Here are my interpretation and answer to your question.
Let $k$ be a field and $(k^n)^*$ be the dual of the vector space $k^n$.
For any $S\subset (k^n)^*$ denote by $(S)\subset k[T_1,...,T_n]$ the ideal generated by $S$ and by $Lin(S) \subset (k^n)^*$ the subspace generated by $S$.
For $S,S_0\subset (k^n)^*$ you then have:
Result 1 $$Lin(S)=Lin(S_0) \iff (S)=(S_0)$$ This is an easy calculation, using just the the definitions.
Call $Sol(S)=Sol(S;k)\subset k^n$ the set of $x\in k^n$ such that for all $s\in S$ we have $s(x)=0$.
Notice that $Sol(S)=Sol(Lin(S))$
The second result is then $$Lin(S)=Lin(S_0)\iff Sol(S)=Sol(S_0)$$
This is a standard result in linear algebra , where $Sol(S)$ is generally denoted by a symbol like $S^0$ or $S^\perp$.
For a specific reference, look at the second Corollary to Theorem 16, Chapter 3, page 102 of Hoffman-Kunze's Linear Algebra.
Conclusion $$ (S)=(S_0)\iff Sol(S)=Sol(S_0) $$
Remark The consideration of arbitrary $k$-algebras $K$ is irrelevant and only confuses the issue.