Difference between revisions of "Bezout's Lemma"
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Since <math>\gcd(x_1,y_1)=1</math>, <math>\text{lcm}(x_1,y_1)=x_1y_1</math>. So <math>\alpha=y_1</math> is smallest positive <math>\alpha</math> for which <math>x\alpha\equiv 0\pmod{y}</math>. Now if for all integers <math>0\le a,b<y_1</math>, we have that <math>x_1a\not\equiv x_1b\pmod{y_1}</math>, then one of those <math>y_1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>x_1a\equiv x_1b\pmod{y_1}</math>, and WLOG let <math>b>a</math>. Then, <math>x_1(b-a)\equiv 0\pmod {y_1}</math>, and so as we saw above this means <math>b-a\ge y_1</math> but this is impossible since <math>0\le a,b<y_1</math>. Thus there exists an <math>\alpha</math> such that <math>x_1\alpha\equiv 1\pmod{y_1}</math>. | Since <math>\gcd(x_1,y_1)=1</math>, <math>\text{lcm}(x_1,y_1)=x_1y_1</math>. So <math>\alpha=y_1</math> is smallest positive <math>\alpha</math> for which <math>x\alpha\equiv 0\pmod{y}</math>. Now if for all integers <math>0\le a,b<y_1</math>, we have that <math>x_1a\not\equiv x_1b\pmod{y_1}</math>, then one of those <math>y_1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>x_1a\equiv x_1b\pmod{y_1}</math>, and WLOG let <math>b>a</math>. Then, <math>x_1(b-a)\equiv 0\pmod {y_1}</math>, and so as we saw above this means <math>b-a\ge y_1</math> but this is impossible since <math>0\le a,b<y_1</math>. Thus there exists an <math>\alpha</math> such that <math>x_1\alpha\equiv 1\pmod{y_1}</math>. | ||
− | Therefore <math>y_1|(x_1\alpha-1)</math>, and so there exists an integer <math>\beta</math> such that <math>x_1\alpha - 1 = y_1\beta</math>, and so <math>x_1\alpha + y_1\beta = 1</math>. Now multiplying through by <math>g</math> gives, <math>gx_1\alpha + gy_1\ | + | Therefore <math>y_1|(x_1\alpha-1)</math>, and so there exists an integer <math>\beta</math> such that <math>x_1\alpha - 1 = y_1\beta</math>, and so <math>x_1\alpha + y_1\beta = 1</math>. Now multiplying through by <math>g</math> gives, <math>gx_1\alpha + gy_1\beta = g</math>, or <math>x\alpha+y\beta = g</math>. |
Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. | Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. |
Revision as of 11:06, 26 June 2019
Bezout's Lemma states that if and
are nonzero integers and
, then there exist integers
and
such that
. In other words, there exists a linear combination of
and
equal to
.
Furthermore, is the smallest positive integer that can be expressed in this form, i.e.
.
In particular, if and
are relatively prime then there are integers
and
for which
.
Proof
Let ,
, and notice that
.
Since ,
. So
is smallest positive
for which
. Now if for all integers
, we have that
, then one of those
integers must be 1 from the Pigeonhole Principle. Assume for contradiction that
, and WLOG let
. Then,
, and so as we saw above this means
but this is impossible since
. Thus there exists an
such that
.
Therefore , and so there exists an integer
such that
, and so
. Now multiplying through by
gives,
, or
.
Thus there does exist integers and
such that
.
Now to prove is minimum, consider any positive integer
. As
we get
, and as
and
are both positive integers this gives
. So
is indeed the minimum.
Generalization to Principal Ideal Domains
Bezout's Lemma can be generalized to principal ideal domains.
Let be a principal ideal domain, and consider any
. Let
. Then there exist elements
for which
. Furthermore,
is the minimal such element (under divisibility), i.e. if
then
.
Note that this statement is indeed a generalization of the previous statement, as the ring of integers, is a principal ideal domain.
Proof
Consider the ideal . As
is a principal ideal domain,
must be principle, that is it must be generated by a single element, say
. Now from the definition of
, there must exist
such that
. We now claim that
.
First we prove the following simple fact: if , then
. To see this, note that if
, then there must be some
such that
. But now by definition we have
.
Now from this, as , we get that
. Furthermore, consider any
with
. We clearly have that
. So indeed
.
Now we shall prove minimality. Let . Then as
, we have
, as desired.
See also
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