Is this a correct solution to show that the ring $mathbb{R}[x]$ does not embed in $mathbb{C}$
$begingroup$
I know this is not a typical question and I will delete it later.
Here is a question together my answer. Please let me know if my answer is correct. The reason that I am posting this is that I believe my answer is correct but my professor refuses to accept that as a correct answer (with no explanation).
Question: Is it possible to embed the ring $mathbb{R}[x]$ in $mathbb{C}$ ?
My Answer: No. Every element of $mathbb{C}$ is algebraic over $mathbb{R}$ so there is no copy of $mathbb{R}[x]$ in $mathbb{C}$ because $x$ is not algebraic over $mathbb{R}$.
abstract-algebra commutative-algebra extension-field
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
$endgroup$
add a comment |
$begingroup$
I know this is not a typical question and I will delete it later.
Here is a question together my answer. Please let me know if my answer is correct. The reason that I am posting this is that I believe my answer is correct but my professor refuses to accept that as a correct answer (with no explanation).
Question: Is it possible to embed the ring $mathbb{R}[x]$ in $mathbb{C}$ ?
My Answer: No. Every element of $mathbb{C}$ is algebraic over $mathbb{R}$ so there is no copy of $mathbb{R}[x]$ in $mathbb{C}$ because $x$ is not algebraic over $mathbb{R}$.
abstract-algebra commutative-algebra extension-field
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
$endgroup$
1
$begingroup$
You need to be more precise. Your argument shows that there is no embedding as an $mathbb{R}$-algebra. Actually I think there is an embedding as rings if you assume axiom of choice.
$endgroup$
– Tobias Kildetoft
Dec 9 '18 at 14:42
3
$begingroup$
Why do you intend to delete this question?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:51
$begingroup$
@JoséCarlosSantos I thought the moderators will close or delete it anyway. Otherwise I would not.
$endgroup$
– Sara.T
Dec 9 '18 at 15:01
$begingroup$
@TobiasKildetoft I see, so my answer is not correct even if there is no embeding.
$endgroup$
– Sara.T
Dec 9 '18 at 15:03
$begingroup$
@Sara.T As rings, no, as $Bbb{R}$-algebras, yes. So as you've written your question, no.
$endgroup$
– jgon
Dec 9 '18 at 15:06
add a comment |
$begingroup$
I know this is not a typical question and I will delete it later.
Here is a question together my answer. Please let me know if my answer is correct. The reason that I am posting this is that I believe my answer is correct but my professor refuses to accept that as a correct answer (with no explanation).
Question: Is it possible to embed the ring $mathbb{R}[x]$ in $mathbb{C}$ ?
My Answer: No. Every element of $mathbb{C}$ is algebraic over $mathbb{R}$ so there is no copy of $mathbb{R}[x]$ in $mathbb{C}$ because $x$ is not algebraic over $mathbb{R}$.
abstract-algebra commutative-algebra extension-field
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
$endgroup$
I know this is not a typical question and I will delete it later.
Here is a question together my answer. Please let me know if my answer is correct. The reason that I am posting this is that I believe my answer is correct but my professor refuses to accept that as a correct answer (with no explanation).
Question: Is it possible to embed the ring $mathbb{R}[x]$ in $mathbb{C}$ ?
My Answer: No. Every element of $mathbb{C}$ is algebraic over $mathbb{R}$ so there is no copy of $mathbb{R}[x]$ in $mathbb{C}$ because $x$ is not algebraic over $mathbb{R}$.
abstract-algebra commutative-algebra extension-field
abstract-algebra commutative-algebra extension-field
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
asked Dec 9 '18 at 14:26
Sara.TSara.T
16610
16610
1
$begingroup$
You need to be more precise. Your argument shows that there is no embedding as an $mathbb{R}$-algebra. Actually I think there is an embedding as rings if you assume axiom of choice.
$endgroup$
– Tobias Kildetoft
Dec 9 '18 at 14:42
3
$begingroup$
Why do you intend to delete this question?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:51
$begingroup$
@JoséCarlosSantos I thought the moderators will close or delete it anyway. Otherwise I would not.
$endgroup$
– Sara.T
Dec 9 '18 at 15:01
$begingroup$
@TobiasKildetoft I see, so my answer is not correct even if there is no embeding.
$endgroup$
– Sara.T
Dec 9 '18 at 15:03
$begingroup$
@Sara.T As rings, no, as $Bbb{R}$-algebras, yes. So as you've written your question, no.
