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Chinese remainder theorem for polynomials in a field
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(Sorry for posting two proof-verification questions twice in a row within such short span of time - I'm not posting any other proof-verification questions soon. )
I'm learning abstract algebra and there's an exercise which states
Let $F$ be a field, and $q(x) in F[x]$. If $displaystyle q(x) = prod_0 leq i leq k [p_i(x)]^e_i$ where $p_i(x)$ are irreducible over $F[x]$, then prove that $$ displaystyle fracF[x]q(x) simeq fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$$
I've a proof for this, but I'm not sure if it's correct or whether the entire proof doesn't makes sense. Can you check my proof ?
Let $A = fracF[x]q(x)$ and $B = fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$. To show that $A simeq B$, we will construct two rings homomorphisms $phi_1, phi_2$ such that $phi_1$ maps $A$ to $B$ and $phi_2$ maps $B$ to $A$, which would imply the conclusion.
Construction of the ring homomorphism $phi_1$ is relatively easier, for $h(x) in F[x]$, let $f_i(h(x))$ denote the remainder of $h(x)$ when divided by $[p_i(x)]^e_i$. Then it's easy to see the mapping $$ phi_1 oversettextdef:=h(x) rightarrow (f_1(h(x)), f_2(h(x)), cdots, f_k(h(x)))$$ is a ring homomorphism from $A$ to $B$.
For the other way, since $gcd([p_i(x)]^e_i, fracq(x)[p_i(x)]^e_i) = 1$, we can always find polynomials $m_i(x), n_i(x) in F[x]$ such that $m_i(x)fracq(x)[p_i(x)]^e_i+n_i(x)[p_i(x)]^e_i = 1$. Then it's not hard to see the mapping $$displaystyle phi_2 oversettextdef:=(h_1(x), h_2(x), cdots, h_k(x)) rightarrow sum_0 leq ileq k fracq(x)[p_i(x)]^e_i m_i(x)h_i(x) mod q(x) $$ is a ring homomorphism from $B$ to $A$, so we're done.
abstract-algebra proof-verification ring-theory
asked Jul 24 at 18:28
alxchen
402220
add a comment |Â
up vote
1
down vote
favorite
(Sorry for posting two proof-verification questions twice in a row within such short span of time - I'm not posting any other proof-verification questions soon. )
I'm learning abstract algebra and there's an exercise which states
Let $F$ be a field, and $q(x) in F[x]$. If $displaystyle q(x) = prod_0 leq i leq k [p_i(x)]^e_i$ where $p_i(x)$ are irreducible over $F[x]$, then prove that $$ displaystyle fracF[x]q(x) simeq fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$$
I've a proof for this, but I'm not sure if it's correct or whether the entire proof doesn't makes sense. Can you check my proof ?
Let $A = fracF[x]q(x)$ and $B = fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$. To show that $A simeq B$, we will construct two rings homomorphisms $phi_1, phi_2$ such that $phi_1$ maps $A$ to $B$ and $phi_2$ maps $B$ to $A$, which would imply the conclusion.
Construction of the ring homomorphism $phi_1$ is relatively easier, for $h(x) in F[x]$, let $f_i(h(x))$ denote the remainder of $h(x)$ when divided by $[p_i(x)]^e_i$. Then it's easy to see the mapping $$ phi_1 oversettextdef:=h(x) rightarrow (f_1(h(x)), f_2(h(x)), cdots, f_k(h(x)))$$ is a ring homomorphism from $A$ to $B$.
For the other way, since $gcd([p_i(x)]^e_i, fracq(x)[p_i(x)]^e_i) = 1$, we can always find polynomials $m_i(x), n_i(x) in F[x]$ such that $m_i(x)fracq(x)[p_i(x)]^e_i+n_i(x)[p_i(x)]^e_i = 1$. Then it's not hard to see the mapping $$displaystyle phi_2 oversettextdef:=(h_1(x), h_2(x), cdots, h_k(x)) rightarrow sum_0 leq ileq k fracq(x)[p_i(x)]^e_i m_i(x)h_i(x) mod q(x) $$ is a ring homomorphism from $B$ to $A$, so we're done.
abstract-algebra proof-verification ring-theory
asked Jul 24 at 18:28
alxchen
402220
It is ok. As a matter of typing, it is maybe simpler to establish the result for a product $q=uv$, $u,v$ relaively prime, then use induction.
– dan_fulea
Jul 24 at 23:21
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
(Sorry for posting two proof-verification questions twice in a row within such short span of time - I'm not posting any other proof-verification questions soon. )
I'm learning abstract algebra and there's an exercise which states
Let $F$ be a field, and $q(x) in F[x]$. If $displaystyle q(x) = prod_0 leq i leq k [p_i(x)]^e_i$ where $p_i(x)$ are irreducible over $F[x]$, then prove that $$ displaystyle fracF[x]q(x) simeq fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$$
I've a proof for this, but I'm not sure if it's correct or whether the entire proof doesn't makes sense. Can you check my proof ?
