I have the following problem:
Suppose $p$ is a prime and $f$ an irreducible polynomial in the ring $\Bbb{Z}/p\Bbb{Z}[X]$. Define a field $\mathbb{K}=\Bbb{Z}/p\Bbb{Z}[X]/(f)$ with the characteristic $f$. A minimal polynomial for an element $k\in\Bbb{K}$ is the monic polynomial with the smallest positive degree which has $k$ as a root. Prove that each element in $\Bbb{K}$ has a unique minimal polynomial.
As I am only a 1st-year student in mathematics, although I feel somewhat comfortable with proofs, I feel that my proofs are almost never "completely" right. Which is that although I might get the rough sketch correct, frequently, there are details missing or false assumptions being used. And I am not so sure either if this proof that I came up with is acceptable.
Proof attempt: Let $k\in \Bbb{K}$ then there is a polynomial with a positive degree which has $k$ as root in $\Bbb{K}$. Now, $X^{\mid \Bbb{K} \mid}-X$ is one of them since $X^{\mid \Bbb{K} \mid}\equiv X \pmod f $ by Fermat's little theorem ($\Bbb{K}$ is a field).
We first prove that $k$ has a minimal polynomial.
Let $p\in\Bbb{K}$ be polynomial with a smallest positive degree such that $p(k)\equiv 0$ in $\Bbb{K}$. If $a\in\Bbb{Z}/p\Bbb{Z}$ is the leading coefficient of $p$, then $a$ has an inverse $a^{-1}$ in $\Bbb{Z}/p\Bbb{Z}$. Then the polynomial $g=a^{-1}p$ is monic and $g(k)\equiv0$. So $g$ is a minimal polynomial for $k$.
Finally, we prove that if $g$ is a minimal polynomial for $k$ then $g$ is unique.
Let $g$ be a minimal polynomial for $k$ and suppose there is another minimal polynomial $h$ for $k$ (Then clearly $\deg(h)=\deg(g)$). Then $h$ can be written as: $$h=g+r$$ where $r$ positive degree polynomial with $\deg(r)<\deg(g)$. Then $r(k)\equiv0$ so $r$ is a polynomial with $k$ as a zero. But then we can again define another polynomial by multiplying $r$ with the inverse of the leading coefficient of $r$ but then that polynomial is a monic polynomial with $k$ as a zero and its degree is smaller than $\deg(g)$. But this is a contradiction so $r=0$ which implies that $h=g$. $\square$
Now what I am not completely certain is that what is the exact role of indeterminate "$X$"? I know that it "acts" as a placeholder in whatever ring or field that I am working on, as you can see in this example, it is partly being replaced by itself (or the polynomials consisting of $X$'s). For example, what does it mean for "$X$" to be replaced by "X" itself, an indeterminate?