Jtdylktuy

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!

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 $-x$.

Then, the remaining part of this problem as follows:

So, the fact $ker{f}=(11)$ implies

$$mathbb{Z}_{11}(i)congmathbb{Z}_{11}[x]/(x^{2}+1)congmathbb{Z}[x]/(11,x^{2}+1)congmathbb{Z}[i]/(11)congmathbb{F}_{11^{2}}.$$

by 3rd isomorphism theorem for rings.

Therefore, it suffices to find the number of homomorphisms from $mathbb{Z}[i]$ onto $mathbb{Z}_{11}(i)$.

First, since $f(1)=1$, $f$ is completely determined by the value of $f(i)$ in $mathbb{Z}_{11}(i)$.

Now, let $f(i)=x+yi,(x,yinmathbb{Z}_{11})$. Then, we have

$$-1=f(-1)=f(i^{2})=left(f(i)right)^{2}=(x+yi)^{2}=(x^{2}-y^{2})+(2xy)i.$$

Since $2xyequiv0pmod{11}$, we have $xequiv0pmod{11}$ or $yequiv0pmod{11}$.

If $yequiv0pmod{11}$, then $x^{2}equiv-1pmod{11}$.

But, it is impossible since $left(tfrac{-1}{11}right)=-1$, where $(ast)$ is the Legendre symbol.

Therefore, we have $xequiv0pmod{11}$ and $y^{2}equiv1pmod{11}$, and hence, $yequivpm1pmod{11}$.

That is, the only possible values of $f(i)$ are $pm1$ in $mathbb{Z}_{11}(i)$.

Consequently, there are only two homomorphisms from $mathbb{Z}[i]$ onto $mathbb{F}_{11^{2}}$ with respect to the value of $f(i)$.