When $langle f(x),y rangle$ is a radical ideal?
$begingroup$
Let $f(t) in mathbb{C}[t]$.
In the polynomial ring $mathbb{C}[x,y]$, denote by $I_{f(x),y}$
the ideal generated by $f(x)$ and $y$,
$I_{f(x),y}:= langle f(x),y rangle$.
Recall that an ideal $I$ in a ring $R$ is called radical, if $a^n in I$ implies that $a in I$.
When $langle f(x),y rangle$ is a radical ideal? (= for which $f(t) in mathbb{C}[t]$).
Examples:
(1) Take $f(t)=t^2$. Then $I_{x^2,y}=langle x^2,y rangle$
is not a radical ideal, since $x^2 in I_{x^2,y}$ but $x notin I_{x^2,y}$. Similarly, for $f(t)=t^m$, $I_{x^m,y}$ is not a radical ideal.
(2) For $f(t)=t$, the ideal $I_{x,y}$ is radical: $I_{x,y}$ is a maximal ideal (by Hilbert's Nullstellensatz), so it is a prime ideal (every maximal ideal is prime) and clearly a prime ideal is a radical ideal (follows immediately from the definitions).
Any hints and comments are welcome!
ring-theory commutative-algebra ideals
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
$endgroup$
add a comment |
$begingroup$
Let $f(t) in mathbb{C}[t]$.
In the polynomial ring $mathbb{C}[x,y]$, denote by $I_{f(x),y}$
the ideal generated by $f(x)$ and $y$,
$I_{f(x),y}:= langle f(x),y rangle$.
Recall that an ideal $I$ in a ring $R$ is called radical, if $a^n in I$ implies that $a in I$.
When $langle f(x),y rangle$ is a radical ideal? (= for which $f(t) in mathbb{C}[t]$).
Examples:
(1) Take $f(t)=t^2$. Then $I_{x^2,y}=langle x^2,y rangle$
is not a radical ideal, since $x^2 in I_{x^2,y}$ but $x notin I_{x^2,y}$. Similarly, for $f(t)=t^m$, $I_{x^m,y}$ is not a radical ideal.
(2) For $f(t)=t$, the ideal $I_{x,y}$ is radical: $I_{x,y}$ is a maximal ideal (by Hilbert's Nullstellensatz), so it is a prime ideal (every maximal ideal is prime) and clearly a prime ideal is a radical ideal (follows immediately from the definitions).
Any hints and comments are welcome!
ring-theory commutative-algebra ideals
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
$endgroup$
$begingroup$
$I$ is a radical ideal of $R$ if and only if $R/I$ is a reduced ring, i.e. it has no non-zero nilpotent element. Using this, $I$ is a radical ideal if and only if $mathbb{C}[x]/(f(x))$ is a reduced ring.
$endgroup$
– Krish
Dec 13 '18 at 15:22
$begingroup$
Thank you very much for your comment! Is it possible to obtain the exact form of such $f(t)$?
$endgroup$
– user237522
Dec 13 '18 at 15:26
$begingroup$
$f(t)$ splits into linear factors (since $mathbb C$ is algebraically closed). From this it's easy to see that $f(t)$ is of the form $(t-a_1)cdots (t-a_n)$ (assume the degree of $f$ is $n$) where $a_1, cdots , a_n$ are distinct complex numbers.
$endgroup$
– Krish
Dec 13 '18 at 15:58
$begingroup$
Thank you again! I see why, for example, $f(t)=(t-a)(t-b)^2$ then $I$ is not radical, since $((t-a)(t-b))^2=(t-a)^2(t-b)^2 in I$ but $(t-a)(t-b) notin I$. Similarly, more generally, I see why if $f(t)=(t-c_1)^{m_1}cdots(t-c_r)^{m_r}$ with at least one $m_j geq 2$, $1 leq j leq r$, then $I_{f(x),y}$ is not radical. In other words, I see why: If $I_{f(x),y}$ is radical then $f$ is separable. But I am not sure about the other direction, namely, why if $f(t)$ is separable then $I_{f(x),y}$ is radical.
