Example of ring can’t be defined over finite field.

Let $K$ be an infinite field of characteristic $p>0$, and $A$ be a $K$-algebra.

We say that $A$ can be defined over a finite field if there is finite subfield $F\subset K$ and an $F$-algebra $A_0$ such that $A_0\otimes_F K \cong A$ over $K$.

Now is there an easy example of an algebra that can’t be defined over a finite field？

Let $K = \mathbb{F}_p (X)$ and let $A = K [Y] / (Y^n - X)$, where $n$ is a prime other than $p$. So $A$ is a finite field extension of $K$. I say $A$ is not defined over any finite subfield of $K$.

The only finite subfield of $K$ is $\mathbb{F}_p$ itself, and, for dimension reasons, any $\mathbb{F}_p$-algebra $A_0$ such that $A_0 \otimes_{\mathbb{F}_p} K \cong A$ as $K$-algebras must be finite. Furthermore, since $A$ is an integral domain and $A_0$ embeds in $A$ (as a $\mathbb{F}_p$-algebra!), $A_0$ is also an integral domain, hence is a finite field extension of $\mathbb{F}_p$, say $\mathbb{F}_q$. But $\mathbb{F}_q \otimes_{\mathbb{F}_p} K \cong \mathbb{F}_q (X)$, and $\mathbb{F}_q (X) \not\cong A$ because (for example) $\mathbb{F}_p$ is algebraically closed in $A$.

You can take any central division $K$-algebra A of dimension $>1$ over $K$. Then it cannot be defined over a finite field. Otherwise, we would have $A\simeq A_0\otimes_FK$, and $A_0$ would be a central division $F$-algebra. By Wedderburn little theorem, this would imply $A_0\simeq F$ ,and thus $A\simeq K$, a contradiction.

If you want an explicit example, take $K=\mathbb{F}_p(X,Y)$ ($p\neq 2$), and let $A=(X,Y)_K$ be the corresponding quaternion $K$-algebra. Then $Q$ is a division algebra since the norm form is isomorphic to the diagonal form $\langle 1,-X,-Y,XY\rangle$ which is easily seen to be anisotropic.