Field of Definition of an Algebraic Group

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Linear Algebraic Groups- James E. Humphreys Chapter-XII

Let $K$ be an algebraically closed field and $k$ be a arbitrary sub-field of $K.$

A closed set X in $A^n=K\times ...$(n times)$\times K$ is a the zero set of some polynomials in $K[X_1, ... ,X_n]$.

A closed set in $A^n$ is said to be $k$-closed if $X$ is the set of zeros of some collection of polynomials having coefficients in $k$.

The $k$-closed sets are the closed sets of a topology of $A^n$ called the 'Zariski $k$-Topology'

And then Humphreys says-

So far, so good. But there is a subtle difficulty with this notion. To say that $X=\mathcal V(I)$ for some ideal generated by its intersection with $K[X_1, ... ,X_n]$ is not to say that $\mathcal I(X)$ has this property.

Can someone give a example of the phenomena?

And then he defines- In case $\mathcal I(X)$ does happen to be generated by the $k$-polynomials, we say that $X$ is defined over $k$.

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One way this might happen is the following. Let $k$ be a non-perfect subfield (assuming characteristic $p>0$ here), and let $a\in k\setminus k^p$. Then $a$ has a $p$th root $\alpha\in K\setminus k$. Let $X=\{\alpha\}\subset A^1$. As $\alpha$ is the only zero in $K$ of the polynomial $(X-\alpha)^p=X^p-a\in k[X]$ we see that $X=V(I)$ for the ideal $I$ generated by $X^p-a$. Yet $I(X)$ is generated by $X-\alpha$ and clearly cannot be generated by polynomials with coefficients in $k$.