
– Robert Lewis
Jan 9 at 0:29
1
$begingroup$
While $mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y varepsilon+z varepsilon^2$ where $varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135
$endgroup$
– James S. Cook
Jan 9 at 0:48
|
show 1 more comment
1
$begingroup$
math.stackexchange.com/q/216820
$endgroup$
– Jean Marie
Jan 8 at 7:42
1
$begingroup$
In a field, a polynomial of degree $n$ cannot have more than $n$ roots.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:43
1
$begingroup$
Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $Bbb C$.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:49
1
$begingroup$
I added the "field-theory" and "extension-field" tags to your post. Cheers!
$endgroup$
– Robert Lewis
Jan 9 at 0:29
1
$begingroup$
While $mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y varepsilon+z varepsilon^2$ where $varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135
$endgroup$
– James S. Cook
Jan 9 at 0:48
1
1
$begingroup$
math.stackexchange.com/q/216820
$endgroup$
– Jean Marie
Jan 8 at 7:42
$begingroup$
math.stackexchange.com/q/216820
$endgroup$
– Jean Marie
Jan 8 at 7:42
1
1
$begingroup$
In a field, a polynomial of degree $n$ cannot have more than $n$ roots.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:43
$begingroup$
In a field, a polynomial of degree $n$ cannot have more than $n$ roots.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:43
1
1
$begingroup$
Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $Bbb C$.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:49
$begingroup$
Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $Bbb C$.
$endgroup$
– Akiva Weinberger
Jan 8 at 7:49
1
1
$begingroup$
I added the "field-theory" and "extension-field" tags to your post. Cheers!
$endgroup$
– Robert Lewis
Jan 9 at 0:29
$begingroup$
I added the "field-theory" and "extension-field" tags to your post. Cheers!
$endgroup$
– Robert Lewis
Jan 9 at 0:29
1
1
$begingroup$
While $mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y varepsilon+z varepsilon^2$ where $varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135
$endgroup$
– James S. Cook
Jan 9 at 0:48
$begingroup$
While $mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y varepsilon+z varepsilon^2$ where $varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135
$endgroup$
– James S. Cook
Jan 9 at 0:48
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
We can show that $Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $Bbb R$ without assuming such a field contains a subfield isomorphic to $Bbb C$ as follows:
If $Bbb R^3$ were such a field, we would have
$[Bbb R^3:Bbb R] = 3; tag 1$
being an extension field of $Bbb R$, $Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1Bbb R = Bbb R1$ isomorphic to $Bbb R$ in the usual manner, that is
$Bbb R ni r leftrightarrow r1 in 1Bbb R subsetneq Bbb R^3; tag 2$
by virtue of (1), there exists
$mathbf v in Bbb R^3 setminus Bbb R1 tag 3$
such that $1, mathbf v, mathbf v^2, mathbf v^3$ are linearly dependent over $Bbb R1 cong Bbb R$; that is
$exists c_i in Bbb R, ; 0 le i le 3, tag 4$
not all $c_i$ zero, with
$c_3 mathbf v^3 +c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 4$
let us first consider the case
$c_3 = 0; tag 5$
then
$c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 6$
now if
$c_2 = 0, tag 7$
then if
$c_1 = 0 tag 8$
as well, we find
$c_0 = 0, tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 ne 0 tag{10}$
we may write
$mathbf v = -dfrac{c_0}{c_1} in Bbb R 1 cong Bbb R, tag{11}$
which contradicts (3); thus we have that
$c_2 ne 0, tag{12}$
and we may write (6) as
$mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{13}$
where
$b_i = dfrac{c_i}{c_2} in Bbb R; tag{14}$
we write (13) as
$mathbf v^2 + b_1 mathbf v = -b_0, tag{15}$
and complete the square:
$left (mathbf v + dfrac{b_1}{2} right )^2 = mathbf v^2 + b_1 mathbf v + dfrac{b_1^2}{4} = dfrac{b_1^2}{4} - b_0 = d; tag{16}$
if
$d ge 0, tag{17}$
(16) yields
$mathbf v = -dfrac{b_1}{2} pm sqrt d in Bbb R, tag{18}$
in contradiction to (3); thus,
$d < 0, tag{19}$
and (16) becomes
$dfrac{1}{{sqrt{-d}}^2} left (mathbf v + dfrac{b_1}{2} right )^2 = -1, tag{20}$
which shows the existence of an element
$mathbf i in Bbb R^3 tag{21}$
with
$mathbf i^2 = -1, tag {22}$
and in the usual manner we see that the subalgebra
$Bbb R + Bbb R mathbf i = { s + t mathbf i mid s, t in Bbb R } cong Bbb C tag{23}$
is a subfield of $Bbb R^3$ with
$[Bbb C: Bbb R] = 2; tag{24}$
but this is impossible since it implies
$3 = [Bbb R^3:Bbb R] =[Bbb R^3:Bbb C] [Bbb C: Bbb R] = 2[Bbb R^3:Bbb C]; tag{25}$
but $2 not mid 3$; we conclude then that no such $mathbf v$ satisfying (6), (13) can exist in $Bbb R^3$.
