I remember trying this problem a while ago and was unable to prove it. I think my idea was to create a surjective homomorphism from $\mathbb{Z}_{5}[x]$ to $\mathbb{Z}_{5} \times \mathbb{Z}_{5}$ and use the first isomorphism theorem but it wasn't working because none of my maps were well-defined. What is the correct way to approach this problem?


Note that $x^2+1=(x+2)(x+3)$ over $\mathbb{Z}/5\mathbb{Z}$, and so by the Chinese Remainder Theorem $$ (\mathbb{Z}/5\mathbb{Z})[x]/(x^2+1)=(\mathbb{Z}/5\mathbb{Z})[x]/(x+2)(x+3)\simeq (\mathbb{Z}/5\mathbb{Z})[x]/(x+2)\times (\mathbb{Z}/5\mathbb{Z})[x]/(x+3)$$ $$\simeq \mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}$$


For a constructive approach, your idea for a strategy works. Here is how to think through building a well-defined map in this case.

First, instead of starting with the quotient $\mathbb Z_5[x]/(x^2+1)$, it is easier to try starting with $\mathbb Z_5[x]$. In order to get a homomorphism $\mathbb Z_5[x]\rightarrow \mathbb Z_5\times \mathbb Z_5$, we simply map $1\mapsto (1,1)$, and choose any value for $x$; say, $x\mapsto (a,b)$.

In order for the map to factor through $\mathbb Z_5/(x^2+1)$, we need the image of $x$ to satisfy the equation $x^2+1=0$, hence $(a,b)^2+(1,1)=0$. Solving this, we need $a^2=b^2=4$. The equation $z^2=4$ has solutions $z=2$ or $3$ in $\mathbb Z_5$, so $(a,b)=(2,2), (3,3), (2,3),$ or $(3,2)$

If $a=b$, then $f$ is not surjective (since its image only contains $(c,c)$ for $c\in \mathbb Z_5$), so we can assume $a=2$ and $b=3$ or vice versa.

Finally, note that the map $x\mapsto (2,3)$ is surjective: for instance, it is a $\mathbb Z_5$-linear map whose image contains the basis $(1,1),(2,3)$ of $\mathbb Z_5\times \mathbb Z_5$.

Thus an isomorphism $\mathbb Z_5[x]/(x^2+1)\rightarrow \mathbb Z_5\times \mathbb Z_5$ can explicitly be given by $a+bx\mapsto (a+2b,a+3b)$ for $a,b\in \mathbb Z_5$.


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