There is a famous theorem saying that
Let $\Bbb F$ be a field and $f(x)$ an irreducible polynomial in $\Bbb F[X]$. Then there exists a field extension $\Bbb L$ of $\Bbb F$ such that $f(x)$ has a root in $\Bbb L$.
I know there is a well-known argument saying that $$f(x)\in\Bbb F[x]\;\;\text{irreducible}\implies \alpha:=x+\langle f(x)\rangle\in\Bbb K:=\Bbb F[x]/\langle f(x)\rangle\;\text{is a root of}\;f(x)$$
Yes, $\Bbb K:=\Bbb F[x]/\langle f(x)\rangle$ contains a copy of $\Bbb F$. But it is painful for me to regard $$\alpha:=x+\langle f(x)\rangle \in K:=\Bbb F[x]/\langle f(x)\rangle$$ as a root of $f(x)\in\Bbb F[x]$. Can we evaluate a polynomial with anything we want, even something that has nothing to do with the coefficient ring of that polynomial?
What I really want is to construct a field $\Bbb L \supset \Bbb F$ (in the sense of set theory) and find an element $u \in \Bbb F$ which is a root of $f \in \Bbb F[x]$, can anybody show me how to construct such an $\Bbb L$ and find such a $u$ explicitly?
Two comments
By saying that $f(x)$ has a root in $\mathbb{L}$, it is usually meant that the image of $f$ under the inclusion $\mathbb{F}[x] \hookrightarrow \mathbb{L}[x]$ has a root in $\mathbb{L}$. While you can't plug in elements of totally arbitrary sets into polynomials, the instructions comprising a polynomial make sense for any $\mathbb{F}$-algebra.
I think my set-theoretic aesthetics do not match yours, but you might be happier to think of $\mathbb{L}$ as a set to be a vector space over $\mathbb{F}$ with basis $\{1, \alpha, \alpha^2, \ldots, \alpha^{\deg(f) - 1}\}$, so that each element is uniquely given by a polynomial in $\alpha$ of degree less than $\deg(f)$.