Mixing
The ring of symmetric polynomials is isomorphic to the ring of polynomials
$begingroup$
Let $ R $ be a ring, $ R[x_1,..., x_r] $ the ring of polynomials over $ R $ in $ r $ indeterminates. Let $ Sigma=R[p_1,..., p_r] $ denote the subring of $ R[x_1,..., x_r] $ consisting of the symmetric polynomials, where $$ p_1=sum_{1}^r x_i, quad p_2=sum_{i<j}x_ix_j,quad p_3=sum_{i<j<k}x_ix_jx_k,quad cdots, quad p_r=x_1x_2cdots x_r $$
are the elementary symmetric polynomials in $ x_1, ..., x_r $.
Since the elementary symmetric polynomials are algebraically independent over $
R $, then we know that the epimorphism: $$ R[x_1, x_2, ..., x_r]longrightarrowSigma=R[p_1, p_2, ..., p_r] $$ is actually an isomorphism.
My question: Is my deduction above valid?
abstract-algebra proof-verification polynomials
asked Jan 14 at 3:24
user549397user549397
1,5101418
$endgroup$
|
show 3 more comments
$begingroup$
Let $ R $ be a ring, $ R[x_1,..., x_r] $ the ring of polynomials over $ R $ in $ r $ indeterminates. Let $ Sigma=R[p_1,..., p_r] $ denote the subring of $ R[x_1,..., x_r] $ consisting of the symmetric polynomials, where $$ p_1=sum_{1}^r x_i, quad p_2=sum_{i<j}x_ix_j,quad p_3=sum_{i<j<k}x_ix_jx_k,quad cdots, quad p_r=x_1x_2cdots x_r $$
are the elementary symmetric polynomials in $ x_1, ..., x_r $.
Since the elementary symmetric polynomials are algebraically independent over $
R $, then we know that the epimorphism: $$ R[x_1, x_2, ..., x_r]longrightarrowSigma=R[p_1, p_2, ..., p_r] $$ is actually an isomorphism.
My question: Is my deduction above valid?
abstract-algebra proof-verification polynomials
asked Jan 14 at 3:24
user549397user549397
1,5101418
$endgroup$
$begingroup$
Which deduction is that exactly?
$endgroup$
– Matt Samuel
Jan 14 at 3:32
$begingroup$
@MattSamuel The paragraph outside the box.
$endgroup$
– user549397
Jan 14 at 3:34
$begingroup$
When you say "the epimorphism," do you mean $x_imapsto p_i$? That's correct, but $x_imapsto p_{r+1-i}$ works too. That's what it means to be algebraically independent.
$endgroup$
– Matt Samuel
Jan 14 at 3:37
$begingroup$
@MattSamuel Yes, the epimorphism means $ Rmapsto R $ and $ [x_1, x_2, ..., x_r]mapsto [ p_{sigma (1)}, p_{sigma (2)},..., p_{sigma (r)} ] $ where $ sigmain S_r $.
$endgroup$
– user549397
Jan 14 at 3:40
$begingroup$
You are correct. However as you enter this area I urge you to think of the two rings as different. For example, $deg(p_i) =i$, but $deg(x_i) =1$. So they aren't isomorphic as graded rings.
$endgroup$
– Matt Samuel
Jan 14 at 3:43
|
show 3 more comments
$begingroup$
Let $ R $ be a ring, $ R[x_1,..., x_r] $ the ring of polynomials over $ R $ in $ r $ indeterminates. Let $ Sigma=R[p_1,..., p_r] $ denote the subring of $ R[x_1,..., x_r] $ consisting of the symmetric polynomials, where $$ p_1=sum_{1}^r x_i, quad p_2=sum_{i<j}x_ix_j,quad p_3=sum_{i<j<k}x_ix_jx_k,quad cdots, quad p_r=x_1x_2cdots x_r $$
are the elementary symmetric polynomials in $ x_1, ..., x_r $.
Since the elementary symmetric polynomials are algebraically independent over $
R $, then we know that the epimorphism: $$ R[x_1, x_2, ..., x_r]longrightarrowSigma=R[p_1, p_2, ..., p_r] $$ is actually an isomorphism.
My question: Is my deduction above valid?
abstract-algebra proof-verification polynomials
asked Jan 14 at 3:24
user549397user549397
1,5101418
$endgroup$
Let $ R $ be a ring, $ R[x_1,..., x_r] $ the ring of polynomials over $ R $ in $ r $ indeterminates. Let $ Sigma=R[p_1,..., p_r] $ denote the subring of $ R[x_1,..., x_r] $ consisting of the symmetric polynomials, where $$ p_1=sum_{1}^r x_i, quad p_2=sum_{i<j}x_ix_j,quad p_3=sum_{i<j<k}x_ix_jx_k,quad cdots, quad p_r=x_1x_2cdots x_r $$
are the elementary symmetric polynomials in $ x_1, ..., x_r $.
