double quotient ring isomorphic to polynomials ring
$begingroup$
This question is quite specific, so apologies in advance. I don't have a clue of how to approach this. Answers and links to more general versions of this will also be very welcomed.
Let $F$ be a field and $A = F[x,y,z]$. $J=langle xy-z^2 rangle$ is an ideal in $A$. Denote $R=A/J$, and let $I=langle x+J, y+J rangle =langle x+J rangle + langle y+J rangle $ be an ideal in $R$.
I need to prove that $R/I cong F[w]$.
abstract-algebra ring-theory
asked Jan 17 at 20:20
just mejust me
184
$endgroup$
|
show 6 more comments
$begingroup$
This question is quite specific, so apologies in advance. I don't have a clue of how to approach this. Answers and links to more general versions of this will also be very welcomed.
Let $F$ be a field and $A = F[x,y,z]$. $J=langle xy-z^2 rangle$ is an ideal in $A$. Denote $R=A/J$, and let $I=langle x+J, y+J rangle =langle x+J rangle + langle y+J rangle $ be an ideal in $R$.
I need to prove that $R/I cong F[w]$.
abstract-algebra ring-theory
asked Jan 17 at 20:20
just mejust me
184
$endgroup$
$begingroup$
What is $w$ supposed to be?
$endgroup$
– Bernard
Jan 17 at 20:25
1
$begingroup$
$w$ is just another element added to $F$, just like $x,y,z$. I assume another letter is used so it won't be confusing
$endgroup$
– just me
Jan 17 at 20:27
1
$begingroup$
This isn't true, if $F[w]$ is supposed to denote a polynomial ring.
$endgroup$
– Eric Wofsey
Jan 17 at 20:31
3
$begingroup$
I don’t see how this could be true. This is equivalent to taking $F[x,y,z]/K$, where $K$ is generated by $x$, $y$, and $xy-z^2$, which is the same as the ideal generated by $x$, $y$, and $z^2$. But in this quotient, $z+Kneq 0$, since $znotin K$, but $(z+K)^2 = 0$, But $F[w]$ has no nonzero nilpotent elements. When you say you “need to prove it”, does that mean you were told it was true, or does it mean you really, really, really want it to be true, but don’t know if it is?
$endgroup$
– Arturo Magidin
Jan 17 at 20:32
1
$begingroup$
@Servaes: Sigh, fine. “If that is what you were told, then what you were told is incorrect.” (As opposed to what he heard/understood/deduced being incorrect). Happy now? ;-)
$endgroup$
– Arturo Magidin
Jan 17 at 22:26
|
show 6 more comments
$begingroup$
This question is quite specific, so apologies in advance. I don't have a clue of how to approach this. Answers and links to more general versions of this will also be very welcomed.
Let $F$ be a field and $A = F[x,y,z]$. $J=langle xy-z^2 rangle$ is an ideal in $A$. Denote $R=A/J$, and let $I=langle x+J, y+J rangle =langle x+J rangle + langle y+J rangle $ be an ideal in $R$.
I need to prove that $R/I cong F[w]$.
abstract-algebra ring-theory
asked Jan 17 at 20:20
just mejust me
184
$endgroup$
This question is quite specific, so apologies in advance. I don't have a clue of how to approach this. Answers and links to more general versions of this will also be very welcomed.
Let $F$ be a field and $A = F[x,y,z]$. $J=langle xy-z^2 rangle$ is an ideal in $A$. Denote $R=A/J$, and let $I=langle x+J, y+J rangle =langle x+J rangle + langle y+J rangle $ be an ideal in $R$.
I need to prove that $R/I cong F[w]$.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Jan 17 at 20:20
just mejust me
184
asked Jan 17 at 20:20
just mejust me
184
edited Jan 17 at 20:40
just me
asked Jan 17 at 20:20
just mejust me
184
asked Jan 17 at 20:20
just mejust me
184
asked Jan 17 at 20:20
just mejust me
184
184
$begingroup$
What is $w$ supposed to be?
