[proofplan] We prove the stated row-shortening condition by applying Bass's classical stable range theorem, in the precise form that a commutative noetherian unital ring of Krull dimension $d$ has Bass stable range at most $d+2$. The rest of the proof is the verification that the hypotheses of Bass's theorem match the present ring and that Bass's stable range convention gives exactly the displayed shortening of a unimodular row of length $d+3$. Finally, we use the elementary maximal-ideal criterion for unit ideals to translate the conclusion into unimodularity of the shortened row. [/proofplan]

[step:Translate $SR_{d+2}$ into the required row-shortening problem] Let \begin{align*} (a_1,\dots,a_{d+3})\in R^{d+3} \end{align*} be a unimodular row. Thus the ideal \begin{align*} (a_1,\dots,a_{d+3})=R \end{align*} is the unit ideal. Put \begin{align*} b:=a_{d+3} \end{align*} and put \begin{align*} n:=d+2. \end{align*} We must find elements $x_1,\dots,x_n\in R$ such that, for \begin{align*} c_i:=a_i+bx_i \end{align*} with $1\le i\le n$, the ideal \begin{align*} (c_1,\dots,c_n) \end{align*} is equal to $R$. Let $\operatorname{MaxSpec}(R)$ denote the set of maximal ideals of $R$, and let $\mathcal{P}(\operatorname{MaxSpec}(R))$ denote its power set. For any ideal $I\trianglelefteq R$, define the set-valued map \begin{align*} V_{\max}:\{\text{ideals of }R\}&\to \mathcal{P}(\operatorname{MaxSpec}(R)) \end{align*} by \begin{align*} V_{\max}(I)&:=\{\mathfrak m\in \operatorname{MaxSpec}(R): I\subseteq \mathfrak m\}. \end{align*} An ideal of a commutative unital ring is proper if and only if it is contained in a maximal ideal. Hence it is enough to construct $x_1,\dots,x_n$ so that \begin{align*} V_{\max}((c_1,\dots,c_n))=\varnothing. \end{align*} [/step]

[step:Apply Bass's classical stable range theorem]We use Bass's stable range theorem in the following precise external form: if $A$ is a commutative noetherian unital ring of Krull dimension at most $d$, then $A$ has Bass stable range at most $d+2$. Equivalently, for every unimodular row \begin{align*} (r_1,\dots,r_{d+3})\in A^{d+3}, \end{align*} there exist $y_1,\dots,y_{d+2}\in A$ such that \begin{align*} (r_1+r_{d+3}y_1,\dots,r_{d+2}+r_{d+3}y_{d+2}) \end{align*} is a unimodular row. This is Bass, Algebraic K-Theory, Chapter V, Theorem 3.5, specialized to commutative noetherian rings, where the Krull dimension of $A$ is the dimension parameter. The hypotheses of Bass's theorem match the present situation. The ring $R$ is commutative, noetherian, unital, and has Krull dimension $d$. The chosen row \begin{align*} (a_1,\dots,a_{d+3})\in R^{d+3} \end{align*} is unimodular by assumption, meaning \begin{align*} (a_1,\dots,a_{d+3})=R. \end{align*} Applying Bass's theorem with $A=R$ and $r_i=a_i$ for $1\le i\le d+3$ gives elements $x_1,\dots,x_{d+2}\in R$ such that \begin{align*} (a_1+a_{d+3}x_1,\dots,a_{d+2}+a_{d+3}x_{d+2}) \end{align*} is unimodular.[/step]