Now if
$c_3 ne 0, tag{26}$
then $mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = dfrac{c_i}{c_3}, ; 0 le i le 2, tag{27}$
we obtain the real monic cubic
$p(mathbf v) = mathbf v^3 +b_2 mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{28}$
which as is well-known always has a root
$r in Bbb R, tag{29}$
whence
$p(mathbf v) = (mathbf v - r)q(mathbf v) tag{30}$
for some monic real quadratic polynomial $q(mathbf v)$; thus,
$(mathbf v - r)q(mathbf v) =p(mathbf v) = 0; tag{31}$
but
$mathbf v - r ne 0 tag{32}$
since
$mathbf v notin Bbb R; tag{33}$
it follows that
$q(mathbf v) = 0, tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $Bbb R^3$ contains a subfield isomorphic to $Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $Bbb R^3$ is an extension field of $Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $Bbb R^3$ has a subfield isomorphic to $Bbb C$ to show that $Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $mathbf x$ than the usual complex numbers falls once we have $Bbb C subset Bbb R^3$, for then the familiar factorizations in $Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $mathbf x$ must lie in $Bbb C$; we need look no further.
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
$endgroup$
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
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– Silent
Jan 9 at 7:31
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@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
1
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
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@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
add a comment |
$begingroup$
- Just a field containing $Bbb C$. Which has problems all on its own, as $Bbb R^3$ would be a field extension of degree 3, and thus cannot have an intermediate extension of degree $2$, such as $Bbb C$. So there are many reasons this wouldn't work.
- It's not a matter of not having explored enough. We can find three complex roots to that equation, relatively easily (at least with access to modern tools, like computer algebra systems, or wikipedia). There are now two possibilities: Either our $Bbb R^3$ field doesn't give us any numbers $Bbb C$ doesn't already have (which is impossible: $Bbb CcongBbb R^2$ is a strict sub-vectorspace), or some degree-3 (or lower) polynomials have more than three roots, which breaks all kinds of things and is therefore not possible.