$endgroup$
– Bernard
Jan 17 at 20:25
1
$begingroup$
$w$ is just another element added to $F$, just like $x,y,z$. I assume another letter is used so it won't be confusing
$endgroup$
– just me
Jan 17 at 20:27
1
$begingroup$
This isn't true, if $F[w]$ is supposed to denote a polynomial ring.
$endgroup$
– Eric Wofsey
Jan 17 at 20:31
3
$begingroup$
I don’t see how this could be true. This is equivalent to taking $F[x,y,z]/K$, where $K$ is generated by $x$, $y$, and $xy-z^2$, which is the same as the ideal generated by $x$, $y$, and $z^2$. But in this quotient, $z+Kneq 0$, since $znotin K$, but $(z+K)^2 = 0$, But $F[w]$ has no nonzero nilpotent elements. When you say you “need to prove it”, does that mean you were told it was true, or does it mean you really, really, really want it to be true, but don’t know if it is?
$endgroup$
– Arturo Magidin
Jan 17 at 20:32
1
$begingroup$
@Servaes: Sigh, fine. “If that is what you were told, then what you were told is incorrect.” (As opposed to what he heard/understood/deduced being incorrect). Happy now? ;-)
$endgroup$
– Arturo Magidin
Jan 17 at 22:26
|
show 6 more comments
$begingroup$
What is $w$ supposed to be?
$endgroup$
– Bernard
Jan 17 at 20:25
1
$begingroup$
$w$ is just another element added to $F$, just like $x,y,z$. I assume another letter is used so it won't be confusing
$endgroup$
– just me
Jan 17 at 20:27
1
$begingroup$
This isn't true, if $F[w]$ is supposed to denote a polynomial ring.
$endgroup$
– Eric Wofsey
Jan 17 at 20:31
3
$begingroup$
I don’t see how this could be true. This is equivalent to taking $F[x,y,z]/K$, where $K$ is generated by $x$, $y$, and $xy-z^2$, which is the same as the ideal generated by $x$, $y$, and $z^2$. But in this quotient, $z+Kneq 0$, since $znotin K$, but $(z+K)^2 = 0$, But $F[w]$ has no nonzero nilpotent elements. When you say you “need to prove it”, does that mean you were told it was true, or does it mean you really, really, really want it to be true, but don’t know if it is?
$endgroup$
– Arturo Magidin
Jan 17 at 20:32
1
$begingroup$
@Servaes: Sigh, fine. “If that is what you were told, then what you were told is incorrect.” (As opposed to what he heard/understood/deduced being incorrect). Happy now? ;-)
$endgroup$
– Arturo Magidin
Jan 17 at 22:26
$begingroup$
What is $w$ supposed to be?
$endgroup$
– Bernard
Jan 17 at 20:25
$begingroup$
What is $w$ supposed to be?
$endgroup$
– Bernard
Jan 17 at 20:25
1
1
$begingroup$
$w$ is just another element added to $F$, just like $x,y,z$. I assume another letter is used so it won't be confusing
$endgroup$
– just me
Jan 17 at 20:27
$begingroup$
$w$ is just another element added to $F$, just like $x,y,z$. I assume another letter is used so it won't be confusing
$endgroup$
– just me
Jan 17 at 20:27
1
1
$begingroup$
This isn't true, if $F[w]$ is supposed to denote a polynomial ring.
$endgroup$
– Eric Wofsey
Jan 17 at 20:31
$begingroup$
This isn't true, if $F[w]$ is supposed to denote a polynomial ring.
$endgroup$
– Eric Wofsey
Jan 17 at 20:31
3
3
$begingroup$
I don’t see how this could be true. This is equivalent to taking $F[x,y,z]/K$, where $K$ is generated by $x$, $y$, and $xy-z^2$, which is the same as the ideal generated by $x$, $y$, and $z^2$. But in this quotient, $z+Kneq 0$, since $znotin K$, but $(z+K)^2 = 0$, But $F[w]$ has no nonzero nilpotent elements. When you say you “need to prove it”, does that mean you were told it was true, or does it mean you really, really, really want it to be true, but don’t know if it is?