Let $A = fracF[x]q(x)$ and $B = fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$. To show that $A simeq B$, we will construct two rings homomorphisms $phi_1, phi_2$ such that $phi_1$ maps $A$ to $B$ and $phi_2$ maps $B$ to $A$, which would imply the conclusion.
Construction of the ring homomorphism $phi_1$ is relatively easier, for $h(x) in F[x]$, let $f_i(h(x))$ denote the remainder of $h(x)$ when divided by $[p_i(x)]^e_i$. Then it's easy to see the mapping $$ phi_1 oversettextdef:=h(x) rightarrow (f_1(h(x)), f_2(h(x)), cdots, f_k(h(x)))$$ is a ring homomorphism from $A$ to $B$.
For the other way, since $gcd([p_i(x)]^e_i, fracq(x)[p_i(x)]^e_i) = 1$, we can always find polynomials $m_i(x), n_i(x) in F[x]$ such that $m_i(x)fracq(x)[p_i(x)]^e_i+n_i(x)[p_i(x)]^e_i = 1$. Then it's not hard to see the mapping $$displaystyle phi_2 oversettextdef:=(h_1(x), h_2(x), cdots, h_k(x)) rightarrow sum_0 leq ileq k fracq(x)[p_i(x)]^e_i m_i(x)h_i(x) mod q(x) $$ is a ring homomorphism from $B$ to $A$, so we're done.
abstract-algebra proof-verification ring-theory
asked Jul 24 at 18:28
alxchen
402220
(Sorry for posting two proof-verification questions twice in a row within such short span of time - I'm not posting any other proof-verification questions soon. )
I'm learning abstract algebra and there's an exercise which states
Let $F$ be a field, and $q(x) in F[x]$. If $displaystyle q(x) = prod_0 leq i leq k [p_i(x)]^e_i$ where $p_i(x)$ are irreducible over $F[x]$, then prove that $$ displaystyle fracF[x]q(x) simeq fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$$
I've a proof for this, but I'm not sure if it's correct or whether the entire proof doesn't makes sense. Can you check my proof ?
Let $A = fracF[x]q(x)$ and $B = fracF[x](p_1(x))^e_1 oplus fracF[x](p_2(x))^e_2 oplus cdots oplus fracF[x](p_k(x))^e_k$. To show that $A simeq B$, we will construct two rings homomorphisms $phi_1, phi_2$ such that $phi_1$ maps $A$ to $B$ and $phi_2$ maps $B$ to $A$, which would imply the conclusion.
Construction of the ring homomorphism $phi_1$ is relatively easier, for $h(x) in F[x]$, let $f_i(h(x))$ denote the remainder of $h(x)$ when divided by $[p_i(x)]^e_i$. Then it's easy to see the mapping $$ phi_1 oversettextdef:=h(x) rightarrow (f_1(h(x)), f_2(h(x)), cdots, f_k(h(x)))$$ is a ring homomorphism from $A$ to $B$.
For the other way, since $gcd([p_i(x)]^e_i, fracq(x)[p_i(x)]^e_i) = 1$, we can always find polynomials $m_i(x), n_i(x) in F[x]$ such that $m_i(x)fracq(x)[p_i(x)]^e_i+n_i(x)[p_i(x)]^e_i = 1$. Then it's not hard to see the mapping $$displaystyle phi_2 oversettextdef:=(h_1(x), h_2(x), cdots, h_k(x)) rightarrow sum_0 leq ileq k fracq(x)[p_i(x)]^e_i m_i(x)h_i(x) mod q(x) $$ is a ring homomorphism from $B$ to $A$, so we're done.
abstract-algebra proof-verification ring-theory
asked Jul 24 at 18:28
alxchen
402220
asked Jul 24 at 18:28
alxchen
402220
asked Jul 24 at 18:28
alxchen
402220
asked Jul 24 at 18:28
alxchen
402220
402220
It is ok. As a matter of typing, it is maybe simpler to establish the result for a product $q=uv$, $u,v$ relaively prime, then use induction.
– dan_fulea
Jul 24 at 23:21
add a comment |Â
It is ok. As a matter of typing, it is maybe simpler to establish the result for a product $q=uv$, $u,v$ relaively prime, then use induction.
– dan_fulea
Jul 24 at 23:21
It is ok. As a matter of typing, it is maybe simpler to establish the result for a product $q=uv$, $u,v$ relaively prime, then use induction.
– dan_fulea
Jul 24 at 23:21
It is ok. As a matter of typing, it is maybe simpler to establish the result for a product $q=uv$, $u,v$ relaively prime, then use induction.
– dan_fulea
Jul 24 at 23:21
add a comment |Â
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It is ok. As a matter of typing, it is maybe simpler to establish the result for a product $q=uv$, $u,v$ relaively prime, then use induction.
– dan_fulea
Jul 24 at 23:21