Let $ \alpha = \frac{\sqrt 2+i\sqrt 2}2 \in \mathbb C$. Find the minimal polynomial of $\alpha$ over the fields $\mathbb Q$, $\mathbb Q(\sqrt 2)$, $\mathbb Q(i)$, and $\mathbb R$.

I found the same question here but I do not know how he has got the answer.

Abstract Algebra - Minimal Polynomials

please help me with this thanks

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    – Em.
    Mar 15, 2017 at 4:44

2 Answers 2


My impression is that the OP is looking for an "elementary" bare-handed proof of the result. Here is my attempt using as little machinery as possible.

Five preliminary things.

(i) If you plot $\alpha$ in the Argand plane you'll see that it is an $8$-th root of unity and so satisfies $X^4+1$.

(ii) This can be factorised $(X^2+\sqrt{2}X+1)(X^2-\sqrt{2}X+1)$ over $\mathbb{R}$.

(iii) $\alpha^2=\text{i}$, so it is clear that we have $X^4+1=(X^2-\text{i})(X^2+\text{i})$.

(iv) The roots of $X^4+1$ are $\frac{\pm 1 \pm \text{i}}{\sqrt{2}}$.

(v) $\alpha$ is not real.

Now we can use these.

(a) Over $\mathbb{R}$ and over $\mathbb{Q}(\sqrt{2})$ we have that $\alpha$ satisfies $X^2+\sqrt{2}X+1$, and this must be irreducible, or else $\alpha$ would lie in a real field. So this is the minimal polynomial in these cases.

(b) Over $\mathbb{Q}$ the polynomial $X^4+1$ is irreducible. This is because it certainly does not have linear factors, as none of its roots $\frac{\pm 1 \pm \text{i}}{\sqrt{2}}$ is real let alone rational. Nor can it be the product of two rational quadratics; there is no way that the sum of any two of the roots is rational. So in this case the minimal polynomial is $X^4+1$.

(c) Finally, over $\mathbb{Q}(\text{i})$, we have that $\alpha$ satisfies $X^2-\text{i}$. If this were not irreducible we'd have that $\alpha\in\mathbb{Q}(\text{i})$, and so (easily) $\sqrt{2}\in\mathbb{Q}(\text{i})$. In that case, write $\sqrt{2}=p+q\text{i}$, and square up for an immediate contradiction. So in this case the minimal polynomial is $X^2-\text{i}$.


For each base field $\mathbb Q$, $\mathbb Q(\sqrt 2)$, $\mathbb Q(i)$, choose a basis for $\mathbb Q(\sqrt 2,i)$ as a vector space over the base field and express the linear map $x \mapsto \alpha x$ as a matrix with respect to that basis. The minimal polynomial of this matrix is what you seek.

For the base field $\mathbb R$, choose a basis for $\mathbb C$ and do the same.


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