Theorem: Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal

Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$
Then for some $x \in D$:
:$J = \ideal x$
where $\ideal x$ denotes the principal ideal generated by $x$.

Proof:
Let the unity of $D$ be $1_D$.
By definition of principal ideal:
:$\ds \ideal a = \set {\sum_{i \mathop = 1}^n r_i \circ a \circ s_i: n \in \N, r_i, s_i \in D}$
Let $x, y \in J$.
By definition of linear combination:
{{begin-eqn}}
{{eqn | l = x
      | r = \sum_{i \mathop = 1}^n r_i \circ a_i
      | c = for some $n \in \N$ and for some $r_i \in D$ where $i \in \set {1, 2, \dotsc, n}$
}}
{{eqn | r = r_1 \circ a_1 + r_2 \circ a_2 + \dotsb + r_n \circ a_n
      | c = for some $r_1, r_2, \dotsc, r_n \in D$
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = y
      | r = \sum_{i \mathop = 1}^n s_i \circ a_i
      | c = for some $n \in \N$ and for some $s_i \in D$ where $i \in \set {1, 2, \dotsc, n}$
}}
{{eqn | r = s_1 \circ a_1 + s_2 \circ a_2 + \dotsb + s_n \circ a_n
      | c = for some $s_1, s_2, \dotsc, s_n \in R$
}}
{{eqn | ll= \leadsto
      | l = -y
      | r = -\sum_{i \mathop = 1}^n s_i \circ a_i
      | c = 
}}
{{eqn | r = -1_D \times \sum_{i \mathop = 1}^n s_i \circ a_i
      | c = Product with Ring Negative: Corollary
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {-1_D} \times \paren {s_i \circ a_i}
      | c = {{Ring-axiom|D}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {-\paren {s_i \circ a_i} }
      | c = Product with Ring Negative: Corollary
}}
{{eqn | r = \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}
      | c = Product with Ring Negative
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = x + \paren {-y}
      | r = \sum_{i \mathop = 1}^n r_i \circ a_i + \sum_{i \mathop = 1}^n s_i \circ \paren {-a_i}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + s_i \circ \paren {-a_i} }
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-\paren {s_i \circ a_i} } }
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i \circ a_i + \paren {-s_i} \circ a_i}
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {r_i + \paren {-s_i} } \circ a_i
      | c = {{Ring-axiom|D}}
}}
{{eqn | r = \sum_{i \mathop = 1}^n t_i \circ a_i
      | c = where $t_i - r_1 + \paren {-s_i}$
}}
{{eqn | o = \in
      | r = J
      | c = as $t_i \in D$
}}
{{end-eqn}}
Then we have:
{{begin-eqn}}
{{eqn | l = x \circ y
      | r = \paren {\sum_{i \mathop = 1}^n r_i \circ a_i} \circ \paren {\sum_{i \mathop = 1}^n s_i \circ a_i }
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {t_i \circ a_i}
      | c = where $t_i \in D$ for $i \in \set {1, 2, \dotsc, n}$
}}
{{eqn | r = \sum_{i \mathop = 1}^n \paren {a_i \circ t_i}
      | c = as $\circ$ is commutative in an integral domain
}}
{{end-eqn}}
{{explain|There exists (or ought to) some convolution result which proves the above -- I just haven't found it yet.}}
Thus by the Test for Ideal, $J$ is an ideal of $D$.
As $D$ is a principal ideal domain, it follows that $J$ is a principal ideal.
Thus by definition of principal ideal:
:$J = \ideal x$
for some $x \in D$.
{{qed}}