How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
New contributor
|
show 4 more comments
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
New contributor
Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago
I can't prove it directly, which makes me sad:(
– yLccc
2 days ago
What does "diectly" mean in this case?
– Arthur
2 days ago
$mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago
3
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago
|
show 4 more comments
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
New contributor
How to prove $Bbb Z[i]/(1+2i)cong Bbb Z_5$?
My method is
$$(1+2i)=big{a+bi丨a+2b≡0pmod 5big},$$
So any $a+bi$ in $Bbb Z(i)$,we got
$$a+bi=(b-2a)i+a(1+2i).$$
So $Bbb Z[i]/(1+2i)=big{0,[i],[2i],[3i],[4i]big}$.
I know how to prove this ring is isomorphic to $Bbb Z_5$, but how can I prove that $Bbb Z[i]/(1+2i)$ equals to $Bbb Z_5$ directly? Any suggestion ia appreciated.
abstract-algebra ring-theory ideals finite-rings gaussian-integers
abstract-algebra ring-theory ideals finite-rings gaussian-integers
New contributor
New contributor
edited 2 days ago
Zvi
5,050430
5,050430
New contributor
asked 2 days ago
yLccc
283
283
New contributor
New contributor
Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago
I can't prove it directly, which makes me sad:(
– yLccc
2 days ago
What does "diectly" mean in this case?
– Arthur
2 days ago
$mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago
3
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago
|
show 4 more comments
Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago
I can't prove it directly, which makes me sad:(
– yLccc
2 days ago
What does "diectly" mean in this case?
– Arthur
2 days ago
$mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago
3
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago
Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago
Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago
I can't prove it directly, which makes me sad:(
– yLccc
2 days ago
I can't prove it directly, which makes me sad:(
– yLccc
2 days ago
What does "diectly" mean in this case?
– Arthur
2 days ago
What does "diectly" mean in this case?
– Arthur
2 days ago
$mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago
$mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago
3
3
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago
|
show 4 more comments
4 Answers
4
active
oldest
votes
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
add a comment |
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
add a comment |
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
add a comment |
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
add a comment |
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4 Answers
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active
oldest
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4 Answers
4
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I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
add a comment |
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
add a comment |
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
I think it is easier to do it this way. Let $varphi:Bbb Z[x]to Bbb Z[i]$ sending $xmapsto i$. Then the preimage of the ideal $I=(1+2i)=(i-2)$ of $Bbb Z[i]$ under $varphi$ is the ideal $J=(x^2+1,x-2)$ of $Bbb Z[x]$. Thus $varphi$ induces the isomorphism $$Bbb{Z}[i]/Icong Bbb{Z}[x]/J.$$
Now observe that
$$J=(x^2+1,x-2)=(x^2+1,x-2,x^2-4)=(5,x-2).$$
This shows that$$frac{Bbb Z[i]}{(1+2i)}=frac{Bbb{Z}[i]}{(i-2)}cong frac{Bbb Z[x]}{(x^2+1,x-2)}=frac{Bbb{Z}[x]}{(5,x-2)}congfrac{Bbb{Z}[x]/(x-2)}{(5,x-2)/(x-2)}.$$
Because under the isomorphism $Bbb{Z}[x]/(x-2)cong Bbb Z$ (which sends $p(x)+{color{red}{(x-2)}}$ to $p(2)$), $(5,x-2)/(x-2)$ is the ideal $5Bbb{Z}$ of $Bbb Z$, we get
$$frac{Bbb Z[i]}{(1+2i)}congfrac{Bbb Z}{5Bbb Z},$$
as required.
edited 2 days ago
answered 2 days ago
Zvi
5,050430
5,050430
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
add a comment |
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Could u explain why (5,x-2)/(x-2) is the ideal 5Z of Z for me? I don't understand it
– yLccc
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
Anything in $(5,x-2)$ is of the form $5p(x)+(x-2)q(x)$. So $(5,x-2)/(x-2)$ contains elements of the form $5p(x)+(x-2)q(x)+{color{red}{(x-2)}}=5p(x)+{color{red}{(x-2)}}$. Therefore, under the given isomorphism, $5p(x)+{color{red}{(x-2)}}$ is mapped to $5p(2)$ which is an integer multiple of $5$. Clearly $5$ is in the image (by taking $p$ to be the constant polynomial $p=1$). Therefore, $(5,x-2)/(x-2)$ is precisely $5Bbb Z$.
