Is the following polynomial reducible? I need to check whether the following polynomial $g(x)=x^2+t^2−1∈\Bbb{C}(t)$ (where $\Bbb{C}(t)$ is the field of all rational expressions over $\Bbb{C}$) is reducible. My attempt: I assume by contradiction that it's reducible, and I write it as:
$$(x+p(t))(x+q(t))=x^2+t^2−1$$
So we get that:
$$p(t)+q(t)=0$$
$$p(t)q(t)=t^2−1$$
Since $\Bbb{C}(t)$ is a field, it's also a UFD. So either $p(t)q(t)=(t−1)(t+1)$ or WLOG, $p(t)=λ(t^2−1)$ for some $0≠λ∈\Bbb{C}$. Either way, we get a contradiction to the first condition over $p(t),q(t)$. Is my approach correct or am I missing something?
 A: You have mostly put together a correct argument, but let me fill in a few details.
You have a polynomial $g(x) = x^2 + t^2 - 1$ with coefficients taken from $\mathbb C(t)$, the field of rational functions in $t$ over the complex numbers.  We ask whether $g(x)$ is irreducible in $\mathbb C(t)[x]$.
As you pointed out, to factor this monic polynomial $g(x)$ would mean:
$$ g(x) = (x + p(t))(x + q(t)) = x^2 + t^2 - 1 $$
and thus $p(t)+q(t) = 0$ and $p(t)q(t) = t^2 - 1$ for some rational functions $p(t),q(t)\in \mathbb C(t)$.
Since the first of these relations gives $q(t) = -p(t)$, we can dispense with $q(t)$ and focus on the second relation in terms only of $p=p(t)$, namely:
$$ p^2 + t^2 -1 = 0 $$
It is pretty obvious that no polynomial solution $p \in \mathbb C[t]$ is possible.  But the next thing to ask is whether a rational solution $p \in \mathbb C(t)$ is possible.
Here we use the observation that $\mathbb C[t]$ is a very nice UFD (in fact a Euclidean domain), and a UFD is integrally closed.  That means that if a solution to the equation above for $p$ exists in the field of quotients $\mathbb C(t)$ of $\mathbb C[t]$, then that solution would belong already to the base ring of polynomials $\mathbb C[t]$.  Done!
A: C is algebraically closed so only polynomes of one degree are irreductible
