Mixing
ideal extension in polynomial ring over quotient field
$begingroup$
This is probably easy, but I can't seem to figure it out.
I have a $K$-algebra $B$, which is a domain and an irreducible polynomial $f in K[x]$.
Why does $B[x]/f$ embed in $Q(B)[x]/f$ where $Q(B)$ is the quotient field of $B$ as is said in the proof of Lemma~1 in this nice answer here math.stackexchange.
I would need to show, that $Q(B)[x] f cap B[x] = B[x]f$ wouldn't I? Or do I have forgotten a further assumption?
algebraic-geometry polynomials commutative-algebra ideals integral-domain
asked Jan 24 at 16:44
pyrogenpyrogen
714
$endgroup$
add a comment |
$begingroup$
This is probably easy, but I can't seem to figure it out.
I have a $K$-algebra $B$, which is a domain and an irreducible polynomial $f in K[x]$.
Why does $B[x]/f$ embed in $Q(B)[x]/f$ where $Q(B)$ is the quotient field of $B$ as is said in the proof of Lemma~1 in this nice answer here math.stackexchange.
I would need to show, that $Q(B)[x] f cap B[x] = B[x]f$ wouldn't I? Or do I have forgotten a further assumption?
algebraic-geometry polynomials commutative-algebra ideals integral-domain
asked Jan 24 at 16:44
pyrogenpyrogen
714
$endgroup$
add a comment |
$begingroup$
This is probably easy, but I can't seem to figure it out.
I have a $K$-algebra $B$, which is a domain and an irreducible polynomial $f in K[x]$.
Why does $B[x]/f$ embed in $Q(B)[x]/f$ where $Q(B)$ is the quotient field of $B$ as is said in the proof of Lemma~1 in this nice answer here math.stackexchange.
I would need to show, that $Q(B)[x] f cap B[x] = B[x]f$ wouldn't I? Or do I have forgotten a further assumption?
algebraic-geometry polynomials commutative-algebra ideals integral-domain
asked Jan 24 at 16:44
pyrogenpyrogen
714
$endgroup$
This is probably easy, but I can't seem to figure it out.
I have a $K$-algebra $B$, which is a domain and an irreducible polynomial $f in K[x]$.
Why does $B[x]/f$ embed in $Q(B)[x]/f$ where $Q(B)$ is the quotient field of $B$ as is said in the proof of Lemma~1 in this nice answer here math.stackexchange.
I would need to show, that $Q(B)[x] f cap B[x] = B[x]f$ wouldn't I? Or do I have forgotten a further assumption?
algebraic-geometry polynomials commutative-algebra ideals integral-domain
algebraic-geometry polynomials commutative-algebra ideals integral-domain
asked Jan 24 at 16:44
pyrogenpyrogen
714
asked Jan 24 at 16:44
pyrogenpyrogen
714
asked Jan 24 at 16:44
pyrogenpyrogen
714
asked Jan 24 at 16:44
pyrogenpyrogen
714
asked Jan 24 at 16:44
pyrogenpyrogen
714
714
add a comment |
add a comment |
1 Answer
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We may assume $f$ is monic (if not, divide by its leading coefficient). As you say, it is enough to show that $Q(B)[x] f cap B[x] = B[x]f$. The trick to prove this is to use polynomial long division: we can test whether a polynomial $g$ is divisible by $f$ by performing the division algorithm and seeing whether the remainder is $0$. Note that since $f$ is monic, the division algorithm never involves division of coefficients (just subtraction and multiplication), so it can be performed in the ring $B[x]$ and will produce a quotient and remainder which still have coefficients in $B$. So, if $gin Q(B)[x] fcap B[x]$, the remainder when we divide $g$ by $f$ is $0$, but the quotient is in $B[x]$ since $gin B[x]$, proving that $gin B[x]f$.
Another perspective on this is that for any $K$-algebra $B$, the quotient $B[x]/f$ is a free $B$-module on $1,x,x^2,dots,x^{n-1}$ where $n=deg f$ (this again follows from polynomial division, since these powers of $x$ span the possible remainders you can get when dividing by $f$). This makes it clear that if $B$ is a subalgebra of $C$, then $B[x]/f$ embeds in $C[x]/f$, since you can just extend the scalars to $C$ using the same basis. (Or in fancier language, $K[x]/f$ is flat over $K$, so we can tensor the inclusion $Bto C$ with $K[x]/f$ to get an injection from $Botimes_K K[x]/fcong B[x]/f$ to $Cotimes_K K[x]/fcong C[x]/f$.)
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
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$begingroup$
We may assume $f$ is monic (if not, divide by its leading coefficient). As you say, it is enough to show that $Q(B)[x] f cap B[x] = B[x]f$. The trick to prove this is to use polynomial long division: we can test whether a polynomial $g$ is divisible by $f$ by performing the division algorithm and seeing whether the remainder is $0$. Note that since $f$ is monic, the division algorithm never involves division of coefficients (just subtraction and multiplication), so it can be performed in the ring $B[x]$ and will produce a quotient and remainder which still have coefficients in $B$. So, if $gin Q(B)[x] fcap B[x]$, the remainder when we divide $g$ by $f$ is $0$, but the quotient is in $B[x]$ since $gin B[x]$, proving that $gin B[x]f$.
