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How to prove that $mathbbZ[X]/gmathbbZ[X]$ is not a field, where $g$ is a non constant polynomial in $mathbbZ[X]$?

(The key is to show that $f : mathbbZ[X]/gmathbbZ[X] to mathbbZ/pmathbbZ$ is not injective.)

edit 1:
Now I'm trying to find the characteristic of the given ring.
Consider the map: $g: mathbbZ to mathbbZ[X]/gmathbbZ[X]$, where $m mapsto 1 + 1 + cdots + 1 $ (m times). Since the $ker f$ is an ideal of $mathbbZ$, it has to be of the form $nmathbbZ$, where $n$ is the characteristic. So I guess the characteristic is $0$...no?

edit 2: Thank you for all the replies below! I looked through the answer but it's a little advanced for me... I figured it out eventually using some more basic ideas and I'll put it here when I get some time...Thanks again!

edit 3:

My solution:

Step 1: Prove there is a $a in mathbbZ$ such that $g(a) neq 0, pm 1$ and let $p$ be a prime number that divides $g(a)$.

Suppose $g(X) = sum_i = 1^na_iX^i$, then $g(X) = 0$ has at most $n$ integer solutions, $p(X) = g(X) -1 = 0$ has at most $n$ integer solutions, and so is $q(X) = g(X)+1 = 0$. There are at most $3n$ integers that satisfy $g(a) = 0,pm1$, but $mathbbZ$ is infinite, so there has to be a $a in mathbbZ$ such that $g(a) neq 0, pm 1$.

Step 2: Prove that there is a unique well-defined surjective morphism of rings $f: mathbbZ[X] to mathbbZ/pmathbbZ$ sending $X$ to $a$ mod $p$ and that it yields a homomorphism of rings $phi: mathbbZ[X]/gmathbbZ[X] to mathbbZ/pmathbbZ$.

Observe that $gmathbbZ[X]$ is in the kernel, so the map factor through $mathbbZ[X]/gmathbbZ[X]$.

Step 3: Show that the resulting map is not injective.

Note that the map $phi$ is injective if and only if $ker f = gmathbbZ[X]$, but actually $gmathbbZ[X] subset ker f$. Alternatively, according to my prof, if the map was injective, $mathbbZ[X]/gmathbbZ[X]$ would identify with a subring of $mathbbZ/pmathbbZ$ so it would have characteristic $p$, contradiction!

Step 4: We use that if $g: K to A$ where $K$ is a field, then $g$ is either injective or the zero map.
Since the above map is not a zero map and we proved that it's not injective, $mathbbZ[X]/gmathbbZ[X]$ can't be a field.

My answer expands on JonHales comment. Suppose your polynomial is $g(x)=a_nx^n + cdots + a_0$, with $a_iinBbbZ$, $a_nne 0$, and $n > 0$. Choose a prime $p$ with $pnmid a_n$. Then mod $p$, $g(x)$ still has degree $n$.

Thus $BbbZ[x]/(g,p)simeq BbbF_p[x]/(g)ne 0$, so $(g,p)$ is a proper ideal of $BbbZ[x]$. Moreover, it properly contains $(g)$, since $(g)cap BbbZ=(0)$. Thus $(g)$ is not maximal, and $BbbZ[x]/(g)$ is not a field.

Edit

In response to the OP's edit, saying that they found this solution a bit advanced, I thought I'd translate the solution into more basic ideas for future readers.

Again, if our polynomial is $g(x)=a_nx^n+cdots + a_0$, then if $p$ is a prime integer with $pnmid a_n$, $p$ is nonzero, but not a unit in $BbbZ[x]/(g)$. Therefore $BbbZ[x]/(g)$ is not a field.

Proof.

By contradiction. Assume $p cdot c(x) equiv 1 pmodg(x)$ for some polynomial $c(x)$. Then $pc(x)-1 = g(x)h(x)$ for some polynomial $h(x)$, so $pc(x) = g(x)h(x)-1$. Write $h(x)=pq(x)+r(x)$, where the coefficients of $r$ are either between $0$ and $p-1$ (inclusive).

Then $$pc(x)=g(x)(pq(x)+r(x))-1=pg(x)q(x) + g(x)r(x)-1,$$
and
$$p(c(x)-g(x)q(x))=g(x)r(x)-1.$$
Now if $r(x)=0$, we have that the left hand side is divisible by $p$, and the right hand side is $-1$, so this is impossible.

On the other hand, if $r(x)ne 0$, $g(x)r(x)$ has positive degree, and it's leading term is $a_nb_m$, where $b_m$ is the leading term of $r(x)$. However $p$ must divide $g(x)r(x)-1$, so it must divide the leading term, so $pmid a_nb_m$. However $pnmid a_n$, and $1le b_mle p-1$, so $pnmid b_m$ either. Thus $pnmid a_nb_m$. Contradiction.