Is a polynomial $f$ zero at $(a_1,ldots,a_n)$ iff $f$ lies in the ideal $(X_1-a_1,ldots,X_n-a_n)$?
$begingroup$
This is probably a very silly question:
If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
$$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.
Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
$$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
If this does not work in general, let $R=K$ be a field.
abstract-algebra polynomials ideals roots
edited Apr 1 '17 at 17:49
coproc
977514
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
$endgroup$
add a comment |
$begingroup$
This is probably a very silly question:
If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
$$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.
Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
$$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
If this does not work in general, let $R=K$ be a field.
abstract-algebra polynomials ideals roots
edited Apr 1 '17 at 17:49
coproc
977514
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
$endgroup$
add a comment |
$begingroup$
This is probably a very silly question:
If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
$$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.
Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
$$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
If this does not work in general, let $R=K$ be a field.
abstract-algebra polynomials ideals roots
edited Apr 1 '17 at 17:49
coproc
977514
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
$endgroup$
This is probably a very silly question:
If $R$ is an arbitrary commutative ring with unit and $fin R[X]$ a polynomial, then for any element $ain R$ we have
$$f(a)=0 Longleftrightarrow X-a ~mbox{ divides }~ f Longleftrightarrow fin (X-a)$$
where the last equivalence is clear. The first is probably a little surprising as $R[X]$ is usually not euclidean and it is perhaps not clear how to divide by $X-a$.
Now let $fin R[X_1,ldots, X_n]$ be a polynomial. How can I see for an element $(a_1,ldots,a_n)in R^n$ that
$$f(a_1,ldots,a_n)=0 Longleftrightarrow fin (X_1-a_1,ldots,X_n-a_n) ?$$
If this does not work in general, let $R=K$ be a field.
abstract-algebra polynomials ideals roots
abstract-algebra polynomials ideals roots
edited Apr 1 '17 at 17:49
coproc
977514
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
edited Apr 1 '17 at 17:49
coproc
977514
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
edited Apr 1 '17 at 17:49
coproc
977514
edited Apr 1 '17 at 17:49
coproc
977514
edited Apr 1 '17 at 17:49
coproc
977514
977514
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
asked Jul 7 '15 at 13:54
user8463524user8463524
577315
577315
add a comment |
add a comment |
2 Answers
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$begingroup$
Yes. Another way to see this is by the following:
We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
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add a comment |
$begingroup$
Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
$$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
and
$$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
where
$$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Yes. Another way to see this is by the following:
We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
$endgroup$
add a comment |
$begingroup$
Yes. Another way to see this is by the following:
We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
$endgroup$
add a comment |
$begingroup$
Yes. Another way to see this is by the following:
We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
$endgroup$
Yes. Another way to see this is by the following:
We can assume without loss of generality that $X_n$ appears in $f$. Then $$f=varphi_0+varphi_1X_n+cdots + varphi_kX_n^k$$
for some $varphiin R[X_1,...,X_{n-1}]$ with $varphi_k(a_1,..,a_{n-1})neq 0$. This reduces to the case you are okay with!
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
edited Dec 18 '18 at 20:39
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
answered Jul 7 '15 at 14:10
EoinEoin
4,5211927
4,5211927
add a comment |
add a comment |
$begingroup$
Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
$$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
and
$$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
where
$$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
$endgroup$
add a comment |
$begingroup$
Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
$$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
and
$$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
where
$$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
$endgroup$
add a comment |
$begingroup$
Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
$$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
and
$$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
where
$$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
$endgroup$
Consider the case of $(a_1,ldots,a_n)=(0,ldots,0)$ first, where it's obvious. Then observe that
$$f(a_1,ldots,a_n)=0iff g(0,ldots,0)=0$$
and
$$fin (x_1-a_1,ldots,x_n-a_n)iff gin (x_1,ldots,x_n)$$
where
$$g(x_1,ldots,x_n)=f(x_1+a_1,ldots,x_n+a_n)$$
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
answered Jul 7 '15 at 14:02
Zev ChonolesZev Chonoles
110k16231430
110k16231430
add a comment |
add a comment |
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