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Difference between revisions of "Bezout's Lemma"

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'''Bezout's Lemma''' states that if <math>x</math> and <math>y</math> are nonzero integers and <math>g = \gcd(x,y)</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>g</math>.
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'''Bezout's Lemma''' states that if <math>x</math> and <math>y</math> are nonzero [[Integer|integers]] and <math>g = \gcd(x,y)</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>g</math>.
  
 
Furthermore, <math>g</math> is the smallest positive integer that can be expressed in this form, i.e. <math>g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}</math>.
 
Furthermore, <math>g</math> is the smallest positive integer that can be expressed in this form, i.e. <math>g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}</math>.
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Now to prove <math>g</math> is minimum, consider any positive integer <math>g' = x\alpha'+y\beta'</math>. As <math>g|x,y</math> we get <math>g|x\alpha'+y\beta' = g'</math>, and as <math>g</math> and <math>g'</math> are both positive integers this gives <math>g\le g'</math>. So <math>g</math> is indeed the minimum.
 
Now to prove <math>g</math> is minimum, consider any positive integer <math>g' = x\alpha'+y\beta'</math>. As <math>g|x,y</math> we get <math>g|x\alpha'+y\beta' = g'</math>, and as <math>g</math> and <math>g'</math> are both positive integers this gives <math>g\le g'</math>. So <math>g</math> is indeed the minimum.
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==Generalization to Principal Ideal Domains==
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Bezout's Lemma can be generalized to [[Principal ideal domain|principal ideal domains]].
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Let <math>R</math> be a principal ideal domain, and consider any <math>x,y\in R</math>. Let <math>g = \gcd(x,y)</math>. Then there exist elements <math>r_1,r_2\in R</math> for which <math>xr_1+yr_2 = g</math>. Furthermore, <math>g</math> is the minimal such element (under divisibility), i.e. if <math>g' = xr_1'+yr_2'</math> then <math>g|g'</math>.
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Note that this statement is indeed a generalization of the previous statement, as the [[ring]] of integers, <math>\mathbb Z</math> is a principal ideal domain.
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==Proof==
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Consider the [[ideal]] <math>I = (x,y) = \{xr_1+yr_2|r_1,r_2\in R\}</math>. As <math>R</math> is a principal ideal domain, <math>I</math> must be principle, that is it must be generated by a single element, say <math>I = (g)</math>. Now from the definition of <math>I</math>, there must exist <math>r_1,r_2\in R</math> such that <math>g = xr_1+yr_2</math>. We now claim that <math>g = \gcd(x,y)</math>.
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First we prove the following simple fact: if <math>z\in I</math>, then <math>g|z</math>. To see this, note that if <math>z\in I = (g)</math>, then there must be some <math>r\in R</math> such that <math>z = rg</math>. But now by definition we have <math>g|z</math>.
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Now from this, as <math>x,y\in I</math>, we get that <math>g|x,y</math>. Furthermore, consider any <math>s\in R</math> with <math>s|x,y</math>. We clearly have that <math>s|xr_1+yr_2 = g</math>. So indeed <math>g = \gcd(x,y)</math>.
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Now we shall prove minimality. Let <math>g' = xr_1'+yr_2'</math>. Then as <math>g|x,y</math>, we have <math>g|xr_1'+yr_2' = g'</math>, as desired.
  
 
==See also==
 
==See also==
 
[[Category:Number theory]]
 
[[Category:Number theory]]
 
{{stub}}
 
{{stub}}

Revision as of 02:01, 21 March 2009

Bezout's Lemma states that if $x$ and $y$ are nonzero integers and $g = \gcd(x,y)$, then there exist integers $\alpha$ and $\beta$ such that $x\alpha+y\beta=g$. In other words, there exists a linear combination of $x$ and $y$ equal to $g$.

Furthermore, $g$ is the smallest positive integer that can be expressed in this form, i.e. $g = \min\{x\alpha+y\beta|\alpha,\beta\in\mathbb Z, x\alpha+y\beta > 0\}$.

In particular, if $x$ and $y$ are relatively prime then there are integers $\alpha$ and $\beta$ for which $x\alpha+y\beta=1$.

Proof

Let $x = gx_1$, $y = gy_1$, and notice that $\gcd(x_1,y_1) = 1$.

Since $\gcd(x_1,y_1)=1$, $\text{lcm}(x_1,y_1)=x_1y_1$. So $\alpha=y_1$ is smallest positive $\alpha$ for which $x\alpha\equiv 0\pmod{y}$. Now if for all integers $0\le a,b<y_1$, we have that $x_1a\not\equiv x_1b\pmod{y_1}$, then one of those $y_1$ integers must be 1 from the Pigeonhole Principle. Assume for contradiction that $x_1a\equiv x_1b\pmod{y_1}$, and WLOG let $b>a$. Then, $x_1(b-a)\equiv 0\pmod y_1$, and so as we saw above this means $b-a\ge y_1$ but this is impossible since $0\le a,b<y_1$. Thus there exists an $\alpha$ such that $x_1\alpha\equiv 1\pmod{y_1}$.

Therefore $y_1|(x_1\alpha-1)$, and so there exists an integer $\beta$ such that $x_1\alpha - 1 = y_1\beta$, and so $x_1\alpha + y_1\beta = 1$. Now multiplying through by $g$ gives, $gx_1\alpha + gy_1\alpha = g$, or $x\alpha+y\beta = g$.

Thus there does exist integers $\alpha$ and $\beta$ such that $x\alpha+y\beta=g$.

Now to prove $g$ is minimum, consider any positive integer $g' = x\alpha'+y\beta'$. As $g|x,y$ we get $g|x\alpha'+y\beta' = g'$, and as $g$ and $g'$ are both positive integers this gives $g\le g'$. So $g$ is indeed the minimum.

Generalization to Principal Ideal Domains

Bezout's Lemma can be generalized to principal ideal domains.

Let $R$ be a principal ideal domain, and consider any $x,y\in R$. Let $g = \gcd(x,y)$. Then there exist elements $r_1,r_2\in R$ for which $xr_1+yr_2 = g$. Furthermore, $g$ is the minimal such element (under divisibility), i.e. if $g' = xr_1'+yr_2'$ then $g|g'$.

Note that this statement is indeed a generalization of the previous statement, as the ring of integers, $\mathbb Z$ is a principal ideal domain.

Proof

Consider the ideal $I = (x,y) = \{xr_1+yr_2|r_1,r_2\in R\}$. As $R$ is a principal ideal domain, $I$ must be principle, that is it must be generated by a single element, say $I = (g)$. Now from the definition of $I$, there must exist $r_1,r_2\in R$ such that $g = xr_1+yr_2$. We now claim that $g = \gcd(x,y)$.

First we prove the following simple fact: if $z\in I$, then $g|z$. To see this, note that if $z\in I = (g)$, then there must be some $r\in R$ such that $z = rg$. But now by definition we have $g|z$.

Now from this, as $x,y\in I$, we get that $g|x,y$. Furthermore, consider any $s\in R$ with $s|x,y$. We clearly have that $s|xr_1+yr_2 = g$. So indeed $g = \gcd(x,y)$.

Now we shall prove minimality. Let $g' = xr_1'+yr_2'$. Then as $g|x,y$, we have $g|xr_1'+yr_2' = g'$, as desired.

See also

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