set of polynomials whose coefficient sum is zero is an ideal.

Question

Show that the set of polynomials whose coefficient sum is zero is an ideal.

Proof: Let $ p(x) = a_nx^n + ... +ax + a_0$ and $q(x) = b_ny^n + ...+ by + b_0$.

Then suppose $a_n + ...+ a_0 = 0$ and $b_n + ...+ b_0 = 0$

Notice that this happens when $p(1)=0$ and $q(1)=0$ , that is $ p(1) = a_n(1)^n + ... +a(1) + a_0 = a_n + ....+ a_0 = 0$. Similarly for $q(1) = 0$.

Then $(p + q)(1) = p(1) + q(1) = 0$ and $p(1)q(1) = 0$ and $q(1)p(1) = 0$. So the set is an ideal.

Can someone please verify this? any suggestion might help. Thanks

Answer 1

This is just the kernel of the evaluation homomorphism when you evaluate at $1$. Kernels of homomorphisms are always ideals.

Answer 2

This is obviously closed under subtraction, but the tricky part is seeing it is closed under multiplication. You should note that for any $bx^k$ and $p(x) = \sum\limits_{i=0}^n a_ix^i$ in the possible ideal, $I$, we have that $$bx^k \sum\limits_{i=0}^n a_ix^i = \sum\limits_{i=0}^n ba_i x^{i+k}.$$ The sum of the coefficients will be $$\sum\limits_{i=0}^n ba_i = b\sum\limits_{i=0}^n a_i = b\cdot 0 = 0.$$ So for any $p(x), q(x) \in I$, we have $$p(x)q(x) = \sum\limits_{k=0}^n \left( a_kx^k \sum\limits_{i=0}^m b_ix^i\right),$$ and the sum of the coefficients will be $$\sum\limits_{k=0}^n \left( a_k\sum\limits_{i=0}^m b_i\right) = 0$$ since the sum of the coefficients of $q(x)$ will be $0$.

Same argument for the left/right multiplication using an arbitrary $r(x) \in \mathbb{Z}[x]$.