The embedding dimension of a commutative $k$-algebra is the minimum $n$ such that it is a quotient of the polynomial algebra in $n$-variables.
The embedding dimension of $\mathbb{F}_2\times \mathbb{F}_2$ is $1$ because it is a quotient of $\mathbb{F}_2[x]$ by the product of the ideals $(x)$ and $(x+1)$.
The embedding dimension of $\mathbb{F}_2\times \mathbb{F}_2\times \mathbb{F}_2$ is $2$ because it is not a quotient of $\mathbb{F}_2[x]$ but it is a quotient of $\mathbb{F}_2[x, y]$ by the product of the ideals $(x, y)$, $(x+1, y)$ and $(x, y+1)$.
Let $\mathbb{F}_{q^n}/\mathbb{F}_q$ be an extension of finite fields. Can the embedding dimension of a finite local $\mathbb{F}_{q^n}$-algebra be different if we consider it as an $\mathbb{F}_q$-algebra?
For example $\mathbb{F}_4[x]/(x^2)$ has embedding dimension $1$ as a $\mathbb{F}_4$-algebra and it has embedding dimension $1$ as a $\mathbb{F}_2$-algebra since $\mathbb{F}_4[x]/(x^2)\cong \mathbb{F}_2[x]/(x^4+x^2+1)$.
P.S. As YCor points out this happens for $\mathbb{F}_{q^n}$ itself so let us consider only the case of positive dimension.
