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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!

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  • $\begingroup$ 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. $\endgroup$ – dantopa Jun 18 '19 at 15:40
  • $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$ – quangtu123 Jun 18 '19 at 15:48
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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

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  • $\begingroup$ 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! $\endgroup$ – Sausage_Devourer Jun 18 '19 at 15:56
  • $\begingroup$ 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...." $\endgroup$ – DonAntonio Jun 18 '19 at 15:58
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    $\begingroup$ I mean that no polynomial of the form $u (t^2 - \sqrt(2)t + 1)$, where $u$ is in $Q$, is a polynomial of $Q[t]$. $\endgroup$ – Sausage_Devourer Jun 18 '19 at 16:21
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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.

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