Let $F[x]$ be be the set of polynomials over a field $F$ and $f_1 , . . . , f_n \in F[x]$ be arbitrary polynomials. I'm using the following definition of an Ideal:
A subset $I \subseteq F[x]$ is called an ideal if $I$ is a subspace of $F[x]$ and $fg \in I$ is true for all $f \in F[x]$ and all $g \in I$. ($fg$ represents the usual multiplication of polynomials)
The ideal generated by the set $\{f_1 , . . . , f_n\}$ is the set $f_1F[x] + . . . + f_nF[x]$ (also a subspace of $F[x]$), where $f_iF[x]$ is defined to be the ideal $\{f_ig: g\in F[x]\}$.
Now the task/problem is to show that this genereated ideal is equal to the intersection of all ideals of $F[x]$ which contain $\{f_1 , . . . , f_n \}$. I'm trying to solve this problem by inclusion of the two sets, and ive managed to show the first inclusion "generated $\subseteq$ intersection", which is fairly straightforward. I'm having troubles however with the other inclusion.
Any input is greatly appreciated.
As @DanielSchepler points out, this is almost immediate, because the ideal $\langle f_1,\dots,f_n\rangle$ is an ideal containing $\{f_1,\dots,f_n\}$. Thus you have the reverse inclusion.