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Suppose that $S$ and $S_0$ are two systems of linear equations over a field $k$. Show that $(S) = (S_0)$ if and only if $Sol(S; k) = Sol(S_0; k)$.

Note that $(S)=(S_0)$ if and only if $Sol(S; K) = Sol(S_0; K)$ for any $k$-algebra $K$. So the only if direction is easy. But I don't know how to solve the converse.

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What does the notation $(S)$ mean? –  Andrew Aug 1 '12 at 15:43
    
@Andrew It is the ideal generated by $S$ in $k[T]=k[T_1,\cdots,T_n]$. –  Gobi Aug 1 '12 at 22:33

1 Answer 1

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.

Result 2
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.

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