Justification of an isomorphism between $mathbb{Z}[t]/(t,3)$ and $mathbb{Z}/3$
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
asked Dec 2 at 3:23
J. Wang
524
add a comment |
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
asked Dec 2 at 3:23
J. Wang
524
add a comment |
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
asked Dec 2 at 3:23
J. Wang
524
I am wondering if the following is true:
$$mathbb{Z}[t]/(t,3)congmathbb{Z}/3.$$
I can tell that the ideal $(t,3)$ is maximal. In contrast, the ideal $(t^2,2t)$ is not even irreducible. So it's less likely that we can simplify $mathbb{Z}[t]/(t^2,2t)$ further.
If the aforementioned isomorphism is true, that means $mathbb{Z}[t]/(t,3)cong(mathbb{Z}[t]/t)/(3)congmathbb{Z}/3$, where the $(3)$ is an ideal in $mathbb{Z}[t]/t$. My questions are:
1) Is this argument legitimate?
2) If it is true, is there any theorem justifying the aforementioned isomorphism?
abstract-algebra ring-theory
abstract-algebra ring-theory
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
asked Dec 2 at 3:23
J. Wang
524
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
asked Dec 2 at 3:23
J. Wang
524
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
edited Dec 2 at 9:24
Asaf Karagila♦
301k32424755
301k32424755
asked Dec 2 at 3:23
J. Wang
524
asked Dec 2 at 3:23
J. Wang
524
asked Dec 2 at 3:23
J. Wang
524
524
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
It's legitimate and a consequence of the isomorphism theorems.
answered Dec 2 at 4:24
CyclotomicField
2,1921313
add a comment |
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2 Answers
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2 Answers
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Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
Yes your intuition is correct and I try to explain necessary steps:
Claim. $Bbb{Z}[x]/langle x,3ranglecongBbb{Z}/langle3rangle$
Well, I'll prove the following steps:
$$Bbb{Z}[x]/langle x,3ranglecong frac{Bbb{Z}[x]/langle xrangle}{langle x,3rangle/langle xrangle} congBbb{Z}/langle 3rangle$$
The first isomorphism is due to the 2nd Isomorphism Theorem of Ring theory.
I think $Bbb{Z}[x]/langle xrangle cong Bbb{Z}$ is clear. (Hint: consider the homomorphism $f(x) mapsto f(0)$ form $Bbb{Z}[x]$ to $Bbb{Z}$ and use 1st Isomorphism Theorem on Rings.
Now for the equality $langle x,3rangle/langle xranglecong langle 3rangle$ Try to define the homomorphism as follows:
$phi :langle x,3rangle to langle 3rangle$ by mapping any element $xf(x)+3g(x)in langle x,3 rangle$ to $3g(0) in langle 3rangle$ (N.B. Here, $f,g in Bbb{Z}[x]$), i.e. $$phi: xf(x)+3g(x) mapsto 3g(0)$$
See that this is an epimorphism. Also $ker phi=langle xrangle$. Then by 1st Isomorphism theorem $langle x,3rangle/ker phi cong langle 3rangle$. Which justifies all your steps...!!
N.B. To find out the $ker phi$, remember for any $g(x)in Bbb{Z}[x]$ if we have $g(0)=0$ that means $x$ is a factor of $g$ and so $gin langle xrangle$.
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
edited Dec 3 at 3:57
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
answered Dec 2 at 4:23
Indrajit Ghosh
1,0441718
1,0441718
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
$>$ is an operator; &rangle$ is a grouping symbol. They have different formatting rules, and are the reason for all the weird spaces in your answer.
– Bob Krueger
Dec 2 at 8:26
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
@BobKrueger...Thanks for mentioning that....I'll take care of it next time....:)
– Indrajit Ghosh
Dec 2 at 8:36
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
What's wrong with using the edit function to fix it this time as well? Is your time somehow more important than that of someone who might have edited this?
– Asaf Karagila♦
Dec 2 at 9:22
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
I’ll note that I didn’t edit it so that the poster would recognize the mistake. @IndrajitGhosh happy to help even if it’s not expressly math.
– Bob Krueger
Dec 2 at 15:28
add a comment |
It's legitimate and a consequence of the isomorphism theorems.
answered Dec 2 at 4:24
CyclotomicField
2,1921313
add a comment |
It's legitimate and a consequence of the isomorphism theorems.
answered Dec 2 at 4:24
CyclotomicField
2,1921313
add a comment |
It's legitimate and a consequence of the isomorphism theorems.
answered Dec 2 at 4:24
CyclotomicField
2,1921313
It's legitimate and a consequence of the isomorphism theorems.
answered Dec 2 at 4:24
CyclotomicField
2,1921313
answered Dec 2 at 4:24
CyclotomicField
2,1921313
answered Dec 2 at 4:24
CyclotomicField
2,1921313
answered Dec 2 at 4:24
CyclotomicField
2,1921313
2,1921313
add a comment |
add a comment |
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