answered Jan 8 at 7:25
ArthurArthur
113k7113195
$endgroup$
add a comment |
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$begingroup$
We can show that $Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $Bbb R$ without assuming such a field contains a subfield isomorphic to $Bbb C$ as follows:
If $Bbb R^3$ were such a field, we would have
$[Bbb R^3:Bbb R] = 3; tag 1$
being an extension field of $Bbb R$, $Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1Bbb R = Bbb R1$ isomorphic to $Bbb R$ in the usual manner, that is
$Bbb R ni r leftrightarrow r1 in 1Bbb R subsetneq Bbb R^3; tag 2$
by virtue of (1), there exists
$mathbf v in Bbb R^3 setminus Bbb R1 tag 3$
such that $1, mathbf v, mathbf v^2, mathbf v^3$ are linearly dependent over $Bbb R1 cong Bbb R$; that is
$exists c_i in Bbb R, ; 0 le i le 3, tag 4$
not all $c_i$ zero, with
$c_3 mathbf v^3 +c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 4$
let us first consider the case
$c_3 = 0; tag 5$
then
$c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 6$
now if
$c_2 = 0, tag 7$
then if
$c_1 = 0 tag 8$
as well, we find
$c_0 = 0, tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 ne 0 tag{10}$
we may write
$mathbf v = -dfrac{c_0}{c_1} in Bbb R 1 cong Bbb R, tag{11}$
which contradicts (3); thus we have that
$c_2 ne 0, tag{12}$
and we may write (6) as
$mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{13}$
where
$b_i = dfrac{c_i}{c_2} in Bbb R; tag{14}$
we write (13) as
$mathbf v^2 + b_1 mathbf v = -b_0, tag{15}$
and complete the square:
$left (mathbf v + dfrac{b_1}{2} right )^2 = mathbf v^2 + b_1 mathbf v + dfrac{b_1^2}{4} = dfrac{b_1^2}{4} - b_0 = d; tag{16}$
if
$d ge 0, tag{17}$
(16) yields
$mathbf v = -dfrac{b_1}{2} pm sqrt d in Bbb R, tag{18}$
in contradiction to (3); thus,
$d < 0, tag{19}$
and (16) becomes
$dfrac{1}{{sqrt{-d}}^2} left (mathbf v + dfrac{b_1}{2} right )^2 = -1, tag{20}$
which shows the existence of an element
$mathbf i in Bbb R^3 tag{21}$
with
$mathbf i^2 = -1, tag {22}$
and in the usual manner we see that the subalgebra
$Bbb R + Bbb R mathbf i = { s + t mathbf i mid s, t in Bbb R } cong Bbb C tag{23}$
is a subfield of $Bbb R^3$ with
$[Bbb C: Bbb R] = 2; tag{24}$
but this is impossible since it implies
$3 = [Bbb R^3:Bbb R] =[Bbb R^3:Bbb C] [Bbb C: Bbb R] = 2[Bbb R^3:Bbb C]; tag{25}$
but $2 not mid 3$; we conclude then that no such $mathbf v$ satisfying (6), (13) can exist in $Bbb R^3$.
Now if
$c_3 ne 0, tag{26}$
then $mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = dfrac{c_i}{c_3}, ; 0 le i le 2, tag{27}$
we obtain the real monic cubic
$p(mathbf v) = mathbf v^3 +b_2 mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{28}$
which as is well-known always has a root
$r in Bbb R, tag{29}$
whence
$p(mathbf v) = (mathbf v - r)q(mathbf v) tag{30}$
for some monic real quadratic polynomial $q(mathbf v)$; thus,
$(mathbf v - r)q(mathbf v) =p(mathbf v) = 0; tag{31}$
but
$mathbf v - r ne 0 tag{32}$
since
$mathbf v notin Bbb R; tag{33}$
it follows that
$q(mathbf v) = 0, tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $Bbb R^3$ contains a subfield isomorphic to $Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $Bbb R^3$ is an extension field of $Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $Bbb R^3$ has a subfield isomorphic to $Bbb C$ to show that $Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $mathbf x$ than the usual complex numbers falls once we have $Bbb C subset Bbb R^3$, for then the familiar factorizations in $Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $mathbf x$ must lie in $Bbb C$; we need look no further.
