Does there exists a surjective homomorphism from $\mathbb Q[X]/(X^2 + 4)$ to $\mathbb Q \times \mathbb Q$? The field $\mathbb Q[X]/(X^2 + 4) := \{ q(x) + (X^2 + 4)\mid q(x) \in \mathbb Q[X]\}$.
And $(X^2 + 4)$ is an ideal of $\mathbb Q[X]$.
My proof:
Consider $$\phi: \mathbb Q[X]/(X^2 + 4) \rightarrow \mathbb Q \times \mathbb Q, f(x) \mapsto (f(1) , f(-1))$$
I think I am not proving the homomorphism correctly
$$\phi(f(x)g(x)) = (f(1)g(1), f(-1)g(-1)= \phi(f(x)) \phi(g(x))$$
$$\phi(f(x)+g(x)) = (f(1)+g(1), f(-1)+g(-1) = \phi(f(x)) + \phi(g(x))$$
Would $\phi: \mathbb Q[X]/(X^2 + 4) \rightarrow \mathbb Q \times \mathbb Q, f(x) \mapsto (f(1), f(-1))$ be surjective?
How do I prove the existence of $f(x) \in \mathbb Q[X]/(X^2 + 4)$? Let $y \in  \mathbb Q \times \mathbb Q$ , say $y = (a, b)$, but we know that there exists a polynomial $f(x) \in \mathbb Q[X] / (X^2 + 4)$ such that $f(1) = a$ and $f(-1) = b$. Therefore $\phi$ is onto? How do I prove the existence of $f(x)$?
I am making a big mess out of this one, please help.
 A: Suppose that there exists a surjective map from $f:\mathbb{Q}[X]/(X^2+4)\rightarrow \mathbb{Q}\times \mathbb{Q}$.
The kernel of $f$ will be an ideal of the form $\mathfrak{a}/(X^2+4)$ where $\mathfrak{a}$ is an ideal of $\mathbb{Q}[X]$ containing $(X^2+4)$.
Since $\mathbb{Q}[X]$ is PID, we have $\mathfrak{a}=(g)$ and $g\mid X^2+4$, but $X^2+4$ is irreducible so we have $g$ has to be $X^2+4$ (up to some units).
So this surjective morphism is an isomorphism, but this is not true as $\mathbb{Q}[X]/(X^2+4)$ is integral domain and $\mathbb{Q}\times\mathbb{Q}$ is not so we have a contradiction, so such surjective map doesn't exist.
A: I assume that by "homomorphism" you mean ring homomorphism.
Then the answer is "no, there isn't a surjective homomorphism from $\Bbb Q[X]/(X^2 + 4) \rightarrow \Bbb Q \times \Bbb Q$.
Suppose there exists such a ring homomorphism $f$. Then $f$ is also a homomorphism of $\Bbb Q$-vector spaces.
Since both sides have dimension $2$ over $\Bbb Q$, the surjectivity of $f$ implies that $f$ is in fact bijective.
Therefore $f$ must be an isomorphism, which is impossible because $\Bbb Q[X]/(X^2 + 4)$ is a field while $\Bbb Q\times \Bbb Q$ is not integral.
A: As another answer says there is no ring-isomorphism between $\mathbb Q[X]/(X^2+4)$ and $\mathbb Q\times\mathbb Q$. But there exists an $\mathbb Q$-vector space isomorphism between them. (This means that you have to specify isomorphism between which objects.)
I'll construct that, $\mathbb Q$-vector space isomorphism.
There exists an $\mathbb Q$-linear map $\varphi:\mathbb Q[X]\to\mathbb Q\times\mathbb Q$ as $\varphi(p(X))=(\Re(p(2i)),\Im(p(2i)))$.
And its kernel is absolutely $(X^2+4)$.
Also, $\varphi$ is surjective because for every $(a,b)\in\mathbb Q\times\mathbb Q$, there exists $p(X)\in\mathbb Q[X]$ such that $\varphi(p(X))=(a,b)$ (specifically, $p(X)=a+\frac b2X$).
So, by first isomorphism theorem, $\bar\varphi:\mathbb Q[X]/(X^2+4)\to\mathbb Q\times\mathbb Q$ as $\bar\varphi([p(X)])=\varphi(p(X))$ is well-defined $\mathbb Q$-vector space isomorphism.
