# Show that a polynomial is irreducible in $\mathbb { Z } [ i \sqrt { 5 } ]$ but not in $\mathbb { Q } [ i \sqrt { 5 } ]$

To prove : Show that the polynomial is irreducible in $$3 + 2 t + 2 t ^ { 2 }$$ over $$\mathbb { Z } [ i \sqrt { 5 } ]$$ but not over $$\mathbb { Q } [ i \sqrt { 5 } ]$$.

Solution : The roots of the polynomial are $$\frac { - 1 \pm i \sqrt { 5 } } { 2 }$$. They are elements of $$\mathbb { Q } [ i \sqrt { 5 } ]$$. Thus $$3 + 2 t + 2 t ^ { 2 } = 2 \left( t + \frac { 1 + i \sqrt { 5 } } { 2 } \right) \left( t + \frac { 1 - i \sqrt { 5 } } { 2 } \right)$$ and the dominating coefficient is $$2$$. $$2$$ is irreducible in $$\mathbb { Z } [ i \sqrt { 5 } ]$$ thus $$3 + 2 t + 2 t ^ { 2 }$$ is irreducible over $$\mathbb { Z } [ i \sqrt { 5 } ]$$.

A factorisation of $$3 + 2 t + 2 t ^ { 2 }$$ would be $$( 2 t - u ) ( t - v )$$ for $$u , v \in \mathbb { Z } [ i \sqrt { 5 } ]$$. The precedent calcul show that this is not possible.

My question : Why $$2$$ irreducible over $$\mathbb { Z } [ i \sqrt { 5 } ]$$ implies that the polynomial is irreducible over $$\mathbb { Q } [ i \sqrt { 5 } ]$$? Also how can we conclude that a factorisation of the polynomial would be of the form $$( 2 t - u ) ( t - v )$$? Why would it implies that the polynomial is reducible in $$\mathbb { Q } [ i \sqrt { 5 } ]$$?

You've asked three questions:

(1) If $$2$$ were reducible, i.e., if $$2 = ab$$ for $$a,b \in \Bbb Z[i\sqrt 5]$$, then there is also a possibility that there are also $$c,d$$ such that $$ac = 1+i\sqrt 5$$ and $$bd = 1-i\sqrt 5$$, in which case you would have $$2\left(t + \frac{1+i\sqrt5}2\right) \left(t + \frac{1-i\sqrt5}2\right)= ab\left(t + \frac {ac}{ab}\right)\left(t + \frac {bd}{ab}\right) = (bt+c)(at+d)$$

But since $$2$$ is not reducible (which I hope has already been demonstrated, since it is hardly any more elementary than the rest of this), we know that is not possible, even without checking the reducibility of $$1\pm i\sqrt 5$$.

(2) If a quadratic polynomial is factorable into a product of two lesser polynomials, then their degrees must both be $$1$$. I.e., the quadratic would be equal to $$(at - u)(bt - v)$$ for some $$a, b, u, v \in \Bbb Z[i\sqrt 5]$$. Their product is $$abt^2 - (av+bu)t + uv$$. So if $$3 + 2t+2t^2$$ is factorable, we must have $$ab = 2$$. Since $$2$$ is not reducible, this requires $$a = 2, b = 1$$ (or vice versa, but we can fix that with a relabeling). Thus the only possible factorizations of $$3 + 2t + t^2$$ have to have form $$(2t - u)(t - v)$$.

(3) This does not imply that $$3 + 2t+2t^2$$ is reducible in $$\Bbb Q[i\sqrt 5]$$, nor did the solution in any way suggest such a thing. What shows this is reducible in $$\Bbb Q[i\sqrt 5]$$ is the demonstrated reduction:

$$3 + 2t+2t^2 = 2\left(t + \frac{1+i\sqrt5}2\right) \left(t + \frac{1-i\sqrt5}2\right)$$ All three factors on the right are polynomials over $$\Bbb Q[i\sqrt 5]$$ of lesser degree.

But note that by the irreducibility of $$2$$ in $$\Bbb Z[i\sqrt 5]$$, we can only multiply one of the two factors through by $$2$$, so we can only write $$3+2t+2t^2 = \left(2t + 1+i\sqrt5\right) \left(t + \frac{1-i\sqrt5}2\right) = \left(t + \frac{1+i\sqrt5}2\right) \left(2t + 1-i\sqrt5\right)$$ while keeping the coefficients of $$t$$ in $$\Bbb Z[i\sqrt 5]$$. Since neither $$\frac{1+i\sqrt5}2$$ nor $$\frac{1-i\sqrt5}2$$ is in $$\Bbb Z[i\sqrt 5]$$ (which while not being quite as obvious as it looks, is still not hard to show), we again find that $$3+2t+2t^2$$ cannot be factors over $$\Bbb Z[i\sqrt 5]$$.