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Prove $mathbb{Q}(zeta)$ is a field when $zeta$ is algebraic
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Let $zeta$ be a root of the irreducible rational polynomial $P(x)$ with degree $n$. I need to prove that the set of numbers of the form $S = { R(zeta) | R(x) in mathbb{Q}[x] }$ is a field.
I'm not sure of my solution.
It's easily seen by piecewise addition/multiplication that $S$ is a ring, so to prove that $S$ is a field, we need to prove that inverses exist.
Pick any polynomial $R(x) in mathbb{Q}[x]$ - we will show the existence of $T(x) in mathbb{Q}[x]$ such that $T(zeta)R(zeta) = 1$.
For a polynomial $T(x) in mathbb{Q}[x]$, let $T'(x)$ be the polynomial with degree not more than $n-1$ such that $T'(x) equiv T(x) mod P(x)$.
Since $R(zeta) = R'(zeta)$, we can WLOG assume that $deg[R] < n$.
By writing $T'(x) = displaystyle sum_{1 leq i leq n} a_ix^i$, consider the injection $f: mathbb{Q}[x] mapsto mathbb{Q}^n$ such that $f(T(x)) = (a_0, a_1, cdots, a_n)$.
For a polynomial $S(x)$ with degree not exceeding $n-1$, define the function $g_S: mathbb{Q}^n mapsto mathbb{Q^n}$ such that $displaystyle g_S(a_0, cdots, a_n) = f((sum_{1 leq i leq n} a_ix^i) S(x))$
Claim $g_S$ is a linar map with dimension equal to $n$
Proof:
It's easy to see that $g_S$ is a linear mapping. Now we show that $dim(g_S) = n$.
It suffices to show that kernel contains exactly one element, viz $0$. Assume $g_S(T) = (0, 0, cdots, 0)$. Then $T(x)S(x) equiv 0 mod P(x) Rightarrow R(x) | T(x)S(x)$. Now since $P(x)$ is irreducible, either $P(x) | S(x)$ or $R(x) | T(x)$. Since $ deg[S] < deg[R]$, the first case isn't possible, and the second isn't possible either if $T$ is nonzero. But it's true, so $T$ is zero polynomial, and we're done. $blacksquare$
Now it follows that $g_s$ is invertible. Simply pick $T(x) = g^{-1}_S(1, 0, cdots, 0)$ and we're done !
linear-algebra number-theory proof-verification algebraic-number-theory
asked 16 mins ago
alxchen
493321
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Let $zeta$ be a root of the irreducible rational polynomial $P(x)$ with degree $n$. I need to prove that the set of numbers of the form $S = { R(zeta) | R(x) in mathbb{Q}[x] }$ is a field.
I'm not sure of my solution.
It's easily seen by piecewise addition/multiplication that $S$ is a ring, so to prove that $S$ is a field, we need to prove that inverses exist.
Pick any polynomial $R(x) in mathbb{Q}[x]$ - we will show the existence of $T(x) in mathbb{Q}[x]$ such that $T(zeta)R(zeta) = 1$.
For a polynomial $T(x) in mathbb{Q}[x]$, let $T'(x)$ be the polynomial with degree not more than $n-1$ such that $T'(x) equiv T(x) mod P(x)$.
Since $R(zeta) = R'(zeta)$, we can WLOG assume that $deg[R] < n$.
By writing $T'(x) = displaystyle sum_{1 leq i leq n} a_ix^i$, consider the injection $f: mathbb{Q}[x] mapsto mathbb{Q}^n$ such that $f(T(x)) = (a_0, a_1, cdots, a_n)$.
For a polynomial $S(x)$ with degree not exceeding $n-1$, define the function $g_S: mathbb{Q}^n mapsto mathbb{Q^n}$ such that $displaystyle g_S(a_0, cdots, a_n) = f((sum_{1 leq i leq n} a_ix^i) S(x))$
Claim $g_S$ is a linar map with dimension equal to $n$
Proof:
It's easy to see that $g_S$ is a linear mapping. Now we show that $dim(g_S) = n$.