[guided]The point of this step is to identify the exact external theorem being used, rather than hiding the main argument inside an unnamed lemma. Bass's theorem says: if $A$ is a commutative noetherian unital ring of Krull dimension at most $d$, then every unimodular row of length $d+3$ over $A$ can be shortened by adding multiples of the last entry to the preceding $d+2$ entries. In symbols, for every unimodular row \begin{align*} (r_1,\dots,r_{d+3})\in A^{d+3}, \end{align*} there exist elements \begin{align*} y_1,\dots,y_{d+2}\in A \end{align*} such that \begin{align*} (r_1+r_{d+3}y_1,\dots,r_{d+2}+r_{d+3}y_{d+2}) \end{align*} is unimodular. This is Bass, Algebraic K-Theory, Chapter V, Theorem 3.5, in the commutative noetherian case. We now verify the hypotheses one by one. The theorem statement gives that $R$ is commutative, noetherian, and unital. It also gives that the Krull dimension of $R$ is the finite integer $d\in\mathbb N\cup\{0\}$, so the dimension hypothesis of Bass's theorem is satisfied with $A=R$. The row already fixed in this proof is \begin{align*} (a_1,\dots,a_{d+3})\in R^{d+3}, \end{align*} and it is unimodular, which means precisely that the ideal generated by its entries is the unit ideal: \begin{align*} (a_1,\dots,a_{d+3})=R. \end{align*} Thus Bass's theorem applies with $r_i=a_i$ for each integer $i$ with $1\le i\le d+3$. It produces elements \begin{align*} x_1,\dots,x_{d+2}\in R \end{align*} such that the shortened row \begin{align*} (a_1+a_{d+3}x_1,\dots,a_{d+2}+a_{d+3}x_{d+2}) \end{align*} is unimodular. This is exactly the row-shortening assertion required by the displayed formulation of $SR_{d+2}$.[/guided]

[step:Record the shortened row produced by Bass's theorem]The preceding application of Bass's theorem gives elements \begin{align*} x_1,\dots,x_n\in R \end{align*} since $n=d+2$. Define \begin{align*} c_i:=a_i+bx_i \end{align*} for each integer $i$ with $1\le i\le n$. Because $b=a_{d+3}$, Bass's conclusion says that the row \begin{align*} (c_1,\dots,c_n) \end{align*} is unimodular. Equivalently, \begin{align*} (c_1,\dots,c_n)=R. \end{align*} Therefore no maximal ideal of $R$ contains this ideal, and hence \begin{align*} V_{\max}((c_1,\dots,c_n))=\varnothing. \end{align*}[/step]

[guided]We now translate Bass's conclusion into the notation introduced at the start of the proof. Since \begin{align*} n=d+2, \end{align*} the first $n$ entries are precisely \begin{align*} a_1,\dots,a_{d+2}, \end{align*} and the last entry is \begin{align*} b=a_{d+3}. \end{align*} Bass's theorem supplies elements \begin{align*} x_1,\dots,x_n\in R. \end{align*} For each integer $i$ with $1\le i\le n$, define the modified entry \begin{align*} c_i:=a_i+bx_i. \end{align*} Substituting $b=a_{d+3}$ into Bass's conclusion gives that \begin{align*} (c_1,\dots,c_n) \end{align*} is a unimodular row, which means that the generated ideal is the unit ideal: \begin{align*} (c_1,\dots,c_n)=R. \end{align*} A maximal ideal is proper, so it cannot contain the unit ideal. Hence no maximal ideal of $R$ contains $(c_1,\dots,c_n)$, or, in the notation of $V_{\max}$, \begin{align*} V_{\max}((c_1,\dots,c_n))=\varnothing. \end{align*}[/guided]

[step:Conclude that the shortened row is unimodular] Since no maximal ideal contains the ideal \begin{align*} (c_1,\dots,c_n), \end{align*} this ideal is not proper. Therefore \begin{align*} (c_1,\dots,c_n)=R. \end{align*} Recalling that $n=d+2$, $b=a_{d+3}$, and $c_i=a_i+a_{d+3}x_i$, we have found elements $x_1,\dots,x_{d+2}\in R$ such that \begin{align*} (a_1+a_{d+3}x_1,\dots,a_{d+2}+a_{d+3}x_{d+2}) \end{align*} is unimodular. This is exactly the Bass stable range condition $SR_{d+2}$. [/step]