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This question already has an answer here:

In other discussions, e.g. here it is claimed that a polynomial in a field $K$ with degree greater than $1$, having a root in $K$, must be reducible. So by this criterion $X^3-1$ would be reducible over $\mathbb{Q}$ since it has a root $1 \in \mathbb{Q}$.

However other sources, e.g. here say that a polynomial in $\mathbb{Q}[X]$ is reducible only if it can be factored into two non-constant polynomials also in $\mathbb{Q}[X]$. So by this criterion it looks as though $X^3-1$ is actually irreducible, because although one factor is $(X-1)$, the other factors are $(X-e^{2\pi i/3})$ and $(X-e^{4\pi i/3})$ which are clearly not in $\mathbb{Q}$.

Is either of these definitions of irreducibility either wrong or non-standard, or are they in fact equivalent? If the latter, what have I overlooked?

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marked as duplicate by Jyrki Lahtonen field-theory Apr 17 at 17:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ The other factor is $x^2+x+1=(x-e^{2\pi i/3})(x-e^{4\pi i/3})$. Its coefficients are rational. $\endgroup$ – Jyrki Lahtonen Apr 17 at 17:26
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We can write $x^3-1=(x-1)(x^2+x+1)$. This is indeed a product of two non constant polynomials in $\mathbb{Q}[x]$. Nobody said the polynomials in the product must be linear.

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But, $x^3-1 = (x-1)(x^2+x+1)$. Both nonconstant polynomials. So $x^3-1$ is indeed reducible over $\mathbb{Q}$.

[It however does not **split in ${\mathbb{Q}}$, as $(x^2+x+1)$ has degree > 1 and is irreducible in ${\mathbb{Q}}$.]

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