$endgroup$
– user237522
Dec 13 '18 at 19:01
$begingroup$
I have just noticed math.stackexchange.com/questions/92391/…. But there the ideal is principal and here it is not.
$endgroup$
– user237522
Dec 13 '18 at 19:08
add a comment |
$begingroup$
Let $f(t) in mathbb{C}[t]$.
In the polynomial ring $mathbb{C}[x,y]$, denote by $I_{f(x),y}$
the ideal generated by $f(x)$ and $y$,
$I_{f(x),y}:= langle f(x),y rangle$.
Recall that an ideal $I$ in a ring $R$ is called radical, if $a^n in I$ implies that $a in I$.
When $langle f(x),y rangle$ is a radical ideal? (= for which $f(t) in mathbb{C}[t]$).
Examples:
(1) Take $f(t)=t^2$. Then $I_{x^2,y}=langle x^2,y rangle$
is not a radical ideal, since $x^2 in I_{x^2,y}$ but $x notin I_{x^2,y}$. Similarly, for $f(t)=t^m$, $I_{x^m,y}$ is not a radical ideal.
(2) For $f(t)=t$, the ideal $I_{x,y}$ is radical: $I_{x,y}$ is a maximal ideal (by Hilbert's Nullstellensatz), so it is a prime ideal (every maximal ideal is prime) and clearly a prime ideal is a radical ideal (follows immediately from the definitions).
Any hints and comments are welcome!
ring-theory commutative-algebra ideals
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
$endgroup$
Let $f(t) in mathbb{C}[t]$.
In the polynomial ring $mathbb{C}[x,y]$, denote by $I_{f(x),y}$
the ideal generated by $f(x)$ and $y$,
$I_{f(x),y}:= langle f(x),y rangle$.
Recall that an ideal $I$ in a ring $R$ is called radical, if $a^n in I$ implies that $a in I$.
When $langle f(x),y rangle$ is a radical ideal? (= for which $f(t) in mathbb{C}[t]$).
Examples:
(1) Take $f(t)=t^2$. Then $I_{x^2,y}=langle x^2,y rangle$
is not a radical ideal, since $x^2 in I_{x^2,y}$ but $x notin I_{x^2,y}$. Similarly, for $f(t)=t^m$, $I_{x^m,y}$ is not a radical ideal.
(2) For $f(t)=t$, the ideal $I_{x,y}$ is radical: $I_{x,y}$ is a maximal ideal (by Hilbert's Nullstellensatz), so it is a prime ideal (every maximal ideal is prime) and clearly a prime ideal is a radical ideal (follows immediately from the definitions).
Any hints and comments are welcome!
ring-theory commutative-algebra ideals
ring-theory commutative-algebra ideals
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
edited Dec 13 '18 at 15:20
user237522
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
asked Dec 13 '18 at 15:07
user237522user237522
2,1601617
2,1601617
$begingroup$
$I$ is a radical ideal of $R$ if and only if $R/I$ is a reduced ring, i.e. it has no non-zero nilpotent element. Using this, $I$ is a radical ideal if and only if $mathbb{C}[x]/(f(x))$ is a reduced ring.
$endgroup$
– Krish
Dec 13 '18 at 15:22
$begingroup$
Thank you very much for your comment! Is it possible to obtain the exact form of such $f(t)$?
$endgroup$
– user237522
Dec 13 '18 at 15:26
$begingroup$
$f(t)$ splits into linear factors (since $mathbb C$ is algebraically closed). From this it's easy to see that $f(t)$ is of the form $(t-a_1)cdots (t-a_n)$ (assume the degree of $f$ is $n$) where $a_1, cdots , a_n$ are distinct complex numbers.
