Proving that a L-algebraic Set is also a K-algebraic set The following question is from assignment 1 of my algebraic geometry course. I have been following only my class notes.

Question: Let L|K be a field extension and $V\subseteq  L^n$ be an L-algebraic set. Then show that the set $V_K = V\cap K^n$ of all K-rational points of V is an K-algebraic set.

V is an L-algebraic set implies that V={ $a =(a_1,...,a_n) \in L^n |  f_i(a) =0 $ for all i from 1 to j } and each $f _i$ is a polynomial with coefficients in L.
The K-rational points of V is defined as, are $V_K =${$a =(a_1,...,a_n) \in K^n | f_i(a) =0 $ for all i from 1 to k } and each  $ f_i$ is a polynomial with coefficients in K.
I have been given $V_K= V\cap K^n$, all elements of $V \cap K^n$( say a) are elements of $K^n$ for which  $f_i(a) =0$  for all i from 1 to j  and each $f _i$ is a polynomial with coefficients in L.
To prove that V is a K-algebraic set in $K^n$, I have to show that all the coefficients in L of $f_i(a) =0$ are actually in K.

But I am at loss of ideas on which theorem should I use?

Can you please shed some light on this on how should I prove it?
 A: This the Exercise I.1.4 by E. Kuntz (1985) Introduction to Commutative Algebra and Algebraic Geometry. Birkhäuser.
For my personal comfort: I change the notations.

Let $\mathbb{K}$ be a field, let $\mathbb{F}$ be an extension field of $\mathbb{K}$ and let $V$ be and algebraic affine variety in $\mathbb{A}^n_{\mathbb{F}}$ (the $n$-dimensional $\mathbb{F}$-affine space).
If $V\cap\mathbb{A}^n_{\mathbb{K}}=\emptyset$ there is nothing to prove; otherwise...
By definition:
$$
\exists f_1,\dots,f_r\in\mathbb{F}[x_1,\dotsc,x_n]:V=\{P\in\mathbb{A}^n_{\mathbb{F}}\mid\forall l\in\{1,\dotsc,r\},f_l(P)=0\}\equiv\mathcal{Z}(f_1\dotsc,f_r);
$$
let $\mathcal{I}(V)=(f_1,\dotsc,f_r)$ be the ideal generated by these polynomials, considering
\begin{gather*}
\mathbb{K}[x_1,\dotsc,x_n]\xrightarrow{\overline{\epsilon}}\mathbb{F}[x_1,\dotsc,x_n]\xrightarrow{\pi}\mathbb{F}[x_1,\dotsc,x_n]_{\displaystyle/\mathcal{I}(V)}\to0,\,\varphi=\pi\circ\overline{\epsilon},\\
\ker(\varphi)=\left\{p\in\mathbb{K}[x_1,\dotsc,x_n]\mid\pi\left(\overline{\epsilon}(p)\right)=0\right\}=\dotsc=\{p\in\mathbb{K}[x_1,\dotsc,x_n]\mid\forall X\in V,\,p(X)=0\},
\end{gather*}
and $\ker(\varphi)$ is an ideal of $\mathbb{K}[x_1,\dotsc,x_n]$; by the Hilbert's Basis Theorem, this ideal is finitely generated, so there exist $g_1,\dotsc,g_s\in\mathbb{K}[x_1,\dotsc,x_n]$ such that $\ker(\varphi)=(g_1,\dotsc,g_s)$.
By previous step, one has:
$$
V\cap\mathbb{A}^n_{\mathbb{K}}\subseteq\mathcal{Z}(\ker(\varphi))\equiv V_0
$$
where $V_0$ is an affine algebraic variety in $\mathbb{A}^n_{\mathbb{K}}$; without loss of generality, one could assume $V_0$ irreducible.
Since:
$$
V=\mathcal{Z}(f_1,\dotsc,f_r)=\bigcap_{l=1}^r\mathcal{Z}(f_l)\equiv\bigcap_{l=1}^rV_l,
$$
without loss of generality one could assume $r=1$. And by previous steps:
$$
\forall k\in\{1,\dotsc,s\},\,\overline{\epsilon}(g_s)\in\mathcal{I}(V)=(f_1)\equiv(f),
$$
this is equivalent to state:
$$
\exists q_1,\dotsc,q_s\in\mathbb{L}[x_1,\dotsc,x_n]\mid\forall k\in\{1,\dotsc,s\},fq_k=g_k.
$$
By all this:
$$
V_0=\mathcal{Z}(g_1,\dotsc,g_s)=\dotsc=\left(V\cap\mathbb{A}^n_{\mathbb{K}}\right)\cup\left(\mathcal{Z}(q_1,\dotsc,q_s)\cap\mathbb{A}^n_{\mathbb{K}}\right),
$$
passing to closure operator:
$$
V_0=\overline{V}_0=\dotsc=\overline{\left(V\cap\mathbb{A}^n_{\mathbb{K}}\right)}\cup\overline{\left(\mathcal{Z}(q_1,\dotsc,q_s)\cap\mathbb{A}^n_{\mathbb{K}}\right)};
$$
by irreducibility of $V_0$ and by hypothesis:
$$
\overline{\left(\mathcal{Z}(q_1,\dotsc,q_s)\cap\mathbb{A}^n_{\mathbb{K}}\right)}=\emptyset\Rightarrow\left(\mathcal{Z}(q_1,\dotsc,q_s)\cap\mathbb{A}^n_{\mathbb{K}}\right)=\emptyset
$$
and therefore $V_0=\left(V\cap\mathbb{A}^n_{\mathbb{K}}\right)$.
