# Roots of Polynomials in Fields

I have a conceptual question about roots of polynomials in fields.

Consider the field $$Q$$. In that field, the polynomial $$t^4 + 1$$ is irreducible. The argument my prof. uses to prove this is to say that in $$R$$ it decomposes into $$(t^2 - \sqrt{2}t + 1) (t^2 + \sqrt{2}t + 1)$$ then say that none of these polynomials are in $$Q$$.

This seems intuitively obvious but I'm not sure I get the formal explanation for this. Why would the decomposition of the polynomial be the same in $$Q$$ as it is in $$R$$? Isn't the whole point of this course that the same polynomial can have different roots in different fields?

Thanks for the help!

• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa Jun 18 '19 at 15:40
• The point is there exists a canonical inclusion $\mathbb{Q}\subset\mathbb{R}$. Hence you can take the version of $\mathbb{R}$ that contains your initial $\mathbb{Q}$. – quangtu123 Jun 18 '19 at 15:48

Suppose there's a non-trivial decomposition $$\;t^4+1=p(t)q(t)\;,\;\;p,q\in\Bbb Q[t]\;$$ . We already can say both $$\;p,\,q\;$$ are quadratics, since $$\;t^4+1\;$$ has no roots in $$\;\Bbb Q\;$$.

But then $$\;t^4+1=p(t)q(t)\;$$ as polynomials in $$\;\Bbb R[t]\;$$! Thus, as $$\;p(t)\;$$ is irreducible, it then must divide either $$\;t^2-\sqrt 2\,t+1\;$$ or the other factor (Why? Here prime is the same as irreducible...). Finish now the argument

• Right, it divides one of these but is in $Q[t]$, and there is no unit in $Q[t]$ by which we can multiply either factor to get a polynomial in $Q[t]$. Thank you! – Sausage_Devourer Jun 18 '19 at 15:56
• You're welcome...though I didn't understand that sentence "there is no unit in $\;\Bbb Q[t]\;$ by which we can multiply either factor...." – DonAntonio Jun 18 '19 at 15:58
• I mean that no polynomial of the form $u (t^2 - \sqrt(2)t + 1)$, where $u$ is in $Q$, is a polynomial of $Q[t]$. – Sausage_Devourer Jun 18 '19 at 16:21

Clearly $$t^4 +1$$ has no root in $$\mathbb{Q}$$ and hence no linear factor.

Hence, to check irreducibility, we just need to show that it’s not possible for the polynomial $$t^4 + 1$$ to be factored in to the product of two quadratic polynomials with $$\textit{rational}$$ coefficients.

Assume that $$t^4 + 1 = (t^2 + at + b)(t^2 + ct + d)$$, with $$a,b,c,d \in \mathbb{Q}$$.

Compare coefficients on each side and this yields a contradiction.