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Let $J$ consist of all the polynomials $a_0+a_1x+...+a_nx^n$ in $A[x]$ where $A$ is a field, such that $a_0+a_1+a_2+...+a_n=0$. Prove tht $J$ is an ideal of $A[x]$.

So I know that a kernel is also the ideal, so I was wondering if $a_0+a_1+a_2+...+a_n=0$ then that would be the kernel (I believe). So since J includes the kernel then J is an ideal. Does this follow?

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Consider the map $\phi : A[x] \rightarrow A$ where $\phi(p) = p(1)$ for all $p \in A[x] $.

Notice that: $$ \phi(p+q) = (p+q)(1) = p(1) + q(1) = \phi(p) + \phi(q) $$

$$ \phi(pq) = (pq)(1) = p(1)q(1) = \phi(p) \phi(q) $$

$$ \phi(1) = 1 $$

Thus $\phi$ is a ring homomorphism with kernel $\{p \in A[x] : \ p(1) = 0\} = \{\sum_{j=0}^n a_j x^j : a_0 + \ldots + a_n = 0 \} = J$.

Therefore, as the kernel of a homomorphism is an ideal (like you said) $J$ is an ideal of $A[x]$.

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Hint: Think this function $\phi:A[x]\rightarrow A$, such that $\phi(p)=p(1)$. Find the kernel of $\phi$.

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