I am studying algebraic elements over a field. The context started with the following discussions $:$
Let $\Bbb K$ be a field. Let $A$ be a $\Bbb K$-algebra. An element $x \in A$ is called algebraic over $\Bbb K$ if the substitution homomorphism $$\varepsilon_x : \Bbb K[X] \longrightarrow A$$ defined by $F(X) \mapsto F(x)$ has a non-zero kernel i.e. $\exists$ some non-constant polynomial $F \in \Bbb K[X]$ such that $F(x)=0.$ Since $\Bbb K$ is a field, $\Bbb K[X]$ ( the polynomial ring over $\Bbb K$ ) is a PID. So $Ker\ (\varepsilon_x)$ being an ideal of $\Bbb K[X]$ is principal. Therefore $\exists$ a unique non-constant monic polynomial $\mu_x \in \Bbb K[X]$ such that $Ker\ (\varepsilon_x) = \left \langle \mu_x \right \rangle.$ $\mu_x$ is called the minimal polynomial of $x$ over $\Bbb K.$ So by the first isomorphism theorem $\exists$ a $\Bbb K$-algebra isomorphism between $\Bbb K[X]/\langle \mu_x \rangle$ and $\text {Im}\ (\varepsilon_x) = \Bbb K [x]$ i.e. $$\Bbb K[X]/\langle \mu_x \rangle \cong \Bbb K[x].$$ Since $\Bbb K$-algebra isomorphisms are also $\Bbb K$-vector space isomorphism so we can say that $$\text {Dim}_{\Bbb K} \left (\Bbb K[X] / \langle \mu_x \rangle \right ) = \text{Dim}_{\Bbb K}\left (\Bbb K[x] \right ).$$ But we know that $$\text {Dim}_{\Bbb K} \left (\Bbb K[X] / \langle \mu_x \rangle \right ) = \text{deg} \left (\mu_x \right ).$$ So we have $$\text{Dim}_{\Bbb K} \left (\Bbb K[x] \right ) = \text{deg} \left (\mu_x \right ).$$
Now for any $x \in A,$ $\Bbb K[x]$ is a $\Bbb K$-subalgebra of $A.$ So in particular if $A$ is a finite dimensional $\Bbb K$-algebra then $\Bbb K[x]$ is also a finite dimensional $\Bbb K$-subalgebra of $A.$ Therefore for any $x \in A$ we have $$\text{Dim}_{\Bbb K} \left (\Bbb K[x] \right ) < \infty.$$ Therefore for any $x \in A$ $$\text {deg} \left (\mu_x \right) < \infty.$$
This shows that if $A$ is a finite dimensional $\Bbb K$-algebra then every element of $A$ is algebraic over $\Bbb K.$
Now $\Bbb K^n$ is a finite dimensional $\Bbb K$-algebra. Let $a=(a_1,a_2,\cdots,a_n) \in \Bbb K^n.$ Then by our previous discussions it is quite clear that $a$ should be algebraic over $\Bbb K.$ Then my question is $:$
How do I find out the minimal polynomial of $a$ over $\Bbb K$?
Any help will be highly appreciated. Thank you very much.
If you put just the "direct product of unital rings" algebra structure on $K^n$, then the minimal polynomial is the minimal monic one that has each of the $a_i$ as root; this is so by definition, since the evaluation of the polynomial in $(a_1,\ldots,a_n)$ is just the $n$-tuple of evaluations in each of the $a_i$. As mentioned in one of the comments, this polynomial can be written as $\prod_{b\in\{ a_1,\ldots,a_n\}}(X-b)$ (its degree is the number of distinct values among the$~a_i$).