What is the dimension of the affine variety $\mathbb{F}_p^1$?

I'm self studying commutative algebra. Here is the question:

Edit(Background and Definition): Let $k$ be a field, we define an affine variety $V\subseteq k^n$ as the set of common zeroes of some polynomials $f_1,\dots, f_m\in k[x_1,\dots, x_n]$, define its coordinate ring as $k[x_1,\dots, x_n]/\mathbb{I}(V)$, and define its dimension as the Krull dimension of the ring $k[x_1,\dots, x_n]/\mathbb{I}(V)$.

(As pointed out in the comments, It seems just a definition issue. I'm not sure when we talk about these definitions, shall we assume that $k$ to be of characteristic zero, or at least infinite?)

In the context of classical algebraic geometry, what is the dimension of $X:=\mathbb{F}_p^1$ as an affine variety over $\mathbb{F}_p$? (In my context the dimension of an affine variety is defined as the dimension of its coordinate ring)

My reasoning is as follows: On the one hand, $x^p-x\in \mathbb{I}(X)$, on the other hand, for any $f\in \mathbb{I}(X)$, it must divide $x-a$ for any $a\in \mathbb{F}_p$, thus divides $x^p-x$. Therefore $$ \mathbb{I}(X)=\langle x^p-x\rangle = \bigcap_{a\in \mathbb{F}_p} \langle x-a\rangle $$ which is an intersection of $p$ maximal ideals. By Chinese Remainder Theorem, its coordinate ring is $\mathbb{F}_p^p$, which is Artinian. Therefore it has dimension zero by Hopkins theorem.

But intuitively, it is a 'line', so its dimension should be one? Am I wrong? Or in modern algebraic geometry some definitions differ?

Thanks for any help!