Fastest method to verify $x^4-2x^2+9$ is irreducible so minimal

Question

I was tasked with finding $\min(\mathbb{Q},i+\sqrt[]{2})$

I found the polynomial in $\mathbb{Z}[X]$ to be $X^4-2X^2+9$

I know it must be minimal over $\mathbb{Q}$ since (used brute force) any arbitrary qubic, quadratic, or linear polynomial has coefficients in $\mathbb{C}-\mathbb{Q}$

My question is is there an easier method?

(Edit: question initially asked about irreducibility, but since minimal implies irreducibility, we make the change)

Answer 1

I fully endorse Michael's method. If you lack a key result and can't follow it, you can do one of the following.

1. Show that $f(x)=x^4-2x^2+9$ is irreducible by observing that it has no rational roots, and eliminating a possible factorization into quadratics by concluding from $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ that first (compare cubic terms) we must have $c=-a$, then (compare linear terms) also $b=d$. Therefore $b=d=\pm3$, and checking that neither sign works is easy.
2. Examining the field generated by $\alpha=i+\sqrt2$. We have $$\frac1\alpha=\frac{\sqrt2-i}3,$$ so you can write both $\sqrt2$ and $i$ as $\Bbb{Q}$-linear combinations of $\alpha$ and $1/\alpha$. Hence $\alpha$ is a generator of the field $K=\Bbb{Q}(i,\sqrt2)$. As $[K:\Bbb{Q}]=4$ the minimal polynomial of $\alpha$ must have degree four.

The choice that you find acceptable depends heavily on what tools have been covered already in class. At this point in a typical algebra course the theory moves forward swiftly, so it is quite difficult for us to determine what really helps you. And in a few weeks time a tool not yet covered may be in frequent use.

Answer 2

Eisenstein works very well!

Let $X=\frac{2}{x}+1.$

Thus, $$X^4-2X^2+9=\frac{8(x^4+2x^2+4x+2)}{x^4}.$$ Now, take $p=2$.

Because if $X^4-2X^2+9$ is reducible then $x^4+2x^2+4x+2$ is reducible, which is a contradiction by Eisenstein.

Answer 3

I will use a tower property in this case:

We know $K=\mathbb{Q}(i,\sqrt[]{2})$ is a field extension of $L=\mathbb{Q}(\sqrt[]{2})$ which is a field extension of $F=\mathbb{Q}$

In a previous result, we showed that $[K:L]=2$ and since $[L:Q]=2$ it follows $[K:Q]=4$

Therefore, the minimum polynomial must have degree 4

Answer 4

A variant of Michael Rozenberg’s method, broken down into several steps:

Step 1: $f(x)=x^4 -2x^2+9$ is irreducible if and only if $\,f(x+1)=x^4+4x^3+4x^2+8$ is irreducible. Call this $g(x)$.

Step 2: $g$ above is irreducible if and only if $8x^4+4x^2+4x+1$ is irreducible. Call this polynomial $h(x)$.

Step 3: $h$ above is irreducible if and only if $h(x/2)=\frac12x^4+x^2+2x+1$ is irreducible. And when you multiply this by $2$, you get an Eisenstein polynomial, irreducible.

So the original $f(x)$ is irreducible.

(Maybe I should say, for the benefit of the cognoscenti that if you know Newton Polygon, you can see irreducibility at Step 1, ’cause the Polygon passes through no integral points other than $(0,3)$ and $(4,0)$. This is exactly the phenomenon you use when you prove Eisenstein.)