A simple fact with a simple proof, both of which I wished to see in one place:
Let $R$ be a commutative ring with unit, and let
$I \subset R[x] \tag 1$
be an ideal in $R[x]$. Prove that
$I \cap R \subset R \tag 2$
is an ideal in $R$.
Note Added in Edit, Tuesday 12 May 2020 5:24 PM PST: This note is written in response to the comments on the questiion made by Gone, in which he points out that the result here may be very naturally extended to the case where $R$ is a subring of some arbitrary unital commutative ring $E$,
$R \subset E, \tag 3$
where we do not necessarily have
$E = R[x]; \tag 4$
Gone points out that if we replace $R[x]$ with $E$, and assume
$I \subset E \tag 5$
is an ideal, then the proof given in the answer goes through unchanged, yielding a noteworthy generalization to the result as stated by Yours Truly in the above formulation of this problem. To wit, following Gone's suggestion we show that if $R$ is a subring of any ring $E$, $I \cap R$ is an ideal of $R$. End of Note.
Observe that
$I \cap R \ne \emptyset, \tag 0$
since
$0 \in I \cap R. \tag{0.5}$
Now if
$r_1, r_2 \in I \cap R, \tag 1$
then since
$r_1, r_2 \in I, \tag 2$
an ideal, we have
$r_1 - r_2 \in I; \tag 3$
since
$r_1, r_2 \in R, \tag 4$
a ring,
$r_1 - r_2 \in R; \tag 5$
thus
$r_1 - r_2 \in I \cap R; \tag 6$
with
$s \in I \cap R \tag 7$
and
$r \in R, \tag 8$
we also have
$rs \in I, \tag 9$
again since $I$ is an ideal; furthermore,
$rs \in R, \tag{10}$
since $R$ is a ring; thus,
$rs \in I \cap R, \tag{11}$
and we conclude that $I \cap R$ is an ideal in $R$.