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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 $\mathbb { Q } [ i \sqrt { 5 } ]$?

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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 $\Bbb Z[i\sqrt 5]$.

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