If $F$ is a field (so therefore a ring), and we have $\{ f(x) + g(x) \text{ s.t. } f(x), g(x) \in F[x] \} = I$. I am trying to show that this is an ideal.
I am confused about what we need to show to be an ideal: can I just take any element in the field and multiply it by this set and show that it is in the set?
This is confusing because the way you are describing the subset of $F$ that you wish to show is an ideal. It looks like elements of this set are of the form $f(x)$, so I'm assuming you mean that they are polynomials. Thus you mean that $F$ is a field of polynomials. In this case, it is common practice to write $F$ as $F[x]$.
to check that your subset, $I=\{f(x)+g(x) : f(x),g(x) \in F[x]\}$, is an ideal of $F[x]$, you need to check two things:
$1): \forall x,y \in I$, we have that $x+y \in I$.
$2): \forall f(x) \in F[x]$ and $\forall m \in S$, we have that $mf(x) \in I$.
To show $(1)$, lets let $m,n \in I$. Thus $m = f(x) + g(x)$ and $n = r(x) + s(x)$ for $f(x),g(x),r(x),s(x) \in F[x]$.
Since $F[x]$ is a field, we have that $f(x)+g(x) \in F[x]$ and $r(x) + s(x) \in F[x]$. Again, since $F[x]$ is a field, we have that $(f(x)+g(x)) + (r(x)+s(x)) \in F[x]$. This means that $m+n \in I$.
To show $(2)$, let $m \in I$ and $f(x) \in F[x]$. Then again, we have $m = r(x)+s(x)$ for $r(x),s(x) \in F[x]$ by the definition of $m$ being in $I$. Thus $mf(x)=r(x)f(x)+s(x)f(x)$ by the distributive property of $F[x]$. Furthermore, since $F[x]$ is a field, we have that $r(x)f(x),s(x)f(x) \in F[x]$. Thus $mf(x) \in I$ by the defintion of $I$. Thus $I$ is an ideal of $F[x]$.
Notice that $I = F[x] + F[x]$. In general, if $R$ is a ring, then $R+R$ is an ideal of $R$.