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
$endgroup$
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
$endgroup$
– Silent
Jan 9 at 7:31
$begingroup$
@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
1
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
$begingroup$
@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
add a comment |
$begingroup$
We can show that $Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $Bbb R$ without assuming such a field contains a subfield isomorphic to $Bbb C$ as follows:
If $Bbb R^3$ were such a field, we would have
$[Bbb R^3:Bbb R] = 3; tag 1$
being an extension field of $Bbb R$, $Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1Bbb R = Bbb R1$ isomorphic to $Bbb R$ in the usual manner, that is
$Bbb R ni r leftrightarrow r1 in 1Bbb R subsetneq Bbb R^3; tag 2$
by virtue of (1), there exists
$mathbf v in Bbb R^3 setminus Bbb R1 tag 3$
such that $1, mathbf v, mathbf v^2, mathbf v^3$ are linearly dependent over $Bbb R1 cong Bbb R$; that is
$exists c_i in Bbb R, ; 0 le i le 3, tag 4$
not all $c_i$ zero, with
$c_3 mathbf v^3 +c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 4$
let us first consider the case
$c_3 = 0; tag 5$
then
$c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 6$
now if
$c_2 = 0, tag 7$
then if
$c_1 = 0 tag 8$
as well, we find
$c_0 = 0, tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 ne 0 tag{10}$
we may write
$mathbf v = -dfrac{c_0}{c_1} in Bbb R 1 cong Bbb R, tag{11}$
which contradicts (3); thus we have that
$c_2 ne 0, tag{12}$
and we may write (6) as
$mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{13}$
where
$b_i = dfrac{c_i}{c_2} in Bbb R; tag{14}$
we write (13) as
$mathbf v^2 + b_1 mathbf v = -b_0, tag{15}$
and complete the square:
$left (mathbf v + dfrac{b_1}{2} right )^2 = mathbf v^2 + b_1 mathbf v + dfrac{b_1^2}{4} = dfrac{b_1^2}{4} - b_0 = d; tag{16}$
if
$d ge 0, tag{17}$
(16) yields
$mathbf v = -dfrac{b_1}{2} pm sqrt d in Bbb R, tag{18}$
in contradiction to (3); thus,
$d < 0, tag{19}$
and (16) becomes
$dfrac{1}{{sqrt{-d}}^2} left (mathbf v + dfrac{b_1}{2} right )^2 = -1, tag{20}$
which shows the existence of an element
$mathbf i in Bbb R^3 tag{21}$
with
$mathbf i^2 = -1, tag {22}$
and in the usual manner we see that the subalgebra
$Bbb R + Bbb R mathbf i = { s + t mathbf i mid s, t in Bbb R } cong Bbb C tag{23}$
is a subfield of $Bbb R^3$ with
$[Bbb C: Bbb R] = 2; tag{24}$
but this is impossible since it implies
$3 = [Bbb R^3:Bbb R] =[Bbb R^3:Bbb C] [Bbb C: Bbb R] = 2[Bbb R^3:Bbb C]; tag{25}$
but $2 not mid 3$; we conclude then that no such $mathbf v$ satisfying (6), (13) can exist in $Bbb R^3$.
Now if
$c_3 ne 0, tag{26}$
then $mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = dfrac{c_i}{c_3}, ; 0 le i le 2, tag{27}$
we obtain the real monic cubic
$p(mathbf v) = mathbf v^3 +b_2 mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{28}$
which as is well-known always has a root
$r in Bbb R, tag{29}$
whence
$p(mathbf v) = (mathbf v - r)q(mathbf v) tag{30}$
for some monic real quadratic polynomial $q(mathbf v)$; thus,
$(mathbf v - r)q(mathbf v) =p(mathbf v) = 0; tag{31}$
but
$mathbf v - r ne 0 tag{32}$
since
$mathbf v notin Bbb R; tag{33}$
it follows that
$q(mathbf v) = 0, tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $Bbb R^3$ contains a subfield isomorphic to $Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $Bbb R^3$ is an extension field of $Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $Bbb R^3$ has a subfield isomorphic to $Bbb C$ to show that $Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $mathbf x$ than the usual complex numbers falls once we have $Bbb C subset Bbb R^3$, for then the familiar factorizations in $Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $mathbf x$ must lie in $Bbb C$; we need look no further.