$endgroup$
– jgon
Dec 9 '18 at 15:06
add a comment |
1
$begingroup$
You need to be more precise. Your argument shows that there is no embedding as an $mathbb{R}$-algebra. Actually I think there is an embedding as rings if you assume axiom of choice.
$endgroup$
– Tobias Kildetoft
Dec 9 '18 at 14:42
3
$begingroup$
Why do you intend to delete this question?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:51
$begingroup$
@JoséCarlosSantos I thought the moderators will close or delete it anyway. Otherwise I would not.
$endgroup$
– Sara.T
Dec 9 '18 at 15:01
$begingroup$
@TobiasKildetoft I see, so my answer is not correct even if there is no embeding.
$endgroup$
– Sara.T
Dec 9 '18 at 15:03
$begingroup$
@Sara.T As rings, no, as $Bbb{R}$-algebras, yes. So as you've written your question, no.
$endgroup$
– jgon
Dec 9 '18 at 15:06
1
1
$begingroup$
You need to be more precise. Your argument shows that there is no embedding as an $mathbb{R}$-algebra. Actually I think there is an embedding as rings if you assume axiom of choice.
$endgroup$
– Tobias Kildetoft
Dec 9 '18 at 14:42
$begingroup$
You need to be more precise. Your argument shows that there is no embedding as an $mathbb{R}$-algebra. Actually I think there is an embedding as rings if you assume axiom of choice.
$endgroup$
– Tobias Kildetoft
Dec 9 '18 at 14:42
3
3
$begingroup$
Why do you intend to delete this question?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:51
$begingroup$
Why do you intend to delete this question?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:51
$begingroup$
@JoséCarlosSantos I thought the moderators will close or delete it anyway. Otherwise I would not.
$endgroup$
– Sara.T
Dec 9 '18 at 15:01
$begingroup$
@JoséCarlosSantos I thought the moderators will close or delete it anyway. Otherwise I would not.
$endgroup$
– Sara.T
Dec 9 '18 at 15:01
$begingroup$
@TobiasKildetoft I see, so my answer is not correct even if there is no embeding.
$endgroup$
– Sara.T
Dec 9 '18 at 15:03
$begingroup$
@TobiasKildetoft I see, so my answer is not correct even if there is no embeding.
$endgroup$
– Sara.T
Dec 9 '18 at 15:03
$begingroup$
@Sara.T As rings, no, as $Bbb{R}$-algebras, yes. So as you've written your question, no.
$endgroup$
– jgon
Dec 9 '18 at 15:06
$begingroup$
@Sara.T As rings, no, as $Bbb{R}$-algebras, yes. So as you've written your question, no.
$endgroup$
– jgon
Dec 9 '18 at 15:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Just to fill in the details of Tobias Kildetoft's comment. Not sure if they'll be useful to you, but I thought it was interesting.
Let's construct an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
Step 1 Observe that the transcendence degree of $Bbb{C}$ over $Bbb{Q}$ is $mathfrak{c}$ (though note that for our argument, it doesn't matter exactly what it is as long as it's infinite).
This follows from simple cardinality arguments. Any field algebraic over a field with an infinite cardinality has the same cardinality. If a field has cardinality $alpha$ and we adjoin a cardinality $beta$'s worth of indeterminates, the result will have cardinality $max{alpha,beta}$.
Step 2 Let $T$ be a transcendence basis for $Bbb{C}$ over $Bbb{Q}$. Since $Bbb{C}$ is algebraically closed and is algebraic over the subfield to $Bbb{Q}(T)$, it is isomorphic to the algebraic closure of $Bbb{Q}(T)$. Choose a proper subset $S$ of $T$ of the same cardinality as $T$. Then the algebraic closure of $Bbb{Q}(S)$ in $Bbb{C}$ is clearly isomorphic to $Bbb{C}$ as rings, since $Bbb{Q}(S)cong Bbb{Q}(T)$, but it is necessarily a proper subset of $Bbb{C}$, and doesn't contain any of the elements in $Tsetminus S$.
Step 3 By step 2, we have a (necessarily injective) ring homomorphism $phi :Bbb{C}toBbb{C}$ which isn't surjective. Let $t in Tsetminus S$. The ring homomorphism $phi$ extends to an injective ring homomorphism
$psi : Bbb{C}[x]to Bbb{C}$ sending $x$ to $t$. Restricting to the obvious subset $Bbb{R}[x]subset Bbb{C}[x]$ we have found an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
$endgroup$
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
1
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
add a comment |
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1 Answer
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$begingroup$
Just to fill in the details of Tobias Kildetoft's comment. Not sure if they'll be useful to you, but I thought it was interesting.
