Decomposing $x^4-5x^2+6$ over some fields

Question

My book asks me to decompose

$$x^4-5x^2+6$$

over:

$K = \mathbb{Q},\\ K = \mathbb{Q[\sqrt{2}]},\\ K = \mathbb{R}$

For $K = \mathbb{Q}$, I substituted $x² = a$ to get:

$$a²-5a+6 = (a-3)(a-2)$$

So Getting back to $a = x²$ we get:

$$x^4-5x^2+6 = (x²-3)(x²-2)$$

Also, for $\mathbb{R}$ we can simply factor $x²-3 = (x-\sqrt{3})(x+\sqrt{3})$ and $(x²-2) = (x-\sqrt{2})(x+\sqrt{2})$ so we have: $$x^4-5x^2+6 = (x-\sqrt{3})(x+\sqrt{3})(x-\sqrt{2})(x+\sqrt{2})$$

but what about $$K = \mathbb{Q[\sqrt{2}]}\ ?$$

Answer 1

Your work is correct; for $\mathbb{Q}[\sqrt{2}]$, you surely have $$ x^4-5x^2+6=(x-\sqrt{2})(x+\sqrt{2})(x^2-3) $$ Is $x^2-3$ reducible over $\mathbb{Q}[\sqrt{2}]$? This would mean…