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
$endgroup$
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
$endgroup$
– Silent
Jan 9 at 7:31
$begingroup$
@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
1
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
$begingroup$
@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
add a comment |
$begingroup$
We can show that $Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $Bbb R$ without assuming such a field contains a subfield isomorphic to $Bbb C$ as follows:
If $Bbb R^3$ were such a field, we would have
$[Bbb R^3:Bbb R] = 3; tag 1$
being an extension field of $Bbb R$, $Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1Bbb R = Bbb R1$ isomorphic to $Bbb R$ in the usual manner, that is
$Bbb R ni r leftrightarrow r1 in 1Bbb R subsetneq Bbb R^3; tag 2$
by virtue of (1), there exists
$mathbf v in Bbb R^3 setminus Bbb R1 tag 3$
such that $1, mathbf v, mathbf v^2, mathbf v^3$ are linearly dependent over $Bbb R1 cong Bbb R$; that is
$exists c_i in Bbb R, ; 0 le i le 3, tag 4$
not all $c_i$ zero, with
$c_3 mathbf v^3 +c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 4$
let us first consider the case
$c_3 = 0; tag 5$
then
$c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 6$
now if
$c_2 = 0, tag 7$
then if
$c_1 = 0 tag 8$
as well, we find
$c_0 = 0, tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 ne 0 tag{10}$
we may write
$mathbf v = -dfrac{c_0}{c_1} in Bbb R 1 cong Bbb R, tag{11}$
which contradicts (3); thus we have that
$c_2 ne 0, tag{12}$
and we may write (6) as
$mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{13}$
where
$b_i = dfrac{c_i}{c_2} in Bbb R; tag{14}$
we write (13) as
$mathbf v^2 + b_1 mathbf v = -b_0, tag{15}$
and complete the square:
$left (mathbf v + dfrac{b_1}{2} right )^2 = mathbf v^2 + b_1 mathbf v + dfrac{b_1^2}{4} = dfrac{b_1^2}{4} - b_0 = d; tag{16}$
if
$d ge 0, tag{17}$
(16) yields
$mathbf v = -dfrac{b_1}{2} pm sqrt d in Bbb R, tag{18}$
in contradiction to (3); thus,
$d < 0, tag{19}$
and (16) becomes
$dfrac{1}{{sqrt{-d}}^2} left (mathbf v + dfrac{b_1}{2} right )^2 = -1, tag{20}$
which shows the existence of an element
$mathbf i in Bbb R^3 tag{21}$
with
$mathbf i^2 = -1, tag {22}$
and in the usual manner we see that the subalgebra
$Bbb R + Bbb R mathbf i = { s + t mathbf i mid s, t in Bbb R } cong Bbb C tag{23}$
is a subfield of $Bbb R^3$ with
$[Bbb C: Bbb R] = 2; tag{24}$
but this is impossible since it implies
$3 = [Bbb R^3:Bbb R] =[Bbb R^3:Bbb C] [Bbb C: Bbb R] = 2[Bbb R^3:Bbb C]; tag{25}$
but $2 not mid 3$; we conclude then that no such $mathbf v$ satisfying (6), (13) can exist in $Bbb R^3$.
Now if
$c_3 ne 0, tag{26}$
then $mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = dfrac{c_i}{c_3}, ; 0 le i le 2, tag{27}$
we obtain the real monic cubic
$p(mathbf v) = mathbf v^3 +b_2 mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{28}$
which as is well-known always has a root
$r in Bbb R, tag{29}$
whence
$p(mathbf v) = (mathbf v - r)q(mathbf v) tag{30}$
for some monic real quadratic polynomial $q(mathbf v)$; thus,
$(mathbf v - r)q(mathbf v) =p(mathbf v) = 0; tag{31}$
but
$mathbf v - r ne 0 tag{32}$
since
$mathbf v notin Bbb R; tag{33}$
it follows that
$q(mathbf v) = 0, tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $Bbb R^3$ contains a subfield isomorphic to $Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $Bbb R^3$ is an extension field of $Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $Bbb R^3$ has a subfield isomorphic to $Bbb C$ to show that $Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $mathbf x$ than the usual complex numbers falls once we have $Bbb C subset Bbb R^3$, for then the familiar factorizations in $Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $mathbf x$ must lie in $Bbb C$; we need look no further.