Let's construct an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
Step 1 Observe that the transcendence degree of $Bbb{C}$ over $Bbb{Q}$ is $mathfrak{c}$ (though note that for our argument, it doesn't matter exactly what it is as long as it's infinite).
This follows from simple cardinality arguments. Any field algebraic over a field with an infinite cardinality has the same cardinality. If a field has cardinality $alpha$ and we adjoin a cardinality $beta$'s worth of indeterminates, the result will have cardinality $max{alpha,beta}$.
Step 2 Let $T$ be a transcendence basis for $Bbb{C}$ over $Bbb{Q}$. Since $Bbb{C}$ is algebraically closed and is algebraic over the subfield to $Bbb{Q}(T)$, it is isomorphic to the algebraic closure of $Bbb{Q}(T)$. Choose a proper subset $S$ of $T$ of the same cardinality as $T$. Then the algebraic closure of $Bbb{Q}(S)$ in $Bbb{C}$ is clearly isomorphic to $Bbb{C}$ as rings, since $Bbb{Q}(S)cong Bbb{Q}(T)$, but it is necessarily a proper subset of $Bbb{C}$, and doesn't contain any of the elements in $Tsetminus S$.
Step 3 By step 2, we have a (necessarily injective) ring homomorphism $phi :Bbb{C}toBbb{C}$ which isn't surjective. Let $t in Tsetminus S$. The ring homomorphism $phi$ extends to an injective ring homomorphism
$psi : Bbb{C}[x]to Bbb{C}$ sending $x$ to $t$. Restricting to the obvious subset $Bbb{R}[x]subset Bbb{C}[x]$ we have found an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
$endgroup$
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
1
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
add a comment |
$begingroup$
Just to fill in the details of Tobias Kildetoft's comment. Not sure if they'll be useful to you, but I thought it was interesting.
Let's construct an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
Step 1 Observe that the transcendence degree of $Bbb{C}$ over $Bbb{Q}$ is $mathfrak{c}$ (though note that for our argument, it doesn't matter exactly what it is as long as it's infinite).
This follows from simple cardinality arguments. Any field algebraic over a field with an infinite cardinality has the same cardinality. If a field has cardinality $alpha$ and we adjoin a cardinality $beta$'s worth of indeterminates, the result will have cardinality $max{alpha,beta}$.
Step 2 Let $T$ be a transcendence basis for $Bbb{C}$ over $Bbb{Q}$. Since $Bbb{C}$ is algebraically closed and is algebraic over the subfield to $Bbb{Q}(T)$, it is isomorphic to the algebraic closure of $Bbb{Q}(T)$. Choose a proper subset $S$ of $T$ of the same cardinality as $T$. Then the algebraic closure of $Bbb{Q}(S)$ in $Bbb{C}$ is clearly isomorphic to $Bbb{C}$ as rings, since $Bbb{Q}(S)cong Bbb{Q}(T)$, but it is necessarily a proper subset of $Bbb{C}$, and doesn't contain any of the elements in $Tsetminus S$.
Step 3 By step 2, we have a (necessarily injective) ring homomorphism $phi :Bbb{C}toBbb{C}$ which isn't surjective. Let $t in Tsetminus S$. The ring homomorphism $phi$ extends to an injective ring homomorphism
$psi : Bbb{C}[x]to Bbb{C}$ sending $x$ to $t$. Restricting to the obvious subset $Bbb{R}[x]subset Bbb{C}[x]$ we have found an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
$endgroup$
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
1
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
add a comment |
$begingroup$
Just to fill in the details of Tobias Kildetoft's comment. Not sure if they'll be useful to you, but I thought it was interesting.
Let's construct an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
Step 1 Observe that the transcendence degree of $Bbb{C}$ over $Bbb{Q}$ is $mathfrak{c}$ (though note that for our argument, it doesn't matter exactly what it is as long as it's infinite).
This follows from simple cardinality arguments. Any field algebraic over a field with an infinite cardinality has the same cardinality. If a field has cardinality $alpha$ and we adjoin a cardinality $beta$'s worth of indeterminates, the result will have cardinality $max{alpha,beta}$.
