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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.

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 $\mathbb{F}_4[x]/(x^2)\cong \mathbb{F}_2[x]/(x^4+x^2+1)$.

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.

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lkx
lkx
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Can the embedding dimension of a finite local algebra change after restricting to a finite subfield?

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 $\mathbb{F}_4[x]/(x^2)\cong \mathbb{F}_2[x]/(x^4+x^2+1)$.