$endgroup$
– Arturo Magidin
Jan 17 at 20:32
$begingroup$
I don’t see how this could be true. This is equivalent to taking $F[x,y,z]/K$, where $K$ is generated by $x$, $y$, and $xy-z^2$, which is the same as the ideal generated by $x$, $y$, and $z^2$. But in this quotient, $z+Kneq 0$, since $znotin K$, but $(z+K)^2 = 0$, But $F[w]$ has no nonzero nilpotent elements. When you say you “need to prove it”, does that mean you were told it was true, or does it mean you really, really, really want it to be true, but don’t know if it is?
$endgroup$
– Arturo Magidin
Jan 17 at 20:32
1
1
$begingroup$
@Servaes: Sigh, fine. “If that is what you were told, then what you were told is incorrect.” (As opposed to what he heard/understood/deduced being incorrect). Happy now? ;-)
$endgroup$
– Arturo Magidin
Jan 17 at 22:26
$begingroup$
@Servaes: Sigh, fine. “If that is what you were told, then what you were told is incorrect.” (As opposed to what he heard/understood/deduced being incorrect). Happy now? ;-)
$endgroup$
– Arturo Magidin
Jan 17 at 22:26
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
By basic commutative algebra there are canonical isomorphisms
$$(A/langle xy-z^2rangle)/langle x+J,y+Jranglecong A/langle x,y,xy-z^2rangle=A/langle x,y,z^2ranglecong F[z]/langle z^2rangle,$$
where the middle identity comes frome the simple fact that $langle x,y,xy-z^2rangle=langle x,y,z^2rangle$. It is not hard to prove (exercise for you?) that for any field $F$ you have
$$F[z]/langle z^2ranglenotcong F[w].$$
In particular, what you are asked to prove is false.
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
Assuming you meant $I = langle x + J, z + Jrangle $:
There’s a bijection between $I subseteq A / J$ and $J subseteq I’ subseteq A$ given by $pi:A to A / J$, $x mapsto x + J$, which gives $pi (I’) = I$.
We can verify that $pi (xA + zA) = I$, that is
$$I = I’ / J = (xA + zA) / J$$
By the Third Isomorphism Theorem we have
$$R/I = A/J/I = (A/J)/(A/I’) simeq A /I’ = A/(xA+zA)$$
Finally, we can construct an isomorphism (“cancel out x and z”) and see that
$$A/(xA+zA) = F[x, y, z] / (xF[x, y, z] + zF[x, y, z]) simeq F[w]$$
Thus $R/I simeq F[w]$.
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
answered Jan 17 at 23:14
TroubleTrouble
46429
$endgroup$
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Uselangleandrangle, not<and>. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
add a comment |
$begingroup$
Let's try and clear up things before starting. If $tilde{x}=x+J$, $tilde{y}=y+J$ and $tilde{z}=z+J$ as elements of $R=A/J$, then all three elements are nonzero (easy degree considerations); however, $tilde{z}^2=0+Jin I=langletilde{x},tilde{y}rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/Icong F[w]$ should be interpreted in the sense
prove that $R/I$ is generated by a single element as an $F$-algebra.
It is not possible that $w$ is an indeterminate over $F$, because $R/I$ is not a domain.
Let $A$ be a ring with an ideal $J$. If $R=A/J$ and $I$ is an ideal of $R$, there is a unique ideal $I'$ of $A$ such that $Jsubseteq I'$ and $I=I'/J$.
Also, $R/Icong A/I'$.
What's $I'$ in your case? It is simply $langle x,yrangle+J$ (prove it).
Therefore the ring $A/I'$ is generated, as an $F$-algebra, by a single element, namely the image of $z$.
answered Jan 18 at 8:23
egregegreg
186k1486208
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
By basic commutative algebra there are canonical isomorphisms
$$(A/langle xy-z^2rangle)/langle x+J,y+Jranglecong A/langle x,y,xy-z^2rangle=A/langle x,y,z^2ranglecong F[z]/langle z^2rangle,$$
where the middle identity comes frome the simple fact that $langle x,y,xy-z^2rangle=langle x,y,z^2rangle$. It is not hard to prove (exercise for you?) that for any field $F$ you have
$$F[z]/langle z^2ranglenotcong F[w].$$
In particular, what you are asked to prove is false.
