Why is the radical ideal of the homogenization of an ideal the homogenization of the radical of said ideal?
1
$begingroup$
Let $I$ be an ideal. If $I^{h}$ is the homogenization of that ideal, defined by $I^{h}:= langle f^h: f in Irangle$ that is the ideal generated by the homogenized elements of I, then why is $sqrt{I^h}=sqrt{I}^h$?
I know, that for one single element $(f^h)^n=(f^n)^h$, but I have trouble representing an arbitrary element in $I^h$, because they do not need to be homogeneous anymore.
abstract-algebra algebraic-geometry ideals
edited Dec 9 '18 at 0:23
KReiser
9,58721435
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
$endgroup$
add a comment |
1
$begingroup$
Let $I$ be an ideal. If $I^{h}$ is the homogenization of that ideal, defined by $I^{h}:= langle f^h: f in Irangle$ that is the ideal generated by the homogenized elements of I, then why is $sqrt{I^h}=sqrt{I}^h$?
I know, that for one single element $(f^h)^n=(f^n)^h$, but I have trouble representing an arbitrary element in $I^h$, because they do not need to be homogeneous anymore.
abstract-algebra algebraic-geometry ideals
edited Dec 9 '18 at 0:23
KReiser
9,58721435
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
$endgroup$
1
$begingroup$
Welcome to the website! Does this help to answer your question? math.stackexchange.com/questions/2111660/… Or is there still more which is unclear?
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– Chill2Macht
Dec 9 '18 at 1:28
$begingroup$
Unfortunately I am still stuck. My ideal is an abitrary one and I do not see how computing $f_*$ might help me.
$endgroup$
– Ozymandias
Dec 9 '18 at 11:09
$begingroup$
I guess I just meant that the if the homogenization of the radical is radical, then it is equal to its own radicalization, i.e. the radicalization of the homogenization (of the radicalization). So if you can show that the homogenization of the radicalization is equal to the homogenization, then you are done. Actually that sounds circular to me now that I say it out loud though.
$endgroup$
– Chill2Macht
Dec 9 '18 at 18:28
add a comment |
1
1
1
$begingroup$
Let $I$ be an ideal. If $I^{h}$ is the homogenization of that ideal, defined by $I^{h}:= langle f^h: f in Irangle$ that is the ideal generated by the homogenized elements of I, then why is $sqrt{I^h}=sqrt{I}^h$?
I know, that for one single element $(f^h)^n=(f^n)^h$, but I have trouble representing an arbitrary element in $I^h$, because they do not need to be homogeneous anymore.
abstract-algebra algebraic-geometry ideals
edited Dec 9 '18 at 0:23
KReiser
9,58721435
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
$endgroup$
Let $I$ be an ideal. If $I^{h}$ is the homogenization of that ideal, defined by $I^{h}:= langle f^h: f in Irangle$ that is the ideal generated by the homogenized elements of I, then why is $sqrt{I^h}=sqrt{I}^h$?
I know, that for one single element $(f^h)^n=(f^n)^h$, but I have trouble representing an arbitrary element in $I^h$, because they do not need to be homogeneous anymore.
abstract-algebra algebraic-geometry ideals
abstract-algebra algebraic-geometry ideals
edited Dec 9 '18 at 0:23
KReiser
9,58721435
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
edited Dec 9 '18 at 0:23
KReiser
9,58721435
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
edited Dec 9 '18 at 0:23
KReiser
9,58721435
edited Dec 9 '18 at 0:23
KReiser
9,58721435
edited Dec 9 '18 at 0:23
KReiser
9,58721435
9,58721435
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
asked Dec 8 '18 at 19:08
OzymandiasOzymandias
62
62
1
$begingroup$
Welcome to the website! Does this help to answer your question? math.stackexchange.com/questions/2111660/… Or is there still more which is unclear?
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:28
$begingroup$
Unfortunately I am still stuck. My ideal is an abitrary one and I do not see how computing $f_*$ might help me.
$endgroup$
– Ozymandias
Dec 9 '18 at 11:09
$begingroup$
I guess I just meant that the if the homogenization of the radical is radical, then it is equal to its own radicalization, i.e. the radicalization of the homogenization (of the radicalization). So if you can show that the homogenization of the radicalization is equal to the homogenization, then you are done. Actually that sounds circular to me now that I say it out loud though.
$endgroup$
– Chill2Macht
Dec 9 '18 at 18:28
add a comment |
1
$begingroup$
Welcome to the website! Does this help to answer your question? math.stackexchange.com/questions/2111660/… Or is there still more which is unclear?
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:28
$begingroup$
Unfortunately I am still stuck. My ideal is an abitrary one and I do not see how computing $f_*$ might help me.
$endgroup$
– Ozymandias
Dec 9 '18 at 11:09
$begingroup$
I guess I just meant that the if the homogenization of the radical is radical, then it is equal to its own radicalization, i.e. the radicalization of the homogenization (of the radicalization). So if you can show that the homogenization of the radicalization is equal to the homogenization, then you are done. Actually that sounds circular to me now that I say it out loud though.
$endgroup$
– Chill2Macht
Dec 9 '18 at 18:28
1
1
$begingroup$
Welcome to the website! Does this help to answer your question? math.stackexchange.com/questions/2111660/… Or is there still more which is unclear?
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:28
$begingroup$
Welcome to the website! Does this help to answer your question? math.stackexchange.com/questions/2111660/… Or is there still more which is unclear?
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:28
$begingroup$
Unfortunately I am still stuck. My ideal is an abitrary one and I do not see how computing $f_*$ might help me.
$endgroup$
– Ozymandias
Dec 9 '18 at 11:09
$begingroup$
Unfortunately I am still stuck. My ideal is an abitrary one and I do not see how computing $f_*$ might help me.
$endgroup$
– Ozymandias
Dec 9 '18 at 11:09
$begingroup$
I guess I just meant that the if the homogenization of the radical is radical, then it is equal to its own radicalization, i.e. the radicalization of the homogenization (of the radicalization). So if you can show that the homogenization of the radicalization is equal to the homogenization, then you are done. Actually that sounds circular to me now that I say it out loud though.
$endgroup$
– Chill2Macht
Dec 9 '18 at 18:28
$begingroup$
I guess I just meant that the if the homogenization of the radical is radical, then it is equal to its own radicalization, i.e. the radicalization of the homogenization (of the radicalization). So if you can show that the homogenization of the radicalization is equal to the homogenization, then you are done. Actually that sounds circular to me now that I say it out loud though.
$endgroup$
– Chill2Macht
Dec 9 '18 at 18:28
add a comment |
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1
$begingroup$
Welcome to the website! Does this help to answer your question? math.stackexchange.com/questions/2111660/… Or is there still more which is unclear?
$endgroup$
– Chill2Macht
Dec 9 '18 at 1:28
$begingroup$
Unfortunately I am still stuck. My ideal is an abitrary one and I do not see how computing $f_*$ might help me.
$endgroup$
– Ozymandias
Dec 9 '18 at 11:09
$begingroup$
I guess I just meant that the if the homogenization of the radical is radical, then it is equal to its own radicalization, i.e. the radicalization of the homogenization (of the radicalization). So if you can show that the homogenization of the radicalization is equal to the homogenization, then you are done. Actually that sounds circular to me now that I say it out loud though.
$endgroup$
– Chill2Macht
Dec 9 '18 at 18:28