Minimal polynomial of $x=\sqrt{2}+i\sqrt{3}$

Question

I was asked to calculate the minimal polynomial of $x=\sqrt{2}+i\sqrt{3}$ over the fields \begin{align*} K_1 = \mathbb{Q}, \quad K_2 = \mathbb{Q}(\sqrt{2}), \quad K_3 = \mathbb{Q}(i\sqrt{3}), \quad L = \mathbb{Q}(\sqrt{2}+i\sqrt{3}) \end{align*}

I calculated $x^2,x^3,x^4$ and then concluded that $x$ is a root of $X^4+2X^2+25=:P$.

After that, I calculated the roots of $P$. $$X^4+2X^2+25 = (X+\sqrt{2}-i\sqrt{3})(X-\sqrt{2}+i\sqrt{3})(X+\sqrt{2}+i\sqrt{3})(X-\sqrt{2}-i\sqrt{3})$$

It's obvious that this polynomial is irreducible in $K_1$ and completely reducible in linear factors in $L$.

But isn't it also irreducible in $K_2, K_3$? I can't display the roots in $K_2,K_3$.

I'm mainly asking because this answer seems too easy for me and I think there might be a catch.

Answer 1

• The simplest case is $L$ since $x\in L$ therefore the minimal polynomial is just $P_L(x)=X-x$.

It's coefficients $(-x,1)$ are both elements of $L$.

• I continue with $K_1$ since you solved it, $P_1(X)=X^4+2X^2+25$.

It's coefficients $(25,0,2,0,1)$ are all rationals.

• For $K_2$ you have to proceed like you did for $\mathbb Q$ and get rid of $i\sqrt{3}$.

Notice $(x-\sqrt{2})^2=(i\sqrt{3})^2=-3\in K_2$

Therefore $x$ is root of $P_2(X)=(X-\sqrt{2})^2+3=X^2-2\sqrt{2}X+5$

It's coefficients $(5,-2\sqrt{2},1)$ are all elements of $K_2$.

• For $K_3$ it is similar, get rid of $\sqrt{2}$

$(x-i\sqrt{3})^2=(\sqrt{2})^2=2\in K_3$

Therefore $x$ is root of $P_3(X)=(X-i\sqrt{3})^2-2=X^2-2i\sqrt{3}X-5$

It's coefficients $(-5,-2i\sqrt{3},1)$ are all elements of $K_3$.

Note: $P_2$ and $P_3$ are minimal because of the degree, there is no polynomial of degree $1$ simply because $\sqrt{2}\notin K_3$ and $i\sqrt{3}\notin K_2$