$endgroup$
– Krish
Dec 13 '18 at 15:58
$begingroup$
Thank you again! I see why, for example, $f(t)=(t-a)(t-b)^2$ then $I$ is not radical, since $((t-a)(t-b))^2=(t-a)^2(t-b)^2 in I$ but $(t-a)(t-b) notin I$. Similarly, more generally, I see why if $f(t)=(t-c_1)^{m_1}cdots(t-c_r)^{m_r}$ with at least one $m_j geq 2$, $1 leq j leq r$, then $I_{f(x),y}$ is not radical. In other words, I see why: If $I_{f(x),y}$ is radical then $f$ is separable. But I am not sure about the other direction, namely, why if $f(t)$ is separable then $I_{f(x),y}$ is radical.
$endgroup$
– user237522
Dec 13 '18 at 19:01
$begingroup$
I have just noticed math.stackexchange.com/questions/92391/…. But there the ideal is principal and here it is not.
$endgroup$
– user237522
Dec 13 '18 at 19:08
add a comment |
$begingroup$
$I$ is a radical ideal of $R$ if and only if $R/I$ is a reduced ring, i.e. it has no non-zero nilpotent element. Using this, $I$ is a radical ideal if and only if $mathbb{C}[x]/(f(x))$ is a reduced ring.
$endgroup$
– Krish
Dec 13 '18 at 15:22
$begingroup$
Thank you very much for your comment! Is it possible to obtain the exact form of such $f(t)$?
$endgroup$
– user237522
Dec 13 '18 at 15:26
$begingroup$
$f(t)$ splits into linear factors (since $mathbb C$ is algebraically closed). From this it's easy to see that $f(t)$ is of the form $(t-a_1)cdots (t-a_n)$ (assume the degree of $f$ is $n$) where $a_1, cdots , a_n$ are distinct complex numbers.
$endgroup$
– Krish
Dec 13 '18 at 15:58
$begingroup$
Thank you again! I see why, for example, $f(t)=(t-a)(t-b)^2$ then $I$ is not radical, since $((t-a)(t-b))^2=(t-a)^2(t-b)^2 in I$ but $(t-a)(t-b) notin I$. Similarly, more generally, I see why if $f(t)=(t-c_1)^{m_1}cdots(t-c_r)^{m_r}$ with at least one $m_j geq 2$, $1 leq j leq r$, then $I_{f(x),y}$ is not radical. In other words, I see why: If $I_{f(x),y}$ is radical then $f$ is separable. But I am not sure about the other direction, namely, why if $f(t)$ is separable then $I_{f(x),y}$ is radical.
$endgroup$
– user237522
Dec 13 '18 at 19:01
$begingroup$
I have just noticed math.stackexchange.com/questions/92391/…. But there the ideal is principal and here it is not.
$endgroup$
– user237522
Dec 13 '18 at 19:08
$begingroup$
$I$ is a radical ideal of $R$ if and only if $R/I$ is a reduced ring, i.e. it has no non-zero nilpotent element. Using this, $I$ is a radical ideal if and only if $mathbb{C}[x]/(f(x))$ is a reduced ring.
$endgroup$
– Krish
Dec 13 '18 at 15:22
$begingroup$
$I$ is a radical ideal of $R$ if and only if $R/I$ is a reduced ring, i.e. it has no non-zero nilpotent element. Using this, $I$ is a radical ideal if and only if $mathbb{C}[x]/(f(x))$ is a reduced ring.
$endgroup$
– Krish
Dec 13 '18 at 15:22
$begingroup$
Thank you very much for your comment! Is it possible to obtain the exact form of such $f(t)$?
$endgroup$
– user237522
Dec 13 '18 at 15:26
$begingroup$
Thank you very much for your comment! Is it possible to obtain the exact form of such $f(t)$?
$endgroup$
– user237522
Dec 13 '18 at 15:26
$begingroup$
$f(t)$ splits into linear factors (since $mathbb C$ is algebraically closed). From this it's easy to see that $f(t)$ is of the form $(t-a_1)cdots (t-a_n)$ (assume the degree of $f$ is $n$) where $a_1, cdots , a_n$ are distinct complex numbers.