It suffices to show that kernel contains exactly one element, viz $0$. Assume $g_S(T) = (0, 0, cdots, 0)$. Then $T(x)S(x) equiv 0 mod P(x) Rightarrow R(x) | T(x)S(x)$. Now since $P(x)$ is irreducible, either $P(x) | S(x)$ or $R(x) | T(x)$. Since $ deg[S] < deg[R]$, the first case isn't possible, and the second isn't possible either if $T$ is nonzero. But it's true, so $T$ is zero polynomial, and we're done. $blacksquare$
Now it follows that $g_s$ is invertible. Simply pick $T(x) = g^{-1}_S(1, 0, cdots, 0)$ and we're done !
linear-algebra number-theory proof-verification algebraic-number-theory
asked 16 mins ago
alxchen
493321
add a comment |
up vote
0
down vote
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up vote
0
down vote
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Let $zeta$ be a root of the irreducible rational polynomial $P(x)$ with degree $n$. I need to prove that the set of numbers of the form $S = { R(zeta) | R(x) in mathbb{Q}[x] }$ is a field.
I'm not sure of my solution.
It's easily seen by piecewise addition/multiplication that $S$ is a ring, so to prove that $S$ is a field, we need to prove that inverses exist.
Pick any polynomial $R(x) in mathbb{Q}[x]$ - we will show the existence of $T(x) in mathbb{Q}[x]$ such that $T(zeta)R(zeta) = 1$.
For a polynomial $T(x) in mathbb{Q}[x]$, let $T'(x)$ be the polynomial with degree not more than $n-1$ such that $T'(x) equiv T(x) mod P(x)$.
Since $R(zeta) = R'(zeta)$, we can WLOG assume that $deg[R] < n$.
By writing $T'(x) = displaystyle sum_{1 leq i leq n} a_ix^i$, consider the injection $f: mathbb{Q}[x] mapsto mathbb{Q}^n$ such that $f(T(x)) = (a_0, a_1, cdots, a_n)$.
For a polynomial $S(x)$ with degree not exceeding $n-1$, define the function $g_S: mathbb{Q}^n mapsto mathbb{Q^n}$ such that $displaystyle g_S(a_0, cdots, a_n) = f((sum_{1 leq i leq n} a_ix^i) S(x))$
Claim $g_S$ is a linar map with dimension equal to $n$
Proof:
It's easy to see that $g_S$ is a linear mapping. Now we show that $dim(g_S) = n$.
It suffices to show that kernel contains exactly one element, viz $0$. Assume $g_S(T) = (0, 0, cdots, 0)$. Then $T(x)S(x) equiv 0 mod P(x) Rightarrow R(x) | T(x)S(x)$. Now since $P(x)$ is irreducible, either $P(x) | S(x)$ or $R(x) | T(x)$. Since $ deg[S] < deg[R]$, the first case isn't possible, and the second isn't possible either if $T$ is nonzero. But it's true, so $T$ is zero polynomial, and we're done. $blacksquare$
Now it follows that $g_s$ is invertible. Simply pick $T(x) = g^{-1}_S(1, 0, cdots, 0)$ and we're done !
linear-algebra number-theory proof-verification algebraic-number-theory
asked 16 mins ago
alxchen
493321
Let $zeta$ be a root of the irreducible rational polynomial $P(x)$ with degree $n$. I need to prove that the set of numbers of the form $S = { R(zeta) | R(x) in mathbb{Q}[x] }$ is a field.
I'm not sure of my solution.
It's easily seen by piecewise addition/multiplication that $S$ is a ring, so to prove that $S$ is a field, we need to prove that inverses exist.
Pick any polynomial $R(x) in mathbb{Q}[x]$ - we will show the existence of $T(x) in mathbb{Q}[x]$ such that $T(zeta)R(zeta) = 1$.
For a polynomial $T(x) in mathbb{Q}[x]$, let $T'(x)$ be the polynomial with degree not more than $n-1$ such that $T'(x) equiv T(x) mod P(x)$.
