What Are the Minimal Polynomials of $i\sqrt[4]{2}$ over $\mathbb{Q}$ and $\mathbb{R}$?

Let $\alpha=i\sqrt[4]{2}$, where $\sqrt[4]{2}$ means the positive real fourth root. The minimal polynomial depends on the base field: a polynomial that is irreducible over $\mathbb{Q}$ may factor over $\mathbb{R}$ because real coefficients can include numbers such as $\sqrt{2}$.

The answers are $x^4-2$ over $\mathbb{Q}$ and $x^2+\sqrt{2}$ over $\mathbb{R}$. Here is why each is the minimal polynomial in its field.

Over the rationals

First compute a power of $\alpha$:

$$\alpha^4=(i\sqrt[4]{2})^4=i^4(\sqrt[4]{2})^4=2.$$

Thus $\alpha$ is a root of $x^4-2$. This proves that its minimal polynomial over $\mathbb{Q}$ divides $x^4-2$, but it does not by itself prove that $x^4-2$ is minimal. For that, check irreducibility over $\mathbb{Q}$.

The polynomial $x^4-2$ satisfies Eisenstein’s criterion with the prime $2$: every nonleading coefficient is divisible by $2$, while the constant term $-2$ is not divisible by $2^2=4$. Therefore $x^4-2$ is irreducible over $\mathbb{Q}$.

Since $x^4-2$ is monic, irreducible over $\mathbb{Q}$, and has $\alpha$ as a root, it is the minimal polynomial:

$$\boxed{m_{\alpha,\mathbb{Q}}(x)=x^4-2.}$$

In particular, the degree of $\alpha$ over $\mathbb{Q}$ is $4$.

Over the reals

Over $\mathbb{R}$, use the square of $\alpha$ instead:

$$\alpha^2=(i\sqrt[4]{2})^2=i^2(\sqrt[4]{2})^2=-\sqrt{2}.$$

Consequently, $\alpha$ is a root of $x^2+\sqrt{2}$, whose coefficients are real. This quadratic has no real roots: for every real $x$, $x^2\geq 0$, so $x^2+\sqrt{2}>0$. A quadratic over $\mathbb{R}$ with no real roots is irreducible over $\mathbb{R}$.

Therefore,

$$\boxed{m_{\alpha,\mathbb{R}}(x)=x^2+\sqrt{2}.}$$

Its degree is $2$. This is consistent with $\alpha$ being nonreal: adjoining it to $\mathbb{R}$ gives $\mathbb{C}$, a degree-two extension of $\mathbb{R}$.

How the two answers fit together

The rational minimal polynomial factors when viewed as a polynomial with real coefficients:

$$x^4-2=(x^2-\sqrt{2})(x^2+\sqrt{2}).$$

The factor $x^2+\sqrt{2}$ vanishes at $\alpha$, since $\alpha^2=-\sqrt{2}$. The other factor does not. In fact, $x^2-\sqrt{2}$ further factors over $\mathbb{R}$ as $(x-\sqrt[4]{2})(x+\sqrt[4]{2})$.

The key point is that “minimal polynomial over a field” means the monic polynomial of least positive degree, with coefficients in that field, that has the element as a root. Enlarging the field of coefficients can make a polynomial factor and can lower the degree of the minimal polynomial.

Common mistakes

• Stopping after finding a polynomial that vanishes. The equation $\alpha^4=2$ shows that $x^4-2$ works, but irreducibility is needed to establish it as the minimal polynomial over $\mathbb{Q}$.
• Using the same polynomial over both fields. The polynomial $x^4-2$ is not irreducible over $\mathbb{R}$, so it is not the real minimal polynomial.
• Allowing irrational coefficients over $\mathbb{Q}$. The polynomial $x^2+\sqrt{2}$ has real coefficients, but not rational coefficients, so it cannot be the minimal polynomial over $\mathbb{Q}$.
• Confusing $i\sqrt[4]{2}$ with $\sqrt[4]{2i}$. These denote different numbers. The results above use the product of $i$ and the positive real fourth root of $2$.

A quick check

Substitute $\alpha$ into each claimed polynomial. For the rational polynomial, $\alpha^4-2=0$. For the real polynomial, $\alpha^2+\sqrt{2}=-\sqrt{2}+\sqrt{2}=0$. The irreducibility arguments then show that these are the minimal polynomials, not merely polynomials that happen to vanish.