Is $\mathbb{Z}[x]/\langle x-3\rangle$ a field?
Answer:
Since $x-3$ is irreducible in $\mathbb{Z}[x]$, we have $\mathbb{Z}[x]/\langle x-3\rangle$ a field.
But I know that $\langle x-3\rangle$ is not a maximal ideal.
Thus, $\mathbb{Z}[x]/\langle x-3\rangle$ is not a field.
So I got confused about the different outcomes.
I need help .
Your mistake is thinking that irreducible elements in integral domains give you fields/maximal ideals. This isn't true in general.
Let $R$ be an integral domain, and let $0 \neq \omega \in R$. Then $\omega$ is irreducible if and only if $(\omega)$ is a proper ideal and maximal in the set of principal ideals of $R$. And $\omega$ is prime if and only if $(\omega)$ is a prime ideal. However, such prime ideals need not be maximal.
In any integral domain, a prime element is irreducible. In a unique factorization domain like $\mathbb{Z}[X]$, the converse is true. The polynomial $X - 3$ is irreducible, so it is prime, and so it generates a prime ideal. Therefore, $\mathbb{Z}[X]/(X-3)$ is an integral domain. But it is not a field: the quotient is isomorphic to $\mathbb{Z}$, as you can see from the surjective homomorphism $\mathbb{Z}[X] \rightarrow \mathbb{Z}, f(X) \mapsto f(3)$ with kernel $(X-3)$.
In a principal ideal domain, all the things you want coincide: $\omega$ generates a maximal ideal if and only if it is prime, if and only if it is irreducible. But $\mathbb{Z}[X]$ is not a principal ideal domain. In fact, if $A$ is a ring, then $A[X]$ is a principal ideal domain if and only if $A$ is a field.
$f \in K[x]$ irreducible $\implies$ $K[x]/(f)$ field only works when $K$ is a field.
A loose philosophical way of thinking about this...
When you form $R[x]$ you are taking the ground ring $R$ and appending an extra element $x$ that behaves as freely as possible while still giving a ring: you can take positive powers, scale, add, and multiply in the usual ways, but you introduce no other relations.
When you then form $R[x]/(f)$ you are doing the first construction, but then forcing $f(x)=0$, which causes additional restrictions on $x$ to kick in, making the resulting ring smaller (or not any bigger). You should think this way: $\mathbb{Z}[x]/(x-3)$ takes $\mathbb{Z}$, appends a mystery symbol $x$, but then declares $x-3=0$, or just $x=3$. Thus you have appended something that was already there, so the resulting ring should be the same as $\mathbb{Z}$ itself: $$ \mathbb{Z}[x]/(x-3) \cong \mathbb{Z}. $$ This is exactly what @D_S claims in their answer.
The same philosophy shows why $\mathbb{R}[x]/(x^2+1)$ creates the complex numbers. You take the real field $\mathbb{R}$, append a new symbol $x$, but then force $x$ to act like $x^2+1=0$, or $x^2=-1$. Thus, we really took $\mathbb{R}$ and appended $x=i$ (or $x=\pm i$, but that's overkill) and made the "smallest" ring possible out of this, which is $\mathbb{C}$.
(These aren't rigorous arguments: I am only trying to guide intuition in thinking about these things, something vastly undersold in teaching.)