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
$endgroup$
We can show that $Bbb R^3$ cannot be assigned a multiplication operation which turns it into an extension field of $Bbb R$ without assuming such a field contains a subfield isomorphic to $Bbb C$ as follows:
If $Bbb R^3$ were such a field, we would have
$[Bbb R^3:Bbb R] = 3; tag 1$
being an extension field of $Bbb R$, $Bbb R^3$ contains a multiplicative identity $1$ and a subfield $1Bbb R = Bbb R1$ isomorphic to $Bbb R$ in the usual manner, that is
$Bbb R ni r leftrightarrow r1 in 1Bbb R subsetneq Bbb R^3; tag 2$
by virtue of (1), there exists
$mathbf v in Bbb R^3 setminus Bbb R1 tag 3$
such that $1, mathbf v, mathbf v^2, mathbf v^3$ are linearly dependent over $Bbb R1 cong Bbb R$; that is
$exists c_i in Bbb R, ; 0 le i le 3, tag 4$
not all $c_i$ zero, with
$c_3 mathbf v^3 +c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 4$
let us first consider the case
$c_3 = 0; tag 5$
then
$c_2 mathbf v^2 + c_1 mathbf v + c_0 = 0; tag 6$
now if
$c_2 = 0, tag 7$
then if
$c_1 = 0 tag 8$
as well, we find
$c_0 = 0, tag 9$
contradicting our hypothesis that not all the $c_i = 0$; and if
$c_1 ne 0 tag{10}$
we may write
$mathbf v = -dfrac{c_0}{c_1} in Bbb R 1 cong Bbb R, tag{11}$
which contradicts (3); thus we have that
$c_2 ne 0, tag{12}$
and we may write (6) as
$mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{13}$
where
$b_i = dfrac{c_i}{c_2} in Bbb R; tag{14}$
we write (13) as
$mathbf v^2 + b_1 mathbf v = -b_0, tag{15}$
and complete the square:
$left (mathbf v + dfrac{b_1}{2} right )^2 = mathbf v^2 + b_1 mathbf v + dfrac{b_1^2}{4} = dfrac{b_1^2}{4} - b_0 = d; tag{16}$
if
$d ge 0, tag{17}$
(16) yields
$mathbf v = -dfrac{b_1}{2} pm sqrt d in Bbb R, tag{18}$
in contradiction to (3); thus,
$d < 0, tag{19}$
and (16) becomes
$dfrac{1}{{sqrt{-d}}^2} left (mathbf v + dfrac{b_1}{2} right )^2 = -1, tag{20}$
which shows the existence of an element
$mathbf i in Bbb R^3 tag{21}$
with
$mathbf i^2 = -1, tag {22}$
and in the usual manner we see that the subalgebra
$Bbb R + Bbb R mathbf i = { s + t mathbf i mid s, t in Bbb R } cong Bbb C tag{23}$
is a subfield of $Bbb R^3$ with
$[Bbb C: Bbb R] = 2; tag{24}$
but this is impossible since it implies
$3 = [Bbb R^3:Bbb R] =[Bbb R^3:Bbb C] [Bbb C: Bbb R] = 2[Bbb R^3:Bbb C]; tag{25}$
but $2 not mid 3$; we conclude then that no such $mathbf v$ satisfying (6), (13) can exist in $Bbb R^3$.