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
By basic commutative algebra there are canonical isomorphisms
$$(A/langle xy-z^2rangle)/langle x+J,y+Jranglecong A/langle x,y,xy-z^2rangle=A/langle x,y,z^2ranglecong F[z]/langle z^2rangle,$$
where the middle identity comes frome the simple fact that $langle x,y,xy-z^2rangle=langle x,y,z^2rangle$. It is not hard to prove (exercise for you?) that for any field $F$ you have
$$F[z]/langle z^2ranglenotcong F[w].$$
In particular, what you are asked to prove is false.
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
By basic commutative algebra there are canonical isomorphisms
$$(A/langle xy-z^2rangle)/langle x+J,y+Jranglecong A/langle x,y,xy-z^2rangle=A/langle x,y,z^2ranglecong F[z]/langle z^2rangle,$$
where the middle identity comes frome the simple fact that $langle x,y,xy-z^2rangle=langle x,y,z^2rangle$. It is not hard to prove (exercise for you?) that for any field $F$ you have
$$F[z]/langle z^2ranglenotcong F[w].$$
In particular, what you are asked to prove is false.
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
$endgroup$
By basic commutative algebra there are canonical isomorphisms
$$(A/langle xy-z^2rangle)/langle x+J,y+Jranglecong A/langle x,y,xy-z^2rangle=A/langle x,y,z^2ranglecong F[z]/langle z^2rangle,$$
where the middle identity comes frome the simple fact that $langle x,y,xy-z^2rangle=langle x,y,z^2rangle$. It is not hard to prove (exercise for you?) that for any field $F$ you have
$$F[z]/langle z^2ranglenotcong F[w].$$
In particular, what you are asked to prove is false.
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
answered Jan 17 at 21:57
ServaesServaes
30.6k342101
30.6k342101
add a comment |
add a comment |
$begingroup$
Assuming you meant $I = langle x + J, z + Jrangle $:
There’s a bijection between $I subseteq A / J$ and $J subseteq I’ subseteq A$ given by $pi:A to A / J$, $x mapsto x + J$, which gives $pi (I’) = I$.
We can verify that $pi (xA + zA) = I$, that is
$$I = I’ / J = (xA + zA) / J$$
By the Third Isomorphism Theorem we have
$$R/I = A/J/I = (A/J)/(A/I’) simeq A /I’ = A/(xA+zA)$$
Finally, we can construct an isomorphism (“cancel out x and z”) and see that
$$A/(xA+zA) = F[x, y, z] / (xF[x, y, z] + zF[x, y, z]) simeq F[w]$$
Thus $R/I simeq F[w]$.
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
answered Jan 17 at 23:14
TroubleTrouble
46429
$endgroup$
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Uselangleandrangle, not<and>. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
add a comment |
$begingroup$
Assuming you meant $I = langle x + J, z + Jrangle $:
There’s a bijection between $I subseteq A / J$ and $J subseteq I’ subseteq A$ given by $pi:A to A / J$, $x mapsto x + J$, which gives $pi (I’) = I$.
We can verify that $pi (xA + zA) = I$, that is
$$I = I’ / J = (xA + zA) / J$$
By the Third Isomorphism Theorem we have
$$R/I = A/J/I = (A/J)/(A/I’) simeq A /I’ = A/(xA+zA)$$
Finally, we can construct an isomorphism (“cancel out x and z”) and see that
$$A/(xA+zA) = F[x, y, z] / (xF[x, y, z] + zF[x, y, z]) simeq F[w]$$
Thus $R/I simeq F[w]$.
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
answered Jan 17 at 23:14
TroubleTrouble
46429
$endgroup$
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Uselangleandrangle, not<and>. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
add a comment |
$begingroup$
Assuming you meant $I = langle x + J, z + Jrangle $:
There’s a bijection between $I subseteq A / J$ and $J subseteq I’ subseteq A$ given by $pi:A to A / J$, $x mapsto x + J$, which gives $pi (I’) = I$.
