Is $\mathbb{Z}[x]/\langle x-3\rangle$ a field? Is $\mathbb{Z}[x]/\langle x-3\rangle$ a field?
My thought: This quotient ring means that all polynomials in $\mathbb{Z}[x]$ are evaluated at $x=3$. So this is basically isomorphic to $\mathbb{Z}_3$ and thus it will be field.
         Did I make right justification? Please help anybody.
 A: Hint/Solution:  $ \frac {\bf Z[x]}{<x-3>} \cong \bf Z$ (using the map $f :\bf Z[x]  \to \bf Z$, $f(x) \to f(3)$)
A: No it is not a field. Your reasoning is wrong in that the structure given by  "evaluating at $3$" does not give something isomorphic to $\mathbb{Z}_3$ (neither does it give $\mathbb{Z}_3[X]$ which however would not be a field anyway). 
Just try it evaluating $X$ you get $3$, evaluating $X+1$ you get $4$, evaluating $X-5$ you get $-2$ and so on. You can get any integer you want. The quotient is isomorphic to the integers. This is not a field. 
More generally, and likely not relevant for you the quotient of this ring by a principal ideal can never be a field, as the ring has dimension $2$, and a principal ideal thus is never a maximal ideal.    
A: All polynomials are indeed evaluated at $3$. Observe that as a result of this, all polynomials of the form $n$ where $n\in\mathbb{Z}$ are preserved. In particular the polynomial $3$ is not mapped to zero, as you claim, but to $3$. The resulting ring is indeed $\mathbb{Z}$.
