Show that this set of polynomials is ideal in F[x] In $\mathbb{F}[x]$, where $\mathbb{F}$ is a ﬁeld, let $J$ be the set of elements of polynomials that have coefficients that add to zero (so $a_0 + a_1 + ... + a_n = 0$). Show that $J$ is an ideal of $\mathbb{F}[x]$. I know that the proof of this statement is meant to be very short, but I don't know how to go about it.
 A: Consider the map $\eta:F[x]\to F$ where $P$ is mapped to the sum of its coefficents, that is $P\mapsto P(1)$. You need to show it is a ring homomorphism, and all will follow.
A: You don't even need to go as far as homomorphy to get at this, though that's probably the way the big kids do it.  But for us little kids, there is the following
Lemma:  Let $R$ be a commutative ring with unit, and let $f(x) = \sum_0^{\deg f}f_i x^i \in R[x]$.  Then $\sum_0^{\deg f}f_i = 0$ if and only if $f(1) = 0$.
Proof of Lemma: Weaaalll, it's really pretty simple:  if $f(1) = 0$, then  $\sum_0^{\deg f}f_i = \sum_0^{\deg f}f_i 1^i = f(1) = 0$; and if $\sum_0^{\deg f}f_i = 0$, 
then$ f(1) = \sum_0^{\deg f}f_i 1^i = \sum_0^{\deg f}f_i = 0$.  QED.
Having the above little gem at our disposal, we can prove the following
Proposition:  Let $R$ be a commutative ring with unit, and let $J \subset R[x]$ be defined by $J = \{f(x) = \sum_0^{\deg f}f_i x^i \in R[x] \mid  \sum_0^{\deg f}f_i = 0 \}$; then $J$ is a (two-sided) ideal in $R[x]$.
Proof of Proposition:  to show $J$ is an ideal, we need to show two things:  i.) if $f(x), g(x) \in J$, then $f(x) - g(x) \in J$; and ii.)  if $f(x) \in J$ and $g(x) \in R[x]$, then $g(x)f(x) \in J$.  For (i.), $f(x), g(x) \in J \Rightarrow f(1) = g(1) = 0$, by our lemma.  Thus $f(1) - g(1) = 0$, so $f(x) - g(x) \in J$.  As for (ii.), we have $f(1) = 0$, so $g(1)f(1) = 0$, whence $f(x)g(x) \in J$, which is obviously two-sided since
$R[x]$ is commutative.  QED.
Now taking $R$ to be a field $F$ and applying the proposition finishes things off.  And we are done.  One more time, QED.
Sandbox algebra. ;-)!!!  Also, I thinks I should admit that though I didn't explicitly mention any homomorphisms, in reality I came pretty close!
Note:  I suspect this proposition may hold even if $R$ is not unital, but I putting off further discussion for the moment.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