We can verify that $pi (xA + zA) = I$, that is
$$I = I’ / J = (xA + zA) / J$$
By the Third Isomorphism Theorem we have
$$R/I = A/J/I = (A/J)/(A/I’) simeq A /I’ = A/(xA+zA)$$
Finally, we can construct an isomorphism (“cancel out x and z”) and see that
$$A/(xA+zA) = F[x, y, z] / (xF[x, y, z] + zF[x, y, z]) simeq F[w]$$
Thus $R/I simeq F[w]$.
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
answered Jan 17 at 23:14
TroubleTrouble
46429
$endgroup$
Assuming you meant $I = langle x + J, z + Jrangle $:
There’s a bijection between $I subseteq A / J$ and $J subseteq I’ subseteq A$ given by $pi:A to A / J$, $x mapsto x + J$, which gives $pi (I’) = I$.
We can verify that $pi (xA + zA) = I$, that is
$$I = I’ / J = (xA + zA) / J$$
By the Third Isomorphism Theorem we have
$$R/I = A/J/I = (A/J)/(A/I’) simeq A /I’ = A/(xA+zA)$$
Finally, we can construct an isomorphism (“cancel out x and z”) and see that
$$A/(xA+zA) = F[x, y, z] / (xF[x, y, z] + zF[x, y, z]) simeq F[w]$$
Thus $R/I simeq F[w]$.
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
answered Jan 17 at 23:14
TroubleTrouble
46429
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
edited Jan 18 at 18:03
Arturo Magidin
266k34591922
266k34591922
answered Jan 17 at 23:14
TroubleTrouble
46429
answered Jan 17 at 23:14
TroubleTrouble
46429
answered Jan 17 at 23:14
TroubleTrouble
46429
46429
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Uselangleandrangle, not<and>. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
add a comment |
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Uselangleandrangle, not<and>. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
This is actually a decent guess at what the assigner may have meant...
$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Use
langle and rangle, not < and >. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Use
langle and rangle, not < and >. The former are delimiters, the latter is a binary operator; they get different spacing assigned by TeX and MathJax.$endgroup$
– Arturo Magidin
Jan 18 at 18:02
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
$begingroup$
Noted, thank you!
$endgroup$
– Trouble
Jan 18 at 18:23
add a comment |
$begingroup$
Let's try and clear up things before starting. If $tilde{x}=x+J$, $tilde{y}=y+J$ and $tilde{z}=z+J$ as elements of $R=A/J$, then all three elements are nonzero (easy degree considerations); however, $tilde{z}^2=0+Jin I=langletilde{x},tilde{y}rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/Icong F[w]$ should be interpreted in the sense
prove that $R/I$ is generated by a single element as an $F$-algebra.
It is not possible that $w$ is an indeterminate over $F$, because $R/I$ is not a domain.
Let $A$ be a ring with an ideal $J$. If $R=A/J$ and $I$ is an ideal of $R$, there is a unique ideal $I'$ of $A$ such that $Jsubseteq I'$ and $I=I'/J$.
Also, $R/Icong A/I'$.
What's $I'$ in your case? It is simply $langle x,yrangle+J$ (prove it).
Therefore the ring $A/I'$ is generated, as an $F$-algebra, by a single element, namely the image of $z$.
answered Jan 18 at 8:23
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Let's try and clear up things before starting. If $tilde{x}=x+J$, $tilde{y}=y+J$ and $tilde{z}=z+J$ as elements of $R=A/J$, then all three elements are nonzero (easy degree considerations); however, $tilde{z}^2=0+Jin I=langletilde{x},tilde{y}rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/Icong F[w]$ should be interpreted in the sense
prove that $R/I$ is generated by a single element as an $F$-algebra.
It is not possible that $w$ is an indeterminate over $F$, because $R/I$ is not a domain.
Let $A$ be a ring with an ideal $J$. If $R=A/J$ and $I$ is an ideal of $R$, there is a unique ideal $I'$ of $A$ such that $Jsubseteq I'$ and $I=I'/J$.
