$\dfrac{\mathbb{Z}[x]}{(x^2 +5)}$ is isomorphic to $\mathbb{Z}[\sqrt{-5}]$

Let $H : \mathbb{Z}[x] \rightarrow \mathbb{Z}[\sqrt{-5}]$ be the evaluation homomorphism given by $H(f) = f(\sqrt{-5})$. We know that $H$ is surjective. I wanna to show that $\ker(H) = (x^2 +5)$. I know that is easy, but I don't find a way to show this. To prove the inclusion $(x^2 +5) \subset \ker(H)$ is easy, so I need help in the other inclusion. I proved this problem using Krull dimension, but I wanna to prove this in a most simple way. Any hint will be helpful.

• As Bill pointed out, you can use Gauss' lemma which is proven here. – Pedro Tamaroff Jan 22 '15 at 19:01

Take $f \in \ker H.$ Then $f(x) = (x^2 + 5)g(x) + r(x),$ for some $g(x), r(x) \in \mathbb Z[x]$ and degree of $r(x)$ is at most one. Now $f(\sqrt{-5}) = 0 \Rightarrow r(\sqrt{-5})=0.$ But $r(x)$ is a polynomial of degree at most one with integer co-efficients. Hence $r(x) =0.$

• $\mathbb{Z}[x]$ is not an euclidian domain. So you must "divide" in $\mathbb{Q}[x]$, and prove that $g(x) \in \mathbb{Z}[x]$. – Gauss Jan 22 '15 at 18:32
• @Gauss: It's true that $\mathbb Z[x]$ is not an Euclidean domain. But I'm dividing by $x^2+5,$ whose leading coefficient is $1.$ So there is no problem. This is true more generally. If $A$ is a commutative ring with identity and $f(x), g(x) \in A[x]$ with $f(x)$ monic (or even leading coefficient of $f$ is a unit in $A$) then we can "divide" $g$ by $f$ to get $g=fh +r$ for some $h, r \in A[x]$ with deg $r <$ deg $f.$ – Krish Jan 22 '15 at 18:37
• It is true! I liked your solution! Thanks! – Gauss Jan 22 '15 at 18:43
• @Gauss Just as in high-school, the polynomial long division alorithm always works for polynomials whose leading coefficient $= 1\,$ (or a unit), e.g. see here and see here. – Bill Dubuque Jan 22 '15 at 18:54

It is clear that $(x^2+5)\subseteq \ker H$. Suppose now that $f(\sqrt 5 i)=0$. Since $f$ has integer coefficients, this means that $f(-\sqrt 5 i)=0$ (conjugation fixes $\Bbb R$), so that in $\Bbb C$ $(x-\sqrt 5i)(x+\sqrt 5i)\mid f$ and then $(x^2+5)\mid f$. This means there is a polynomial $g\in\Bbb C[x]$ such that $f(x)=(x^2+5)g(x)$. By what is proven below, we must have $g\in \Bbb Z[x]$, and hence the other inclusion follows.

ADD Let $A\subseteq B$ be rings, and suppose that $h(x)g(x)=f(x)$ in $B[x]$, and $f,g\in A[x]$, $g$ monic. Then $h\in A[x]$.

Proof By induction on $n=\deg f+\deg g$. If $n=0$ then $g=1$ and $f=h\in A[x]$. Suppose $\deg f+\deg g>0$. We may assume that $\deg g<\deg f$. If $f(x)=a_0x^m+\cdots$ and $m-k=\deg g$, then $\tilde f(x)=f(x)-a_0x^kg(x)$ has degree $<\deg f$ and $\tilde f(x)=g(x)(h(x)-a_0x^k )$. By induction, $h(x)-a_0x^k \in A[x]$ so $h(x)\in A[x]$.

• Yes, but… Isn’t the fact that $x^2+5$ is monic part of the story? This is what makes Euclidean division by $x^2+5$ work within $\Bbb Z[x]$. The only way I can see how to know that $g$ has integer coefficients is by using Euc. div. Am I being foolish here? – Lubin Jan 22 '15 at 17:58
• @Lubin I have added a proof. – Pedro Tamaroff Jan 22 '15 at 18:21
• It is true! In other way, I can divide in $\mathbb{Q}[x]$ and to prove with your "ADD" that the quotient belongs to $\mathbb{Z}[x]$. – Gauss Jan 22 '15 at 18:47
• The lemma is already proved in many places, e.g. see this thread. – Bill Dubuque Jan 22 '15 at 19:03