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How does the fact that p divides b leads to the expression $$sp + ta = (p,a)?$$

Mathematicing
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    Please type the text. The "image" that you have posted is impossible (or difficult, at least) to search. – ajotatxe May 01 '15 at 11:22
  • The key phrase is "By the theorem above". So the previous theorem in the book shows that for any integers $p, q$ there exist integers $s, t$ such that $sp+tq=\gcd(p,q)$, and I assume it also shows that only integer multiples of $\gcd(p,q)$ can be written in the form $sp+tq$. – PM 2Ring May 01 '15 at 11:32
  • Edit: Finally understood. – Mathematicing May 02 '15 at 06:04
  • I'm confused again. The proof seeks to show that p divides either a or b. It supposes that p does not divide a and as a corollary, must divide b. Does this not imply that p is the gcd(a,b)? – Mathematicing May 02 '15 at 06:14
  • I find it hard to see how the proof does not begin from $$g(a,b)=p=as+bt$$ – Mathematicing May 02 '15 at 06:21

2 Answers2

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$(p,a)\mid p\,$ so $\,(p,a)\,$ is $\,1\,$ or $\,p,\,$ by $\,p\,$ prime. But $\,(p,a)=p\,\Rightarrow\,p\mid a,\,$ contra hypothesis. Therefore $\,(p,a)=1.\,$ Hence we deduce by the "theorem above" (Bezout's gcd identity) that the gcd may be expressed as an integral linear combination of its arguments, i.e. $$ (p,a)=1\ \Rightarrow\ sp + ta = 1\ \ \text{for some } s,t\in\Bbb Z$$

Scaling the prior Bezout identity by $\,b\,$ we obtain

$$ s\color{#c00}pb + t\color{#c00}{ab} =\color{#0a0} b\qquad $$

Concluding we note that $\,\color{#c00}{p\mid ab}\,\Rightarrow\,{\color{#c00}{p\mid\rm LHS}}\,\Rightarrow\,p\mid\rm\color{#0a0}{ RHS = b}.\ \ $ QED

More generally: $\ \color{#c00}{p\mid p}b, \color{#c00}{ab}\,\overset{\!\rm\color{darkorange}U}\Rightarrow\, p\mid (\color{#c00}pb,\color{#c00}{ab})\overset{\color{#90f}{\rm D}} = (p,a)b = b\,$ by $\,(p,a)= 1,\, $ by $\rm\color{darkorange}U$ = gcd Universal property, where we replace scaled Bezout equation by $\color{#90f}{\rm D}$ = gcd Distributive Law (see here for a few proofs, and see here for a comparison of various forms of Euclid's Lemma in Bezout, gcd and ideal form). This yields a more general proof that works in any gcd domain, e.g. any UFD. But Bezout-based proofs may fail in more general rings where gcds exist but they are not of Bezout linear form, e.g. polynomial rings $\Bbb Z[x]$ or $\,\Bbb Q[x,y],\,$ where $\,(x,y) = 1\,$ but this gcd has no Bezout linear representation $\,xg(x,y) + y f(x,y) = 1,\,$ else evaluation at $\,x = y = 0\,$ yields $\,0 = 1.\,$ Ditto for $(2,x)$ in $\,\Bbb Z[x]$

Alternatively, $\ (\color{}{p,a})=\color{#c00}{\bf 1}\,\Rightarrow\,(\color{#0a0}{p,ab}) = (\color{#c0f}{p,b})\,$ by evaluating the following in two ways

$$\overbrace{(\color{#c0f}p,\ pb}^{\Large \color{#0a0}p},\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\phantom{pb,}\ \color{#0a0}{ab})}_{\quad\Large\color{#c0f}b\,=\, (\!\underbrace{p,a}_{\huge \bf\color{#c00}1}\!)b}\qquad\qquad\qquad\qquad\qquad $$

where we implicitly used the associative and distributive laws of the gcd (see here for more on such gcd "polynomial" arithmetic).

Bill Dubuque
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It's not because $p|b$, it's because $gcd(p,a)=1$. The equation is true even if it does not equal one. This fact must be in your book somewhere before this proof, look for it,

Gregory Grant
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