Also, $R/Icong A/I'$.
What's $I'$ in your case? It is simply $langle x,yrangle+J$ (prove it).
Therefore the ring $A/I'$ is generated, as an $F$-algebra, by a single element, namely the image of $z$.
answered Jan 18 at 8:23
egregegreg
186k1486208
$endgroup$
add a comment |
$begingroup$
Let's try and clear up things before starting. If $tilde{x}=x+J$, $tilde{y}=y+J$ and $tilde{z}=z+J$ as elements of $R=A/J$, then all three elements are nonzero (easy degree considerations); however, $tilde{z}^2=0+Jin I=langletilde{x},tilde{y}rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/Icong F[w]$ should be interpreted in the sense
prove that $R/I$ is generated by a single element as an $F$-algebra.
It is not possible that $w$ is an indeterminate over $F$, because $R/I$ is not a domain.
Let $A$ be a ring with an ideal $J$. If $R=A/J$ and $I$ is an ideal of $R$, there is a unique ideal $I'$ of $A$ such that $Jsubseteq I'$ and $I=I'/J$.
Also, $R/Icong A/I'$.
What's $I'$ in your case? It is simply $langle x,yrangle+J$ (prove it).
Therefore the ring $A/I'$ is generated, as an $F$-algebra, by a single element, namely the image of $z$.
answered Jan 18 at 8:23
egregegreg
186k1486208
$endgroup$
Let's try and clear up things before starting. If $tilde{x}=x+J$, $tilde{y}=y+J$ and $tilde{z}=z+J$ as elements of $R=A/J$, then all three elements are nonzero (easy degree considerations); however, $tilde{z}^2=0+Jin I=langletilde{x},tilde{y}rangle$. Hence $I$ is not a prime ideal. Thus the request to prove that $R/Icong F[w]$ should be interpreted in the sense
prove that $R/I$ is generated by a single element as an $F$-algebra.
It is not possible that $w$ is an indeterminate over $F$, because $R/I$ is not a domain.
Let $A$ be a ring with an ideal $J$. If $R=A/J$ and $I$ is an ideal of $R$, there is a unique ideal $I'$ of $A$ such that $Jsubseteq I'$ and $I=I'/J$.
Also, $R/Icong A/I'$.
What's $I'$ in your case? It is simply $langle x,yrangle+J$ (prove it).
Therefore the ring $A/I'$ is generated, as an $F$-algebra, by a single element, namely the image of $z$.
answered Jan 18 at 8:23
egregegreg
186k1486208
answered Jan 18 at 8:23
egregegreg
186k1486208
answered Jan 18 at 8:23
egregegreg
186k1486208
answered Jan 18 at 8:23
egregegreg
186k1486208
186k1486208
add a comment |
add a comment |
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$begingroup$
What is $w$ supposed to be?
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– Bernard
Jan 17 at 20:25
1
$begingroup$
$w$ is just another element added to $F$, just like $x,y,z$. I assume another letter is used so it won't be confusing
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– just me
Jan 17 at 20:27
1
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This isn't true, if $F[w]$ is supposed to denote a polynomial ring.
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– Eric Wofsey
Jan 17 at 20:31
3
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I don’t see how this could be true. This is equivalent to taking $F[x,y,z]/K$, where $K$ is generated by $x$, $y$, and $xy-z^2$, which is the same as the ideal generated by $x$, $y$, and $z^2$. But in this quotient, $z+Kneq 0$, since $znotin K$, but $(z+K)^2 = 0$, But $F[w]$ has no nonzero nilpotent elements. When you say you “need to prove it”, does that mean you were told it was true, or does it mean you really, really, really want it to be true, but don’t know if it is?
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– Arturo Magidin
Jan 17 at 20:32
1
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@Servaes: Sigh, fine. “If that is what you were told, then what you were told is incorrect.” (As opposed to what he heard/understood/deduced being incorrect). Happy now? ;-)
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– Arturo Magidin
Jan 17 at 22:26