Proof that every element of a Euclidean domain has a $\gcd$

Question

I am required to prove that every element $a$ and $b$ of a Euclidean domain $E$ has a greatest common divisor. Here's my approach. Let $D=\{ax + by: a,b,x,y \in E \} $ We claim that $D$ is an ideal of $E$

For it is clear that $D$ is closed under addition and subtraction and that if $c_1,c_2 \in D$ then so is $c_1c_2 \land c_2c_1$ Also, for every $r \in E$ and $c \in D$ then $cr, rc \in D$ as $cr = (ax + by)r =a(xr)+ b(yr) \in D$ Similar arguments also holds for $rc$. Now we have established that $D$ is an Ideal of $E$. Now since every ideal of a Euclidean domain is a Principal Ideal Domain It follows that $D$ has a generator,say

$D = <d>$ Now, $a \in D$ as for $x=1, y=0 \in D$ $ax + by \in E \implies a\cdot1 + b\cdot0 = a \in D$ And similarly, $a\cdot0 + b\cdot1 = b \in E$ Hence $a \in D, b \in D$ And since $D$ is generated by $<d>$ There are elements $q_1,q_2 \in E$ such that $d = aq_1 \land d= bq_2$ Which implies that $a|d \land b|d$ Now let $c$ be an arbitrary divisor of $a$ and $b$ Then by definition, $c=ax \land c= by$ Also $c$ must divide any linear combination of $a$ and $b$, thus,in particular $c|(ax + by) \implies c|d$ Hence we have proved that $d$ is the greatest common divisor of $a,b$. This completes the proof.

I gathered the ideas for this proof from resources I found on the web. I would like to know if my approach is sound. I wish to present this proof in an exam, the one given to us during the class depends on the constructive proof of the Euclidean algorithm which I found messy and unattractive. If this approach is not sound someone should please show me a better way.

Answer 1

• Edit to correct your title as suggested by J. W. Tanner, instead of answering his comment.
• $D\ne\{ax + by: a,b,x,y \in E \}$ ($a,b$ are fixed) $D=\{ax + by: x,y \in E \}.$
• "if $c_1,c_2 \in D$ then so are $c_1c_2$ and $c_2c_1$" is (weaker hence) redundant with "for every $r \in E$ and $c \in D$ then $cr, rc \in D$".
• "Similar arguments also holds for $rc$" does not work. You have to assume $E$ commutative.
• "Which implies that $a|d \land b|d$" no: $d\mid a$ and $d\mid b.$
• "$c=ax \land c= by$" no: $a=cx$ and $b=cy$ (anyway useless).
• "in particular $c|(ax + by) \implies c|d$" explain more (who are your $x,y$? not the useless previous ones).