Let $\psi:\mathbb Z[x] \rightarrow \mathbb R$ be the homomorphism defined by $\psi(p(x))=p(\sqrt3)$.

Question

Let $\psi:\mathbb Z[x] \rightarrow \mathbb R$ be the homomorphism defined by $\psi(p(x))=p(\sqrt3)$.

a) Prove that the kernel of $\psi$ is a principal ideal.

b) Find the subring $S$ of $\mathbb R$ with which the quotient ring $\mathbb Z[x]/\text{ker}(\psi)$ is isomorphic.

My try: By definition, $\operatorname{ker}(\psi)$=$\{P(x)\in\mathbb{Z}[x]: P(\sqrt{3})=0\}$. Then $\operatorname{ker}(\psi) = x-\sqrt3$, but that polynomial does not belong to the ring; so I must find a polynomial that is in the ring, in this case the polynomial $P(x)=x^2−3$ is in the ring. Now I must find the ideal generated by $P(x)=x^2−3$ and show that it is a principal ideal and using the first isomorphism theorem for rings, find which is the subring $S$ of $\mathbb R$ with which it is isomorphic $\mathbb Z[x]/(x^2−3)$

Answer 1

Any polynomial $p(x) \in \mathbb Z[x]$ can be written as $q(x)(x^2 - 3) + (ax + b)$

$q(x)(x^2 - 3)$ is the kernel of $\phi$

$\phi(p(x)) = \phi(ax+b) = b+a\sqrt 3$

The image of $\phi$ in $\mathbb R$ is isomorphic to $\mathbb Z[\sqrt 3]$

Answer 2

The kernel is indeed $(x^2-3)$.

Note that you can use the division algorithm for $x^2 - 3$, since its leading coefficient is a unit in $\mathbb{Z}$.

Certainly, $(x^2-3)$ is contained in the kernel. Conversely, assume that $g(x)$ is in the kernel. The division algorithm allows you obtain $q(x)$ and $r(x)$ in $\mathbb{Z}[x]$ such that $$g(x) = (x^2-3)q(x) + r(x)$$ where $r(x) = 0$ or $0 \leq deg\text{ }r(x) < deg (x^2-3) = 2$. If $r(x)$ is nonzero, then plugging in $\sqrt{3}$ gives $ r(\sqrt{3}) = 0$. Why is this a contradiction ?

For the other part, the image certainly contains $\mathbb{Z}[\sqrt{3}]$. Why is it exactly equal to $\mathbb{Z}[\sqrt{3}]$?