From this this book.
Given a finite set $\left\{f_1,f_2,\ldots,f_r\right\} \subset R$, the ideal $I$ generated by this set is denoted $f_1, f_2, \ldots , f_r$ and consists of all the sums $f_1h_1 + f_2h_2 + ·· ·+ f_r h_r$ where the $h_j \in R$. (Showing this defines an ideal is an exercise.)
My attempt:
Let $g_1, g_2 \in I$, I have to prove that $g_1 + g_2 = h_1 f_1 + \ldots h_r f_r, \; h_1, \ldots h_r \in R$, we have:
$$ g_i = h_1^i f_1 + \ldots h_r^i f_r \;\; ,h_j^i \in R , i \in \left\{1,2\right\}, j \in \left\{1, \ldots, r\right\} \Rightarrow g_1 + g_2 = \left(h_1^1 + h_1^2 \right) f_1 + \ldots \left(h_r^1 + h_r^2 \right) f_r \;\; $$
But since $R$ is a ring we have $h_j^1 + h_j^2 \in R$ so we proved the first part.
Similarly if $h \in R, g \in I$ we have:
$$ hg = h h_1 f_1 + \ldots h h_r f_r = \left(h h_1 \right) f_1 + \ldots + \left( h h_r \right) f_r $$
But $h h_j \in R \forall j \in \left\{1, \ldots r \right\}$.
Is such proof correct?
Related I have a question, from the same book sometimes it omits to specify the ring used to generate an ideal like in the algorithm 2.7.
Fix a monomial order > on $k[x_1,\ldots, x_n]$ and nonzero polynomials $f_1,\ldots, f_r ∈ k[x_1, . . . , x_n]$. Given $g \in k[x_1, \ldots , x_n]$, we want to determine whether $g \in <f_1, \ldots, f_r> $
In situations like this it is assumed the ring is the set $k[x_1,\ldots x_n]$ right? I just find a bit unclear situations like this, because maybe the set $f_1,\ldots f_n$ could generate a closed subset of the ring I mentioned earlier.