Since $R(zeta) = R'(zeta)$, we can WLOG assume that $deg[R] < n$.
By writing $T'(x) = displaystyle sum_{1 leq i leq n} a_ix^i$, consider the injection $f: mathbb{Q}[x] mapsto mathbb{Q}^n$ such that $f(T(x)) = (a_0, a_1, cdots, a_n)$.
For a polynomial $S(x)$ with degree not exceeding $n-1$, define the function $g_S: mathbb{Q}^n mapsto mathbb{Q^n}$ such that $displaystyle g_S(a_0, cdots, a_n) = f((sum_{1 leq i leq n} a_ix^i) S(x))$
Claim $g_S$ is a linar map with dimension equal to $n$
Proof:
It's easy to see that $g_S$ is a linear mapping. Now we show that $dim(g_S) = n$.
It suffices to show that kernel contains exactly one element, viz $0$. Assume $g_S(T) = (0, 0, cdots, 0)$. Then $T(x)S(x) equiv 0 mod P(x) Rightarrow R(x) | T(x)S(x)$. Now since $P(x)$ is irreducible, either $P(x) | S(x)$ or $R(x) | T(x)$. Since $ deg[S] < deg[R]$, the first case isn't possible, and the second isn't possible either if $T$ is nonzero. But it's true, so $T$ is zero polynomial, and we're done. $blacksquare$
Now it follows that $g_s$ is invertible. Simply pick $T(x) = g^{-1}_S(1, 0, cdots, 0)$ and we're done !
linear-algebra number-theory proof-verification algebraic-number-theory
linear-algebra number-theory proof-verification algebraic-number-theory
asked 16 mins ago
alxchen
493321
asked 16 mins ago
alxchen
493321
asked 16 mins ago
alxchen
493321
asked 16 mins ago
alxchen
493321
asked 16 mins ago
alxchen
493321
493321
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1 Answer
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First of all you have to assume that $R(x)$ isn't divisible by $P(x)$, as otherwise we have that $R(zeta) = 0$ in $mathbb{Q}(zeta)$ so it can't be invertible to begin with.Otherwise the proof seems alright at first look.
Nevertheless the easiest way would be to use the Extended Euclidean Algorithm. If we have that $P(x) nmid R(x)$ then $gcd(P(x),R(x)) = 1$ by the irreducibility of $P(x)$. Then by the Extended Euclidean Algorithm there exist $T,S in mathbb{Q}[x]$ s.t.
$$T(x)R(x) + S(x)P(x) = 1$$
As $P(zeta) = 0$ in $mathbb{Q}(zeta)$ we get that $T(zeta)R(zeta) = 1 in mathbb{Q}(zeta)$ and we are done.
Another way is to use the fact that $(P(x))$ generates a maximal ideal in $mathbb{Q}[x]$ and using that $mathbb{Q}(zeta) cong mathbb{Q}[x]/(P(x))$ we get that $mathbb{Q}(zeta)$ is a field.
answered 4 mins ago
Stefan4024
29.9k53377
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
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down vote
First of all you have to assume that $R(x)$ isn't divisible by $P(x)$, as otherwise we have that $R(zeta) = 0$ in $mathbb{Q}(zeta)$ so it can't be invertible to begin with.Otherwise the proof seems alright at first look.
Nevertheless the easiest way would be to use the Extended Euclidean Algorithm. If we have that $P(x) nmid R(x)$ then $gcd(P(x),R(x)) = 1$ by the irreducibility of $P(x)$. Then by the Extended Euclidean Algorithm there exist $T,S in mathbb{Q}[x]$ s.t.
$$T(x)R(x) + S(x)P(x) = 1$$
As $P(zeta) = 0$ in $mathbb{Q}(zeta)$ we get that $T(zeta)R(zeta) = 1 in mathbb{Q}(zeta)$ and we are done.
