# Show that the following factor ring is a field

Let $f = X^3 + 3X + 2$ in $\mathbb{Z}[X]$

1) Show that the factor ring $\mathbb{Q}[X]/\langle f \rangle$ is a field

2) Show that in the factor ring $\mathbb{Z}[X]/\langle f \rangle$ the element $X$ is nonzero and not invertible and deduce that $\mathbb{Z}[X]/\langle f \rangle$ is not a field.

How do I prove those two statements?

For the second one I was thinking that I should use a proof by contradiction, but I don't completely grasp how to do it.

• non- null = it is not equal to the null element Jun 16, 2017 at 16:13
• @Chill2Macht I'm guessing OP means "non-invertible." Jun 16, 2017 at 16:15
• Yes, thank you! I meant non-invertible. I do not know all the terms in english. I'm sorry for the mistake. Jun 16, 2017 at 16:24

For $(2)\!:\,\ xh\equiv 1\pmod{f}\,\Rightarrow\, xh+fg = 1\ {\rm in}\ \Bbb Z[x]\,\overset{\large x\,=\,0}\Rightarrow\, 2g(0)=1\ {\rm in}\ \Bbb Z\ \Rightarrow\!\Leftarrow$

For $(1)$ recall $R/I$ is a field $\iff I$ is max, and for $R$ a pid we have $(f)$ max $\!\iff f$ irreducible

For 1 try showing that $\langle f \rangle$ is a maximal ideal in Q[X] -- then it follows that $Q[X]/\langle f \rangle$ is a field. See here.

For 2, after showing that $X$ is non-zero and non-invertible, then $\langle f , X \rangle$ is also a proper ideal in $Z[X]$ -- this means in particular that $\langle f \rangle$, which is properly contained in $\langle f, X \rangle$, can not be a maximal ideal. Therefore, by the same theorem as above, $Z[X]/\langle f \rangle$ cannot be a field.

Hint for 1: show the polynomial is irreducible, because it has no roots in $\mathbb{Q}$ (and it has degree $3$).

Hint for 2: by 1 and Gauss’ lemma, the polynomial is also irreducible in $\mathbb{Z}[X]$, so $\mathbb{Z}[X]/\langle f\rangle$ is a domain. Any element of this ring can be written in a unique form as $a_0+a_1X+a_2X^2$ (here $X$ denotes the image in the quotient ring). Can you find an inverse to $X$?

• why does irreducibility involves that $\mathbb{Q}[X]/f$ is a field ? I'm dealing with the same problem Jun 16, 2017 at 17:18
• @EduardValentin If the polynomial $f$ is irreducible, then the ideal it generates is maximal; recall that $\mathbb{Q}[X]$ is a principal ideal domain. Jun 16, 2017 at 17:35