Describing the elements of quotient ring of $mathbbZ[sqrtD]$.Why is $mathbbZ[sqrt-n], nge 3$ not a UFD?Does $mathbbZ[sqrt[3]2]=mathbbZ[sqrt[3]2]^mathrmint$?In which extensions of type $mathbbQ(sqrtd)$ is the number 3 reducible?Why is quadratic integer ring defined in that way?Maximal ideal in a polynomial ring over $mathbb Z$Quotient ring $R/I$, $R= mathbbZ[sqrt-10]$, $I =(sqrt-10)$About the ring $mathbbZ[sqrt4]$Tensor algebra on non free modulesMethod to adjoin elements to a ring - is the symbol $x$ overloaded?Prime ideals in certain quadratic ring
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Describing the elements of quotient ring of $mathbbZ[sqrtD]$.
Why is $mathbbZ[sqrt-n], nge 3$ not a UFD?Does $mathbbZ[sqrt[3]2]=mathbbZ[sqrt[3]2]^mathrmint$?In which extensions of type $mathbbQ(sqrtd)$ is the number 3 reducible?Why is quadratic integer ring defined in that way?Maximal ideal in a polynomial ring over $mathbb Z$Quotient ring $R/I$, $R= mathbbZ[sqrt-10]$, $I =(sqrt-10)$About the ring $mathbbZ[sqrt4]$Tensor algebra on non free modulesMethod to adjoin elements to a ring - is the symbol $x$ overloaded?Prime ideals in certain quadratic ring
$begingroup$
For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.
The images(in K. Conrad's note) below are examples of the method using visualization.
So, i wonder about that:
Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?
I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.
Give some advice! Thank you!
abstract-algebra gaussian-integers
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
$endgroup$
add a comment |
$begingroup$
For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.
The images(in K. Conrad's note) below are examples of the method using visualization.
So, i wonder about that:
Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?
I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.
Give some advice! Thank you!
abstract-algebra gaussian-integers
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
$endgroup$
add a comment |
$begingroup$
For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.
The images(in K. Conrad's note) below are examples of the method using visualization.
So, i wonder about that:
Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?
I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.
Give some advice! Thank you!
abstract-algebra gaussian-integers
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
$endgroup$
For Gaussian integer ring $mathbbZ[i]$, there is a method describing distinct elements of certain quotient ring of $mathbbZ[i]$ using 'the visualization'.
The images(in K. Conrad's note) below are examples of the method using visualization.
So, i wonder about that:
Is there a similar method for quotient ring of $mathbbZ[sqrtD]$, where $D$ is square-free?
I guess it is possible in imaginary quadratic field $mathbbQ[sqrt-d]$, where $d>0$.
Give some advice! Thank you!
abstract-algebra gaussian-integers
abstract-algebra gaussian-integers
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
edited Mar 11 at 11:06
Primavera
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
asked Mar 11 at 11:01
PrimaveraPrimavera
30719
30719
add a comment |
add a comment |
1 Answer
1
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$begingroup$
I'll make a try...
Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.
So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$
and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I'll make a try...
Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.
So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$
and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
$endgroup$
add a comment |
$begingroup$
I'll make a try...
Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.
So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$
and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
$endgroup$
add a comment |
$begingroup$
I'll make a try...
Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.
So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$
and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
$endgroup$
I'll make a try...
Let's say we have $mathbbZ[isqrtk]=a+bisqrtk:a,binmathbbZ$ where $kinmathbbN $ is square-free.
So for the multiples of $(1+2i)$ we see that
$$(1+2i)(a+bisqrtk)=a(1+2i)+b(-2sqrtk+sqrtki)$$
and $(1+2i),(-2sqrtk+sqrtki)$ correspond to perpendicular vectors, so you can proceed as before (now you will not have squares but rectangles).
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
answered Mar 11 at 12:05
giannispapavgiannispapav
1,794324
1,794324
add a comment |
add a comment |
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