1. ## prime

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.

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?

2. Originally Posted by 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?
Please do read now, coldly, what you wrote: can you understand it? Because if you can then you're not explaining it clearly at all, and from things like this marks in exams go away for vacation in Acapulco:

$\displaystyle (x-1)\subset (2,x-1)$

Tonio

3. What does the notation $\displaystyle (2,x-1)$ mean?

4. Originally Posted by Sampras
What does the notation $\displaystyle (2,x-1)$ mean?

Err...the ideal generated by x-1 and 2, also denoted by $\displaystyle 2\mathbb{Z}[x] +(x-1)\mathbb{Z}[x]$

Tonio