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
In my answer and in my previous comment, anything that appears in $color{red}{mathrm{red}}$ is an ideal (i.e., not a polynomial in parentheses).
– Zvi
2 days ago
add a comment |
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
add a comment |
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
add a comment |
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
A ring homomorphism $varphicolonmathbb{Z}[i]tomathbb{Z}_5$ is uniquely determined by assigning $q=varphi(i)$, which should be an element satisfying $q^2=-1$; thus you have two choices: $q=2$ or $q=-2$.
With the second choice, you have $varphi(a+bi)=[a-2b]$. Can you determine $kervarphi$?
answered 2 days ago
egreg
178k1484201
178k1484201
add a comment |
add a comment |
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
add a comment |
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
add a comment |
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
Hint: $(1+2i)(1-2i)=5$. Hence $(1+2i)supset (5)$. Hence $mathbb Z[i]/(1+2i)subset mathbb Z_5[i]/(1+2i)$. But $Z_5[i]/(1+2i)cong mathbb Z_5$, since $a+bi=a+2b+(1+2i)$.
To show there are $5$ distinct elements (cosets), just consider (the equivalence classes of) ${0,1,2,3,4}$. It's easy to see none of them are the same.
edited 2 days ago
answered 2 days ago
Chris Custer
10.9k3824
10.9k3824
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add a comment |
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
add a comment |
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
add a comment |
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
If you know you have five cosets here, then try the representatives $0,1,2,3,4$ which you can prove are not equal. Because $(1+2i)(1-2i)=5$ is in your ideal then the arithmetic with these representatives is just that in $mathbb Z_5$ - you can reduce any integer representative of a coset modulo $5$.
Note: If you want to do the whole thing, it is easy to show that every coset contains an integer because $1+2i$ and $i(1+2i)=i-2$ are both in your ideal. Then $5$ is in your ideal, so every coset contains one of $0,1,2,3,4$.
answered 2 days ago
Mark Bennet
80.6k981179
80.6k981179
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
add a comment |
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
So next step I prove 5 is in my ideal, so 1 is in my ideal, and then I get 2,3,4 are the same, and then prove these integers are in different cosets. Is that right?
– yLccc
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
@yLccc How do you get $1$ in the ideal generated by $1+2i$? If $1$ is in the ideal is the whole ring. The basic thing is you have five cosets, so choose five representatives which look as much like $mathbb Z_5$ as possible. Everything divisible by $5$ gets absorbed in the ideal, just as it does when you are working with the ideal generated by $5$ over the integers.
– Mark Bennet
2 days ago
add a comment |
yLccc is a new contributor. Be nice, and check out our Code of Conduct.
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yLccc is a new contributor. Be nice, and check out our Code of Conduct.
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Why are you unhappy with your solution? It looks perfectly reasonable to me.
– Arthur
2 days ago
I can't prove it directly, which makes me sad:(
– yLccc
2 days ago
What does "diectly" mean in this case?
– Arthur
2 days ago
$mathbb Z[i]/(a+bi)cong mathbb Z/N(a+bi)mathbb Z$ where $(a,b)=1$
– Mustafa
2 days ago
3
$mathbb Z[i] / (1+2i)$ is not equal to $mathbb Z_5$. By definition $mathbb Z_5$ is $mathbb Z / (5)$ and each element of $mathbb Z/(5)$ is a coset of the ideal $(5)$ in the ring $mathbb Z$. But each element of $mathbb Z [i] / (1+2i)$ is a coset of the ideal $(1+2i)$ in the ring $mathbb Z[i]$. No coset of the first type is equal to a coset of the second type. So the best you can hope for is an isomorphism, which you say you already have. Do you want some kind of "special" isomorphism? Can you state what special property you want that isomorphism to have?
– Lee Mosher
2 days ago