Mixing
The number of surjective ring homomorphism from $mathbb{Z}[i]$ to $mathbb{F}_{11^2}$.
up vote
3
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 Nov 12 at 14:17
Primavera
2288
add a comment |
up vote
3
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 Nov 12 at 14:17
Primavera
2288
2
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
Nov 12 at 14:34
add a comment |
up vote
3
down vote
favorite
up vote
3
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 Nov 12 at 14:17
Primavera
2288
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 Nov 12 at 14:17
Primavera
2288
asked Nov 12 at 14:17
Primavera
2288
asked Nov 12 at 14:17
Primavera
2288
asked Nov 12 at 14:17
Primavera
2288
asked Nov 12 at 14:17
Primavera
2288
2288
2
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
Nov 12 at 14:34
add a comment |
2
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
Nov 12 at 14:34
2
2
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
Nov 12 at 14:34
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
Nov 12 at 14:34
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
What 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 Nov 12 at 14:34
Alex J Best
1,69211222
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
What 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 Nov 12 at 14:34
Alex J Best
1,69211222
add a comment |
up vote
2
down vote
accepted
What 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 Nov 12 at 14:34
Alex J Best
1,69211222
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
What 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 Nov 12 at 14:34
Alex J Best
1,69211222
What 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 Nov 12 at 14:34
Alex J Best
1,69211222
edited 17 hours ago
answered Nov 12 at 14:34
Alex J Best
1,69211222
answered Nov 12 at 14:34
Alex J Best
1,69211222
answered Nov 12 at 14:34
Alex J Best
1,69211222
1,69211222
add a comment |
add a comment |
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2
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
Nov 12 at 14:34