$endgroup$
– Krish
Dec 13 '18 at 15:58
$begingroup$
$f(t)$ splits into linear factors (since $mathbb C$ is algebraically closed). From this it's easy to see that $f(t)$ is of the form $(t-a_1)cdots (t-a_n)$ (assume the degree of $f$ is $n$) where $a_1, cdots , a_n$ are distinct complex numbers.
$endgroup$
– Krish
Dec 13 '18 at 15:58
$begingroup$
Thank you again! I see why, for example, $f(t)=(t-a)(t-b)^2$ then $I$ is not radical, since $((t-a)(t-b))^2=(t-a)^2(t-b)^2 in I$ but $(t-a)(t-b) notin I$. Similarly, more generally, I see why if $f(t)=(t-c_1)^{m_1}cdots(t-c_r)^{m_r}$ with at least one $m_j geq 2$, $1 leq j leq r$, then $I_{f(x),y}$ is not radical. In other words, I see why: If $I_{f(x),y}$ is radical then $f$ is separable. But I am not sure about the other direction, namely, why if $f(t)$ is separable then $I_{f(x),y}$ is radical.
$endgroup$
– user237522
Dec 13 '18 at 19:01
$begingroup$
Thank you again! I see why, for example, $f(t)=(t-a)(t-b)^2$ then $I$ is not radical, since $((t-a)(t-b))^2=(t-a)^2(t-b)^2 in I$ but $(t-a)(t-b) notin I$. Similarly, more generally, I see why if $f(t)=(t-c_1)^{m_1}cdots(t-c_r)^{m_r}$ with at least one $m_j geq 2$, $1 leq j leq r$, then $I_{f(x),y}$ is not radical. In other words, I see why: If $I_{f(x),y}$ is radical then $f$ is separable. But I am not sure about the other direction, namely, why if $f(t)$ is separable then $I_{f(x),y}$ is radical.
$endgroup$
– user237522
Dec 13 '18 at 19:01
$begingroup$
I have just noticed math.stackexchange.com/questions/92391/…. But there the ideal is principal and here it is not.
$endgroup$
– user237522
Dec 13 '18 at 19:08
$begingroup$
I have just noticed math.stackexchange.com/questions/92391/…. But there the ideal is principal and here it is not.
$endgroup$
– user237522
Dec 13 '18 at 19:08
add a comment |
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$begingroup$
$I$ is a radical ideal of $R$ if and only if $R/I$ is a reduced ring, i.e. it has no non-zero nilpotent element. Using this, $I$ is a radical ideal if and only if $mathbb{C}[x]/(f(x))$ is a reduced ring.
$endgroup$
– Krish
Dec 13 '18 at 15:22
$begingroup$
Thank you very much for your comment! Is it possible to obtain the exact form of such $f(t)$?
$endgroup$
– user237522
Dec 13 '18 at 15:26
$begingroup$
$f(t)$ splits into linear factors (since $mathbb C$ is algebraically closed). From this it's easy to see that $f(t)$ is of the form $(t-a_1)cdots (t-a_n)$ (assume the degree of $f$ is $n$) where $a_1, cdots , a_n$ are distinct complex numbers.
$endgroup$
– Krish
Dec 13 '18 at 15:58
$begingroup$
Thank you again! I see why, for example, $f(t)=(t-a)(t-b)^2$ then $I$ is not radical, since $((t-a)(t-b))^2=(t-a)^2(t-b)^2 in I$ but $(t-a)(t-b) notin I$. Similarly, more generally, I see why if $f(t)=(t-c_1)^{m_1}cdots(t-c_r)^{m_r}$ with at least one $m_j geq 2$, $1 leq j leq r$, then $I_{f(x),y}$ is not radical. In other words, I see why: If $I_{f(x),y}$ is radical then $f$ is separable. But I am not sure about the other direction, namely, why if $f(t)$ is separable then $I_{f(x),y}$ is radical.
$endgroup$
– user237522
Dec 13 '18 at 19:01
$begingroup$
I have just noticed math.stackexchange.com/questions/92391/…. But there the ideal is principal and here it is not.
$endgroup$
– user237522
Dec 13 '18 at 19:08