Another way is to use the fact that $(P(x))$ generates a maximal ideal in $mathbb{Q}[x]$ and using that $mathbb{Q}(zeta) cong mathbb{Q}[x]/(P(x))$ we get that $mathbb{Q}(zeta)$ is a field.
answered 4 mins ago
Stefan4024
29.9k53377
add a comment |
up vote
0
down vote
First of all you have to assume that $R(x)$ isn't divisible by $P(x)$, as otherwise we have that $R(zeta) = 0$ in $mathbb{Q}(zeta)$ so it can't be invertible to begin with.Otherwise the proof seems alright at first look.
Nevertheless the easiest way would be to use the Extended Euclidean Algorithm. If we have that $P(x) nmid R(x)$ then $gcd(P(x),R(x)) = 1$ by the irreducibility of $P(x)$. Then by the Extended Euclidean Algorithm there exist $T,S in mathbb{Q}[x]$ s.t.
$$T(x)R(x) + S(x)P(x) = 1$$
As $P(zeta) = 0$ in $mathbb{Q}(zeta)$ we get that $T(zeta)R(zeta) = 1 in mathbb{Q}(zeta)$ and we are done.
Another way is to use the fact that $(P(x))$ generates a maximal ideal in $mathbb{Q}[x]$ and using that $mathbb{Q}(zeta) cong mathbb{Q}[x]/(P(x))$ we get that $mathbb{Q}(zeta)$ is a field.
answered 4 mins ago
Stefan4024
29.9k53377
add a comment |
up vote
0
down vote
up vote
0
down vote
First of all you have to assume that $R(x)$ isn't divisible by $P(x)$, as otherwise we have that $R(zeta) = 0$ in $mathbb{Q}(zeta)$ so it can't be invertible to begin with.Otherwise the proof seems alright at first look.
Nevertheless the easiest way would be to use the Extended Euclidean Algorithm. If we have that $P(x) nmid R(x)$ then $gcd(P(x),R(x)) = 1$ by the irreducibility of $P(x)$. Then by the Extended Euclidean Algorithm there exist $T,S in mathbb{Q}[x]$ s.t.
$$T(x)R(x) + S(x)P(x) = 1$$
As $P(zeta) = 0$ in $mathbb{Q}(zeta)$ we get that $T(zeta)R(zeta) = 1 in mathbb{Q}(zeta)$ and we are done.
Another way is to use the fact that $(P(x))$ generates a maximal ideal in $mathbb{Q}[x]$ and using that $mathbb{Q}(zeta) cong mathbb{Q}[x]/(P(x))$ we get that $mathbb{Q}(zeta)$ is a field.
answered 4 mins ago
Stefan4024
29.9k53377
First of all you have to assume that $R(x)$ isn't divisible by $P(x)$, as otherwise we have that $R(zeta) = 0$ in $mathbb{Q}(zeta)$ so it can't be invertible to begin with.Otherwise the proof seems alright at first look.
Nevertheless the easiest way would be to use the Extended Euclidean Algorithm. If we have that $P(x) nmid R(x)$ then $gcd(P(x),R(x)) = 1$ by the irreducibility of $P(x)$. Then by the Extended Euclidean Algorithm there exist $T,S in mathbb{Q}[x]$ s.t.
$$T(x)R(x) + S(x)P(x) = 1$$
As $P(zeta) = 0$ in $mathbb{Q}(zeta)$ we get that $T(zeta)R(zeta) = 1 in mathbb{Q}(zeta)$ and we are done.
Another way is to use the fact that $(P(x))$ generates a maximal ideal in $mathbb{Q}[x]$ and using that $mathbb{Q}(zeta) cong mathbb{Q}[x]/(P(x))$ we get that $mathbb{Q}(zeta)$ is a field.
answered 4 mins ago
Stefan4024
29.9k53377
answered 4 mins ago
Stefan4024
29.9k53377
answered 4 mins ago
Stefan4024
29.9k53377
answered 4 mins ago
Stefan4024
29.9k53377
29.9k53377
add a comment |
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