Step 2 Let $T$ be a transcendence basis for $Bbb{C}$ over $Bbb{Q}$. Since $Bbb{C}$ is algebraically closed and is algebraic over the subfield to $Bbb{Q}(T)$, it is isomorphic to the algebraic closure of $Bbb{Q}(T)$. Choose a proper subset $S$ of $T$ of the same cardinality as $T$. Then the algebraic closure of $Bbb{Q}(S)$ in $Bbb{C}$ is clearly isomorphic to $Bbb{C}$ as rings, since $Bbb{Q}(S)cong Bbb{Q}(T)$, but it is necessarily a proper subset of $Bbb{C}$, and doesn't contain any of the elements in $Tsetminus S$.
Step 3 By step 2, we have a (necessarily injective) ring homomorphism $phi :Bbb{C}toBbb{C}$ which isn't surjective. Let $t in Tsetminus S$. The ring homomorphism $phi$ extends to an injective ring homomorphism
$psi : Bbb{C}[x]to Bbb{C}$ sending $x$ to $t$. Restricting to the obvious subset $Bbb{R}[x]subset Bbb{C}[x]$ we have found an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
$endgroup$
Just to fill in the details of Tobias Kildetoft's comment. Not sure if they'll be useful to you, but I thought it was interesting.
Let's construct an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
Step 1 Observe that the transcendence degree of $Bbb{C}$ over $Bbb{Q}$ is $mathfrak{c}$ (though note that for our argument, it doesn't matter exactly what it is as long as it's infinite).
This follows from simple cardinality arguments. Any field algebraic over a field with an infinite cardinality has the same cardinality. If a field has cardinality $alpha$ and we adjoin a cardinality $beta$'s worth of indeterminates, the result will have cardinality $max{alpha,beta}$.
Step 2 Let $T$ be a transcendence basis for $Bbb{C}$ over $Bbb{Q}$. Since $Bbb{C}$ is algebraically closed and is algebraic over the subfield to $Bbb{Q}(T)$, it is isomorphic to the algebraic closure of $Bbb{Q}(T)$. Choose a proper subset $S$ of $T$ of the same cardinality as $T$. Then the algebraic closure of $Bbb{Q}(S)$ in $Bbb{C}$ is clearly isomorphic to $Bbb{C}$ as rings, since $Bbb{Q}(S)cong Bbb{Q}(T)$, but it is necessarily a proper subset of $Bbb{C}$, and doesn't contain any of the elements in $Tsetminus S$.
Step 3 By step 2, we have a (necessarily injective) ring homomorphism $phi :Bbb{C}toBbb{C}$ which isn't surjective. Let $t in Tsetminus S$. The ring homomorphism $phi$ extends to an injective ring homomorphism
$psi : Bbb{C}[x]to Bbb{C}$ sending $x$ to $t$. Restricting to the obvious subset $Bbb{R}[x]subset Bbb{C}[x]$ we have found an embedding of $Bbb{R}[x]$ into $Bbb{C}$.
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
edited Dec 9 '18 at 18:12
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
answered Dec 9 '18 at 18:07
jgonjgon
14.1k22041
14.1k22041
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
1
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
add a comment |
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
1
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
$begingroup$
thanks. How about $mathbb{R}[[x]]$ ?
$endgroup$
– Sara.T
Dec 9 '18 at 20:00
1
1
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
$begingroup$
@Sara.T Interesting question, but I'm not sure. You could ask a new question. Since you're still somewhat new to the site, I'll just advise you that if you do ask a new question, it's a good idea to explain that it's a follow up question to this one.
$endgroup$
– jgon
Dec 9 '18 at 21:04
add a comment |
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1
$begingroup$
You need to be more precise. Your argument shows that there is no embedding as an $mathbb{R}$-algebra. Actually I think there is an embedding as rings if you assume axiom of choice.
$endgroup$
– Tobias Kildetoft
Dec 9 '18 at 14:42
3
$begingroup$
Why do you intend to delete this question?
$endgroup$
– José Carlos Santos
Dec 9 '18 at 14:51
$begingroup$
@JoséCarlosSantos I thought the moderators will close or delete it anyway. Otherwise I would not.
$endgroup$
– Sara.T
Dec 9 '18 at 15:01
$begingroup$
@TobiasKildetoft I see, so my answer is not correct even if there is no embeding.
$endgroup$
– Sara.T
Dec 9 '18 at 15:03
$begingroup$
@Sara.T As rings, no, as $Bbb{R}$-algebras, yes. So as you've written your question, no.
$endgroup$
– jgon
Dec 9 '18 at 15:06