I've come across an issue while doing an assigment for a field theory subject. The issue seems pretty basic to me but I can't figure it out it's solution by myself.
Let's consider the field extension $\mathbb{Q}(\zeta_3)|\mathbb{Q}$, where $\zeta_3$ is the third primitive root of the unit. During the process of solving my assigment, which consisted of describing all the irreducible polynomials and Galois groups of all subfields of $\mathbb{Q}(\zeta_{15})$, I've come across with this query:
It is known that since $\mathbb{Q}(\zeta_3)|\mathbb{Q}$ is a cyclotomic extension, its irreducible polynomial is $\Phi_3(x)=x^2+x+1$, and the roots are $\alpha=-\frac{1}{2}\pm\frac{i\sqrt{3}}{2}$, so the teahcer happily wrote that $\mathbb{Q}(\zeta_3)=\mathbb{Q}(i\sqrt{3})$. But then I've realised that Irr$(i\sqrt{3},\mathbb{Q})=x^2+3\neq\Phi_3$. The question is, since the two polynomials are not the same and the irreducible polynomial has to be unique, am I confusing equality with isomorphism? In case $\mathbb{Q}(\zeta_3)\neq\mathbb{Q}(i\sqrt{3})$ but $\mathbb{Q}(\zeta_3)\cong\mathbb{Q}(i\sqrt{3})$ and it is an abuse of notation that is confusing me, which representation one would be more suitable for the assignment.
Thank you.
Edit: Typo on the sign of the polynomial.
Actually the minimal/irreducible polynomial for $i\sqrt{3}$ over $\Bbb Q$ is $x^2\color{red}{+}3$. However, $\Bbb Q(\zeta_3)=\Bbb Q(i\sqrt{3})$ is a literal equality if you read these fields as subfields of the same algebraically closed container of $\Bbb Q$, for instance as subfields of $\Bbb C$. Like many things in maths it's easy to interpret the same thing in isomorphic but technically nonequal ways, for instance $\Bbb Q$ itself is ever so slightly ambiguous as there are a few different ways to construct $\Bbb Q$.
But the real point is that there is no one unique minimal polynomial for a field extension. For an element of an algebraic field extension $L/K$, that element of $L$ will have a unique minimal polynomial over $K$. But there are potentially infinitely many different elements which generate the same extension, and no one ever claimed (I hope!) that all these elements necessarily have the same minimal polynomial. You're maybe getting mixed up because if $L=K(\alpha)$ and $\beta$ is a conjugate of $\alpha$, meaning $\beta,\alpha$ have the same minimal polynomial, then $\beta$ also generates $L$ over $K$ (if $\beta\in L$) but it's very possible for $L$ to be generated by some $\gamma$, $\gamma$ not a conjugate of $\alpha$, as the example $\Bbb Q(\zeta_3)=\Bbb Q(i\sqrt{3})$ shows.