Now if
$c_3 ne 0, tag{26}$
then $mathbf v$ satisfies the full cubic (4), and as above setting
$b_i = dfrac{c_i}{c_3}, ; 0 le i le 2, tag{27}$
we obtain the real monic cubic
$p(mathbf v) = mathbf v^3 +b_2 mathbf v^2 + b_1 mathbf v + b_0 = 0, tag{28}$
which as is well-known always has a root
$r in Bbb R, tag{29}$
whence
$p(mathbf v) = (mathbf v - r)q(mathbf v) tag{30}$
for some monic real quadratic polynomial $q(mathbf v)$; thus,
$(mathbf v - r)q(mathbf v) =p(mathbf v) = 0; tag{31}$
but
$mathbf v - r ne 0 tag{32}$
since
$mathbf v notin Bbb R; tag{33}$
it follows that
$q(mathbf v) = 0, tag{34}$
and we have reduced the cubic to the previous (quadratic) case, which we have reduced to the absurd; we thus conclude that $Bbb R^3$ admits no multiplication operation compatible with the field axioms, and we are done.
We close with the observation that our argument requires no assumption that $Bbb R^3$ contains a subfield isomorphic to $Bbb C$; indeed, we have shown that the existence of such a subfield follows from the assertion that $Bbb R^3$ is an extension field of $Bbb R$, from which a contradiction is deduced.
Finally, as for our OP Silent's two closing questions, Apostol's proof indeed makes use of the assumption that $Bbb R^3$ has a subfield isomorphic to $Bbb C$ to show that $Bbb R^3$ can't be made into a field; and the issue that there are "other" roots of the polynomial in $mathbf x$ than the usual complex numbers falls once we have $Bbb C subset Bbb R^3$, for then the familiar factorizations in $Bbb C[x]$ hold, and since a polynomial of degree $n$ over any field has at most $n$ zeroes, we see that all the roots of a real polynomial in $mathbf x$ must lie in $Bbb C$; we need look no further.
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
edited Jan 8 at 23:57
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
answered Jan 8 at 23:06
Robert LewisRobert Lewis
45.5k23065
45.5k23065
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
$endgroup$
– Silent
Jan 9 at 7:31
$begingroup$
@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
1
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
$begingroup$
@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
add a comment |
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
$endgroup$
– Silent
Jan 9 at 7:31
$begingroup$
@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
1
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
$begingroup$
@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
$endgroup$
– Silent
Jan 9 at 7:31
$begingroup$
Thanks for such a good explanation. I can't help but accept this answer
$endgroup$
– Silent
Jan 9 at 7:31
$begingroup$
@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
$begingroup$
@Silent: thanks for the kind words, and for the "acceptance". I find problems like this fascinating; I looked at higher dimensional cases, but it gets a lot harder once $n > 3$. Thanks again!
$endgroup$
– Robert Lewis
Jan 9 at 7:34
1
1
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
$begingroup$
Yes, this becomes harder for larger $n$. An odd $n$ can be excluded easily enough. The field extension is known to be simple, so generated by a zero of a degree $n$ polynomial. But any odd degree polynomial has a real zero by the intermediate value theorem. For even $n$ you can either prove that $Bbb{C}$ is algebraically closed using complex analysis. Or, assuming pieces of Galois theory and group theory, do this.
$endgroup$
– Jyrki Lahtonen
Jan 12 at 10:17
$begingroup$
@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
$begingroup$
@JyrkiLahtonen: thanks for the lead. Will check it out. Looks fascinating!
$endgroup$
– Robert Lewis
Jan 12 at 12:22
add a comment |
$begingroup$
- Just a field containing $Bbb C$. Which has problems all on its own, as $Bbb R^3$ would be a field extension of degree 3, and thus cannot have an intermediate extension of degree $2$, such as $Bbb C$. So there are many reasons this wouldn't work.
- It's not a matter of not having explored enough. We can find three complex roots to that equation, relatively easily (at least with access to modern tools, like computer algebra systems, or wikipedia). There are now two possibilities: Either our $Bbb R^3$ field doesn't give us any numbers $Bbb C$ doesn't already have (which is impossible: $Bbb CcongBbb R^2$ is a strict sub-vectorspace), or some degree-3 (or lower) polynomials have more than three roots, which breaks all kinds of things and is therefore not possible.
answered Jan 8 at 7:25
ArthurArthur
113k7113195
$endgroup$
add a comment |
$begingroup$
- Just a field containing $Bbb C$. Which has problems all on its own, as $Bbb R^3$ would be a field extension of degree 3, and thus cannot have an intermediate extension of degree $2$, such as $Bbb C$. So there are many reasons this wouldn't work.
