Is the following polynomial reducible?

Question

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?

Answer 1

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!

Answer 2

C is algebraically closed so only polynomes of one degree are irreductible