The number of surjective ring homomorphism from $mathbb{Z}[i]$ to $mathbb{F}_{11^2}$.
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Find the number of surjective ring homomorphism from $mathbb{Z}[i]$ onto $mathbb{F}_{11^2}$.
If such a surjective ring homomorphism exists with kernel $(a+bi)$, then $mathbb{Z}[i]/(a+bi)congmathbb{F}_{11^2}$ by 1st isomorphism theorem for rings.
Since $|mathbb{Z}[i]/(a+bi)|=a^2+b^2$, we have $a^{2}+b^{2}=11^{2}$.
Then, the integer pair of solutions of the latter equation are $(a,b)=(11,0),(0,11),(-11,0),(0,-11)$.
So, I conclude that there is one such a surjective ring homomorphism with kernel $(11)$.
Is it true or missing some case?
Give some advice or collection! Thank you!
abstract-algebra ring-homomorphism gaussian-integers
asked 22 hours ago
Primavera
2148
add a comment |
up vote
1
down vote
favorite
Find the number of surjective ring homomorphism from $mathbb{Z}[i]$ onto $mathbb{F}_{11^2}$.
If such a surjective ring homomorphism exists with kernel $(a+bi)$, then $mathbb{Z}[i]/(a+bi)congmathbb{F}_{11^2}$ by 1st isomorphism theorem for rings.
Since $|mathbb{Z}[i]/(a+bi)|=a^2+b^2$, we have $a^{2}+b^{2}=11^{2}$.
Then, the integer pair of solutions of the latter equation are $(a,b)=(11,0),(0,11),(-11,0),(0,-11)$.
So, I conclude that there is one such a surjective ring homomorphism with kernel $(11)$.
Is it true or missing some case?
Give some advice or collection! Thank you!
abstract-algebra ring-homomorphism gaussian-integers
asked 22 hours ago
Primavera
2148
1
This is a good start, but you have a bit more work to do. After all, it is possible for distinct homomorphisms to have the same kernel!
– Daniel Mroz
22 hours ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Find the number of surjective ring homomorphism from $mathbb{Z}[i]$ onto $mathbb{F}_{11^2}$.
If such a surjective ring homomorphism exists with kernel $(a+bi)$, then $mathbb{Z}[i]/(a+bi)congmathbb{F}_{11^2}$ by 1st isomorphism theorem for rings.
Since $|mathbb{Z}[i]/(a+bi)|=a^2+b^2$, we have $a^{2}+b^{2}=11^{2}$.
Then, the integer pair of solutions of the latter equation are $(a,b)=(11,0),(0,11),(-11,0),(0,-11)$.
So, I conclude that there is one such a surjective ring homomorphism with kernel $(11)$.
Is it true or missing some case?
Give some advice or collection! Thank you!
abstract-algebra ring-homomorphism gaussian-integers
asked 22 hours ago
Primavera
2148
Find the number of surjective ring homomorphism from $mathbb{Z}[i]$ onto $mathbb{F}_{11^2}$.
If such a surjective ring homomorphism exists with kernel $(a+bi)$, then $mathbb{Z}[i]/(a+bi)congmathbb{F}_{11^2}$ by 1st isomorphism theorem for rings.
Since $|mathbb{Z}[i]/(a+bi)|=a^2+b^2$, we have $a^{2}+b^{2}=11^{2}$.
Then, the integer pair of solutions of the latter equation are $(a,b)=(11,0),(0,11),(-11,0),(0,-11)$.
So, I conclude that there is one such a surjective ring homomorphism with kernel $(11)$.
Is it true or missing some case?
Give some advice or collection! Thank you!
abstract-algebra ring-homomorphism gaussian-integers
abstract-algebra ring-homomorphism gaussian-integers
asked 22 hours ago
Primavera
2148
asked 22 hours ago
Primavera
2148
asked 22 hours ago
Primavera
2148
asked 22 hours ago
Primavera
2148
asked 22 hours ago
Primavera
2148
2148
1
This is a good start, but you have a bit more work to do. After all, it is possible for distinct homomorphisms to have the same kernel!
– Daniel Mroz
22 hours ago
add a comment |
1
This is a good start, but you have a bit more work to do. After all, it is possible for distinct homomorphisms to have the same kernel!
– Daniel Mroz
22 hours ago
1
1
This is a good start, but you have a bit more work to do. After all, it is possible for distinct homomorphisms to have the same kernel!
– Daniel Mroz
22 hours ago
This is a good start, but you have a bit more work to do. After all, it is possible for distinct homomorphisms to have the same kernel!
– Daniel Mroz
22 hours ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
You have looks good, but you have just counted the possible kernels of maps from $mathbf Z [i]to mathbf F_{11^2}$.
In a general ring we can have more surjective maps than just the number of kernels, for example in $k[x,y] to k[x]$ we could have kernel $(y)$ and let $x$ map to $-x$.
answered 22 hours ago
Alex J Best
1,59711122
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
You have looks good, but you have just counted the possible kernels of maps from $mathbf Z [i]to mathbf F_{11^2}$.
In a general ring we can have more surjective maps than just the number of kernels, for example in $k[x,y] to k[x]$ we could have kernel $(y)$ and let $x$ map to $-x$.
answered 22 hours ago
Alex J Best
1,59711122
add a comment |
up vote
1
down vote
You have looks good, but you have just counted the possible kernels of maps from $mathbf Z [i]to mathbf F_{11^2}$.
In a general ring we can have more surjective maps than just the number of kernels, for example in $k[x,y] to k[x]$ we could have kernel $(y)$ and let $x$ map to $-x$.
answered 22 hours ago
Alex J Best
1,59711122
add a comment |
up vote
1
down vote
up vote
1
down vote
You have looks good, but you have just counted the possible kernels of maps from $mathbf Z [i]to mathbf F_{11^2}$.
In a general ring we can have more surjective maps than just the number of kernels, for example in $k[x,y] to k[x]$ we could have kernel $(y)$ and let $x$ map to $-x$.
answered 22 hours ago
Alex J Best
1,59711122
You have looks good, but you have just counted the possible kernels of maps from $mathbf Z [i]to mathbf F_{11^2}$.
In a general ring we can have more surjective maps than just the number of kernels, for example in $k[x,y] to k[x]$ we could have kernel $(y)$ and let $x$ map to $-x$.
answered 22 hours ago
Alex J Best
1,59711122
answered 22 hours ago
Alex J Best
1,59711122
answered 22 hours ago
Alex J Best
1,59711122
answered 22 hours ago
Alex J Best
1,59711122
1,59711122
add a comment |
add a comment |
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1
This is a good start, but you have a bit more work to do. After all, it is possible for distinct homomorphisms to have the same kernel!
– Daniel Mroz
22 hours ago