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Is Z[x]/3 a field? I am wondering this since a field quotiented with an ideal generated by a polynomial is a field.

user404974
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4 Answers4

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Note that

$$\mathbb{Z}[X]/(3) \cong (\mathbb{Z}/3\mathbb{Z})[X]$$

and thus $\mathbb{Z}[X]/(3)$ is not a field (since a polynomial ring is never a field).

J. De Ro
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Below is a way that doesn't require $\,\Bbb Z[x]/3\Bbb Z[x]\,\cong\, (\Bbb Z/3)[x]$ and $x$ invertible in $R[x]\!\iff\! R = 0$ $$\begin{align} xf(x)&=1\ \ {\rm in}\ \ \Bbb Z[x]/3\Bbb Z[x]\\ \Rightarrow\ \ x f(x) &= 1 + 3 g(x)\ \ {\rm in}\ \ \Bbb Z[x] \\ \Rightarrow\ \ 0 f(0) &= 1 + 3 g(0)\ \ {\rm in}\ \ \Bbb Z \\ \Rightarrow\ \ 0 &= 1\ \ {\rm in} \ \Bbb Z/3 \end{align}\qquad$$

Remark $ $ It has an instructive universal view as in the linked post: if $x$ is invertible in $\Bbb Z[x]/3\Bbb Z[x]$ then so too is every element $\,r\,$ in every ring where $\,3 = 0,\,$ as follows simply by evaluating $\, xf(x) = 1 + 3g(x)\,$ at $\,x = r.\,$ Thus to present a counterexample it suffices to exhibit any nonunit $\,r\,$ in any ring where $\,3=0.\, $ A natural choice is the nonunit $\,\rm r=0\in \Bbb Z/3,\,$ yielding the above proof.

Strangely, these simple elementary consequences of universal properties of polynomial (and quotient) rings are often overlooked in such elementary contexts - where it is common to instead appeal to degree-based arguments. Witness to this is the fact that - embarrassingly - my trivial linked answer is one of my most popular answers. That many readers apparently were not aware of this viewpoint seems to indicate that we need to do a better job highlighting the power of such universal properties. That's why I chose to emphasize this viewpoint. Even readers who have not yet studied such properties can understand simple proofs like that above - which may help plant the germ of the idea and provide some further motivation once one begins study of universal mapping properties.

Bill Dubuque
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We have $(3) \subset (3,X) \subset \mathbb{Z}[X]$. Since both inclusions are strict, $(3)$ is not a maximal ideal of $\mathbb{Z}[X]$.

Indeed, if $X \in (3)$, then $X=3f(X)$ and so $1=3f(1)$, a contradiction.

Similarly, if $1\in(3,X)$, then $1=3f(X)+Xg(X)$ and so $1=3f(0)$, a contradiction.

lhf
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It is not a field, as polynomials are not invertible. Moreover you need to quotient by an irreducible polynomial to get a field. If you quotient by $x^{2}$, then $x*x=0$ in the quotient.

Alessandro
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    An irreducible polynomial is not sufficient with coefficients in $\mathbb Z$. You need a maximal ideal, and irreducible polynomials give you non maximal prime ideals. – Matt Samuel Nov 11 '19 at 18:18
  • I stated a necessary condition, not a sufficent one. I was just addressing the claim that once you quotient by a polynomial, you have a field. I was merely pointing out that the polynomial needs to be irreducible. @MattSamuel – Alessandro Nov 11 '19 at 18:38
  • You need an ideal generated by an irreducible polynomial and a prime integer. There is no single polynomial you could quotient out by to get a field. – Matt Samuel Nov 11 '19 at 18:39
  • Therefore the polynomial necessarily needs to be irreducible. I never claimed it was a sufficient condition. Since the ring was already quotiented by (3) I omitted the last part. @MattSamuel – Alessandro Nov 11 '19 at 19:13