# Is $x^3-1$ reducible over $\mathbb{Q}$ [duplicate]

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?

## marked as duplicate by Jyrki Lahtonen field-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 17 at 17:29

• The other factor is $x^2+x+1=(x-e^{2\pi i/3})(x-e^{4\pi i/3})$. Its coefficients are rational. – Jyrki Lahtonen Apr 17 at 17:26
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.
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 $${\mathbb{Q}}$$.]