I am looking at the homomorphism $\mathbb{Z}[x] \rightarrow \mathbb{Z}$ that sends $x$ to $1$. I need to explain what the Correspondence Theorem when applied to this map says about the ideals of $\mathbb{Z}[x]$. This is what I have so far:
I know that the map is surjective. I now want to use the division algorithm, and I have been told that $f(x) = q(x)(x-1) + r(x)$ with $deg(r(x)) < deg((x-1))=1$. From this I see that $r(x)$ is a constant integer and $f(1)=r(1)$. I know how to prove the rest of the problem, but I am confused with:
- $f(x) = q(x)(x-1) + r(x)$
I am not sure why we have $(x-1)$ multiplied by $q(x)$. Does anyone know why this is the case?
Notice the homomorphisms sends a polynomial to the sum of its coefficients.
What is the kernel of the homomorphism? It is the set of polynomials for which the sum of its coefficients is $0$, call this ideal $I$.
The correspondence theorem tells us the ideals containing $I$ are in bijective correspendence with the ideals of $\mathbb Z$. Which is the correspondance? Assign to any ideal of $\mathbb Z$ its preimage under the projection map.
What do we conclude? The ideals of $\mathbb Z[x]$ containing $I$ are as follows:
Take $n\in \mathbb Z$, then our ideal consists of the polynomials so that the sum of its coefficients is a multiple of $n$. (Notice when $n=0$ we get $I$).