Difference between revisions of "Bezout's Lemma"
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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>. | ||
− | + | jklkjlkjlkjlkjkljljlkjlk | |
In particular, if <math>x</math> and <math>y</math> are [[relatively prime]] then there are integers <math>\alpha</math> and <math>\beta</math> for which <math>x\alpha+y\beta=1</math>. | In particular, if <math>x</math> and <math>y</math> are [[relatively prime]] then there are integers <math>\alpha</math> and <math>\beta</math> for which <math>x\alpha+y\beta=1</math>. | ||
Revision as of 17:44, 25 November 2022
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.
.
jklkjlkjlkjlkjkljljlkjlk
In particular, if
and
are relatively prime then there are integers
and
for which
.
Contents
[hide]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/Extension of Bezout's Lemma
Let be positive integers. Then there exists integers
such that
Also,
is the least positive integer satisfying this property.
Proof
Consider the set . Obviously,
. Thus, because all the elements of
are positive, by the Well Ordering Principle, there exists a minimal element
. So
if and
then
But by the Division Algorithm:
But so this would imply that
which contradicts the assumption that
is the minimal element in
. Thus,
hence,
. But this would imply that
for
because
.
Now, because
for
we have that
. But then we also have that
. Thus, we have that
~qwertysri987
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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