- It's not a matter of not having explored enough. We can find three complex roots to that equation, relatively easily (at least with access to modern tools, like computer algebra systems, or wikipedia). There are now two possibilities: Either our $Bbb R^3$ field doesn't give us any numbers $Bbb C$ doesn't already have (which is impossible: $Bbb CcongBbb R^2$ is a strict sub-vectorspace), or some degree-3 (or lower) polynomials have more than three roots, which breaks all kinds of things and is therefore not possible.
answered Jan 8 at 7:25
ArthurArthur
113k7113195
$endgroup$
add a comment |
$begingroup$
- Just a field containing $Bbb C$. Which has problems all on its own, as $Bbb R^3$ would be a field extension of degree 3, and thus cannot have an intermediate extension of degree $2$, such as $Bbb C$. So there are many reasons this wouldn't work.
- It's not a matter of not having explored enough. We can find three complex roots to that equation, relatively easily (at least with access to modern tools, like computer algebra systems, or wikipedia). There are now two possibilities: Either our $Bbb R^3$ field doesn't give us any numbers $Bbb C$ doesn't already have (which is impossible: $Bbb CcongBbb R^2$ is a strict sub-vectorspace), or some degree-3 (or lower) polynomials have more than three roots, which breaks all kinds of things and is therefore not possible.
answered Jan 8 at 7:25
ArthurArthur
113k7113195
$endgroup$
- Just a field containing $Bbb C$. Which has problems all on its own, as $Bbb R^3$ would be a field extension of degree 3, and thus cannot have an intermediate extension of degree $2$, such as $Bbb C$. So there are many reasons this wouldn't work.
- It's not a matter of not having explored enough. We can find three complex roots to that equation, relatively easily (at least with access to modern tools, like computer algebra systems, or wikipedia). There are now two possibilities: Either our $Bbb R^3$ field doesn't give us any numbers $Bbb C$ doesn't already have (which is impossible: $Bbb CcongBbb R^2$ is a strict sub-vectorspace), or some degree-3 (or lower) polynomials have more than three roots, which breaks all kinds of things and is therefore not possible.
answered Jan 8 at 7:25
ArthurArthur
113k7113195
edited Jan 8 at 7:30
answered Jan 8 at 7:25
ArthurArthur
113k7113195
answered Jan 8 at 7:25
ArthurArthur
113k7113195
answered Jan 8 at 7:25
ArthurArthur
113k7113195
113k7113195
add a comment |
add a comment |
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math.stackexchange.com/q/216820
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– Jean Marie
Jan 8 at 7:42
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In a field, a polynomial of degree $n$ cannot have more than $n$ roots.
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– Akiva Weinberger
Jan 8 at 7:43
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Also: if a quadratic polynomial has no real roots, and a field contains those roots, we can still complete the square and derive that the field contains a square root of $-1$ and thus that it contains $Bbb C$.
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– Akiva Weinberger
Jan 8 at 7:49
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I added the "field-theory" and "extension-field" tags to your post. Cheers!
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– Robert Lewis
Jan 9 at 0:29
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While $mathbb{R}^3$ is cannot be made into a field, it does permit a number of interesting algebraic structures. For example, $x+jy+j^2z$ with $j^3=1$ defines a pretty neat multiplication which we might recognize as the group algebra of the cyclic group of order $3$. Or, look at $x+y varepsilon+z varepsilon^2$ where $varepsilon^3=0$. There are infinitely many multiplications possible, they're just not fields. Lot of zero divisors and sometimes nilpotent elements. Despite the lack of inverses, calculus is still possible. See arxiv.org/abs/1708.04135
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– James S. Cook
Jan 9 at 0:48