Another perspective on this is that for any $K$-algebra $B$, the quotient $B[x]/f$ is a free $B$-module on $1,x,x^2,dots,x^{n-1}$ where $n=deg f$ (this again follows from polynomial division, since these powers of $x$ span the possible remainders you can get when dividing by $f$). This makes it clear that if $B$ is a subalgebra of $C$, then $B[x]/f$ embeds in $C[x]/f$, since you can just extend the scalars to $C$ using the same basis. (Or in fancier language, $K[x]/f$ is flat over $K$, so we can tensor the inclusion $Bto C$ with $K[x]/f$ to get an injection from $Botimes_K K[x]/fcong B[x]/f$ to $Cotimes_K K[x]/fcong C[x]/f$.)
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
We may assume $f$ is monic (if not, divide by its leading coefficient). As you say, it is enough to show that $Q(B)[x] f cap B[x] = B[x]f$. The trick to prove this is to use polynomial long division: we can test whether a polynomial $g$ is divisible by $f$ by performing the division algorithm and seeing whether the remainder is $0$. Note that since $f$ is monic, the division algorithm never involves division of coefficients (just subtraction and multiplication), so it can be performed in the ring $B[x]$ and will produce a quotient and remainder which still have coefficients in $B$. So, if $gin Q(B)[x] fcap B[x]$, the remainder when we divide $g$ by $f$ is $0$, but the quotient is in $B[x]$ since $gin B[x]$, proving that $gin B[x]f$.
Another perspective on this is that for any $K$-algebra $B$, the quotient $B[x]/f$ is a free $B$-module on $1,x,x^2,dots,x^{n-1}$ where $n=deg f$ (this again follows from polynomial division, since these powers of $x$ span the possible remainders you can get when dividing by $f$). This makes it clear that if $B$ is a subalgebra of $C$, then $B[x]/f$ embeds in $C[x]/f$, since you can just extend the scalars to $C$ using the same basis. (Or in fancier language, $K[x]/f$ is flat over $K$, so we can tensor the inclusion $Bto C$ with $K[x]/f$ to get an injection from $Botimes_K K[x]/fcong B[x]/f$ to $Cotimes_K K[x]/fcong C[x]/f$.)
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
We may assume $f$ is monic (if not, divide by its leading coefficient). As you say, it is enough to show that $Q(B)[x] f cap B[x] = B[x]f$. The trick to prove this is to use polynomial long division: we can test whether a polynomial $g$ is divisible by $f$ by performing the division algorithm and seeing whether the remainder is $0$. Note that since $f$ is monic, the division algorithm never involves division of coefficients (just subtraction and multiplication), so it can be performed in the ring $B[x]$ and will produce a quotient and remainder which still have coefficients in $B$. So, if $gin Q(B)[x] fcap B[x]$, the remainder when we divide $g$ by $f$ is $0$, but the quotient is in $B[x]$ since $gin B[x]$, proving that $gin B[x]f$.
Another perspective on this is that for any $K$-algebra $B$, the quotient $B[x]/f$ is a free $B$-module on $1,x,x^2,dots,x^{n-1}$ where $n=deg f$ (this again follows from polynomial division, since these powers of $x$ span the possible remainders you can get when dividing by $f$). This makes it clear that if $B$ is a subalgebra of $C$, then $B[x]/f$ embeds in $C[x]/f$, since you can just extend the scalars to $C$ using the same basis. (Or in fancier language, $K[x]/f$ is flat over $K$, so we can tensor the inclusion $Bto C$ with $K[x]/f$ to get an injection from $Botimes_K K[x]/fcong B[x]/f$ to $Cotimes_K K[x]/fcong C[x]/f$.)
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
$endgroup$
We may assume $f$ is monic (if not, divide by its leading coefficient). As you say, it is enough to show that $Q(B)[x] f cap B[x] = B[x]f$. The trick to prove this is to use polynomial long division: we can test whether a polynomial $g$ is divisible by $f$ by performing the division algorithm and seeing whether the remainder is $0$. Note that since $f$ is monic, the division algorithm never involves division of coefficients (just subtraction and multiplication), so it can be performed in the ring $B[x]$ and will produce a quotient and remainder which still have coefficients in $B$. So, if $gin Q(B)[x] fcap B[x]$, the remainder when we divide $g$ by $f$ is $0$, but the quotient is in $B[x]$ since $gin B[x]$, proving that $gin B[x]f$.
Another perspective on this is that for any $K$-algebra $B$, the quotient $B[x]/f$ is a free $B$-module on $1,x,x^2,dots,x^{n-1}$ where $n=deg f$ (this again follows from polynomial division, since these powers of $x$ span the possible remainders you can get when dividing by $f$). This makes it clear that if $B$ is a subalgebra of $C$, then $B[x]/f$ embeds in $C[x]/f$, since you can just extend the scalars to $C$ using the same basis. (Or in fancier language, $K[x]/f$ is flat over $K$, so we can tensor the inclusion $Bto C$ with $K[x]/f$ to get an injection from $Botimes_K K[x]/fcong B[x]/f$ to $Cotimes_K K[x]/fcong C[x]/f$.)
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
edited Jan 24 at 22:16
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
answered Jan 24 at 22:08
Eric WofseyEric Wofsey
189k14216347
189k14216347
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