# 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.

Hint/Solution: $\frac {\bf Z[x]}{<x-3>} \cong \bf Z$ (using the map $f :\bf Z[x] \to \bf Z$, $f(x) \to f(3)$)

• Z[x]/<x-3>=Z3[x] is it not correct? – user1991 Jan 18 '16 at 14:41
• No its not,see the answer by quid! – Arpit Kansal Jan 18 '16 at 14:43
• @user1991 No, though it is true that $\mathbb{Z}[x]/<3>$ is isomorphic to $\mathbb{Z}_3[x]$ – πr8 Jan 18 '16 at 14:47

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.

• I have one confusion . Is not $\langle x-3 \rangle$ a maximal ideal of $\mathbb Z[x]$ $?$ Commutative ring with identity quotiented by a maximal ideal should be a field . What am I getting wrong $?$ – user118494 Jan 18 '16 at 14:59
• See his last sentence. – Future Jan 18 '16 at 15:01
• @T.S.L : I have not learnt the dimension of ring things at all . Is there any other way of explaining $?$ – user118494 Jan 18 '16 at 15:05
• Can you find a proper ideal containing $(x-3)$? – Future Jan 18 '16 at 15:06
• @user118494 no it is not a maximal ideal. For example, it'd be contained in $\langle x-3, 2 \rangle$ Or, while circular in this context, it is not maximal because the qutient is not a field. In $Q[X]$ though your ideal would be maximal. – quid Jan 18 '16 at 15:07

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 $\mathbb{Z}$.