Ring monomorphism and epimorphism

Question

$$\large{\Phi: \mathbb Z[X] \rightarrow \mathbb R \; s \; \Phi(p(X)) \; := \; p(\sqrt{5})}$$

Hello guys so I have the following problem:

---

I have to prove whether the following mapping is ring homomorphism, and maybe monomorphism or epimorphism.

My attempt at a solution:

Obviously the mapping is well defined, and I have proven that it's ring homomorphism. It is not monomorphism because polynomial $P(x) = 5 - x^2$ is in the kernel so kernel is not trivial. As for the epimorphism part I have to check weather the function is surjective or not. If the codomain were integers then of course for every integer C , I could just take constant polynomial $P(x)= C$ and surjectivity would follow. I am not sure how to prove surjectivity in this case, my hunch is that it's not surjective since I cannot make up a polynomial that gets mapped for example into $\frac{1}{2}$.

Thanks for help

Answer 1

As you remark, $x^2-5\in\ker\Phi$, so $\Phi$ is not injective.

Hint: $\sqrt{2}\notin\operatorname{im}\Phi$.

Second hint: Show that the elements in the image are of the form $a+b\sqrt{5}$, for $a$ and $b$ integers.

Answer 2

This map cannot be onto, because $\mathbb Z[X]$ is countable, while $\mathbb R$ is uncountable.

If you are not familiar with cardinalities, here is another reason. Let $V$ be the $\mathbb Q$ vector space with basis $\{1, \sqrt 5 \}$. Then $im(\phi) \subseteq V$ is a proper subset of $\mathbb R$ as $dim_\mathbb Q (\mathbb R) = \infty$, while $dim_\mathbb Q (V) = 2$.