Show that every Euclidean ring is a principal ring Question: Show that every Euclidean ring is a principal ring.


My attempt: Let $(R, +, \cdot)$ be a Euclidean domain. Let $I \subseteq R$ be a nonzero ideal of $R$. Let $d \in I$ with $d \neq 0$ be any element such that $\delta (d)$ is minimum in $I$. We claim that $I = dR = \{ dx : x \in R \}$. Since $d \in I$ we have that:
$$\begin{align} \quad dR \subseteq I \quad (*) \end{align}$$
Now let $a \in I$ with $a \ne 0$. Since $R$ is a Euclidean domain there exists elements $q,r \in R$ such that $a=dq+r$ and $r=0$ or $\delta (r) &lt; \delta (d)$. Now we have that $r=a−dq$ where $a \in I$ and $−dq \in I$, so $r=a−dq \in I$. From the minimality of $\delta(d)$ we have that:
$$\begin{align} \quad \delta (r) \geq \delta (d) \end{align}$$
Hence, since $\delta(r) ≮  \delta(d)$ we must have that $r=0$. So $0=a−dq$. Hence $a=dq$. So $a \in dR$. Thus:
$$\begin{align} \quad dR \supseteq I \quad (**) \end{align}$$
From $(∗)$ and $(∗∗)$ we conclude that $I=dR$. In other words, $I=&lt;d&gt;$. So every Euclidean domain is a principal ideal domain. 

From here, how to show that every Euclidean ring is a principal ring???
No, the relation $0\leq \delta(r)&lt;\delta(d)$ implies that $\delta(r)=0$ and so $r=0$. Thus each element of $I$ is a multiple of $d$. On the other hand, the multiplies of $d$ form an ideal. Hence, the result follows.
No, the relation $0\leq \delta(r)&lt;\delta(d)$ implies that $\delta(r)=0$ and so $r=0$. Thus each element of $I$ is a multiple of $d$. On the other hand, the multiplies of $d$ form an ideal. Hence, the result follows.