Let $R$ be an integral domain and let $a,b \in R$. Let $aR$ and $bR$ denote the principal ideals of $R$ generated by $a$ and $b$ respectively. (If you prefer, you can use the notation $\langle a \rangle$ and $\langle b \rangle$ instead.)
Prove that $aR = bR$ if and only if $a = bu$ for some unit $u$ of $R$.
Say more explicitly what part a means in the case $R = \mathbb{Z}$.
Say more explicitly what part a means in the case $R = \mathbb{Q}[x]$.
Consider the evaluation homomorphism $\phi_0 : \mathbb{Z}[x] \to \mathbb{Z}$ defined by $\varphi_0(f(x)) = f(0)$. Define the inverse image $N = \phi_0^{-1}[2\mathbb{Z}]$. Then $N$ is an ideal of $\mathbb{Z}[x]$ because $2\mathbb{Z}$ is an ideal of $\mathbb{Z}$ and the inverse image of an ideal under a homomorphism is an ideal.
Is $N$ a principal ideal of $\mathbb{Z}[x]$? Justify your answer.
Warning: there is no such function as $\phi_0^{-1}$ here because $\phi_0$ is not injective, so if you write about a polynomial $\phi_0^{-1}(n)$ where $n$ is an integer, then you are writing nonsense.
Consider the principal ideal $\langle x \rangle$ of the ring $\mathbb{Z}[x]$.
Is it prime?
Is it maximal?
Justify your answers.
Let $R$ be a principal ideal domain (meaning that $R$ is an integral domain, and every ideal of $R$ is principal.) Let $P$ be a nontrivial prime ideal of $R$. Prove that $P$ is maximal.
Hint: let $N$ be an ideal of $R$ such that $P \subseteq N$. By our hypothesis, $P$ and $N$ are both principal, say $P = \langle p \rangle$ and $N = \langle a \rangle$. That should get you started.
Find an ideal $N$ of the polynomial ring $\mathbb{Q}[x]$ such that the factor ring $E = \mathbb{Q}[x]/N$ is a field with more than one cube root of unity. Find all of the cube roots of unity in your field $E$.
There are only five problems this week.