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**Sampras** Show that the principal ideal $\displaystyle (x-1) $ in $\displaystyle \mathbb{Z}[x] $ is prime but not maximal.

To show it is prime consider $\displaystyle ab \in (x-1) $. Then we want to show that $\displaystyle a \in (x-1) $ or $\displaystyle b \in (x-1) $.

Now $\displaystyle (x-1) = \{c(x-1): c \in \mathbb{Z}[x] \} $. So $\displaystyle ab = c[x-1]$ for some polynomial $\displaystyle c \in \mathbb{Z}[x] $. This implies that either $\displaystyle a \in (x-1) $ or $\displaystyle b \in (x-1) $ since we can take the other to be the unit polynomial.

Uuh?? What "other"? And what is "the unit polynomial"? Perhaps this is simpler: a polynomial is divisible by (x-1) iff 1 is one of its roots (this is the residue theorem for (long) division of polynomials), so $\displaystyle ab\in (x-1) \Longleftrightarrow ab(1)=0 \Longleftrightarrow a(1)=0 \;or \;b(1)=0...$

Suppose for contradiction that it was maximal. Then if $\displaystyle (x-1) \subseteq J \subseteq \mathbb{Z}[x] $, either $\displaystyle (x-1) = J $ or $\displaystyle J = \mathbb{Z}[x] $. Consider $\displaystyle (2x) $. Then it doesn't contain $\displaystyle x^2+1 = x(x+1) $. Thus $\displaystyle (x+1) $ is not maximal.

Is this correct?