Since the elementary symmetric polynomials are algebraically independent over $
R $, then we know that the epimorphism: $$ R[x_1, x_2, ..., x_r]longrightarrowSigma=R[p_1, p_2, ..., p_r] $$ is actually an isomorphism.
My question: Is my deduction above valid?
abstract-algebra proof-verification polynomials
abstract-algebra proof-verification polynomials
asked Jan 14 at 3:24
user549397user549397
1,5101418
asked Jan 14 at 3:24
user549397user549397
1,5101418
asked Jan 14 at 3:24
user549397user549397
1,5101418
asked Jan 14 at 3:24
user549397user549397
1,5101418
asked Jan 14 at 3:24
user549397user549397
1,5101418
1,5101418
$begingroup$
Which deduction is that exactly?
$endgroup$
– Matt Samuel
Jan 14 at 3:32
$begingroup$
@MattSamuel The paragraph outside the box.
$endgroup$
– user549397
Jan 14 at 3:34
$begingroup$
When you say "the epimorphism," do you mean $x_imapsto p_i$? That's correct, but $x_imapsto p_{r+1-i}$ works too. That's what it means to be algebraically independent.
$endgroup$
– Matt Samuel
Jan 14 at 3:37
$begingroup$
@MattSamuel Yes, the epimorphism means $ Rmapsto R $ and $ [x_1, x_2, ..., x_r]mapsto [ p_{sigma (1)}, p_{sigma (2)},..., p_{sigma (r)} ] $ where $ sigmain S_r $.
$endgroup$
– user549397
Jan 14 at 3:40
$begingroup$
You are correct. However as you enter this area I urge you to think of the two rings as different. For example, $deg(p_i) =i$, but $deg(x_i) =1$. So they aren't isomorphic as graded rings.
$endgroup$
– Matt Samuel
Jan 14 at 3:43
|
show 3 more comments
$begingroup$
Which deduction is that exactly?
$endgroup$
– Matt Samuel
Jan 14 at 3:32
$begingroup$
@MattSamuel The paragraph outside the box.
$endgroup$
– user549397
Jan 14 at 3:34
$begingroup$
When you say "the epimorphism," do you mean $x_imapsto p_i$? That's correct, but $x_imapsto p_{r+1-i}$ works too. That's what it means to be algebraically independent.
$endgroup$
– Matt Samuel
Jan 14 at 3:37
$begingroup$
@MattSamuel Yes, the epimorphism means $ Rmapsto R $ and $ [x_1, x_2, ..., x_r]mapsto [ p_{sigma (1)}, p_{sigma (2)},..., p_{sigma (r)} ] $ where $ sigmain S_r $.
$endgroup$
– user549397
Jan 14 at 3:40
$begingroup$
You are correct. However as you enter this area I urge you to think of the two rings as different. For example, $deg(p_i) =i$, but $deg(x_i) =1$. So they aren't isomorphic as graded rings.
$endgroup$
– Matt Samuel
Jan 14 at 3:43
$begingroup$
Which deduction is that exactly?
$endgroup$
– Matt Samuel
Jan 14 at 3:32
$begingroup$
Which deduction is that exactly?
$endgroup$
– Matt Samuel
Jan 14 at 3:32
$begingroup$
@MattSamuel The paragraph outside the box.
$endgroup$
– user549397
Jan 14 at 3:34
$begingroup$
@MattSamuel The paragraph outside the box.
$endgroup$
– user549397
Jan 14 at 3:34
$begingroup$
When you say "the epimorphism," do you mean $x_imapsto p_i$? That's correct, but $x_imapsto p_{r+1-i}$ works too. That's what it means to be algebraically independent.
$endgroup$
– Matt Samuel
Jan 14 at 3:37
$begingroup$
When you say "the epimorphism," do you mean $x_imapsto p_i$? That's correct, but $x_imapsto p_{r+1-i}$ works too. That's what it means to be algebraically independent.
$endgroup$
– Matt Samuel
Jan 14 at 3:37
$begingroup$
@MattSamuel Yes, the epimorphism means $ Rmapsto R $ and $ [x_1, x_2, ..., x_r]mapsto [ p_{sigma (1)}, p_{sigma (2)},..., p_{sigma (r)} ] $ where $ sigmain S_r $.
$endgroup$
– user549397
Jan 14 at 3:40
$begingroup$
@MattSamuel Yes, the epimorphism means $ Rmapsto R $ and $ [x_1, x_2, ..., x_r]mapsto [ p_{sigma (1)}, p_{sigma (2)},..., p_{sigma (r)} ] $ where $ sigmain S_r $.
$endgroup$
– user549397
Jan 14 at 3:40
$begingroup$
You are correct. However as you enter this area I urge you to think of the two rings as different. For example, $deg(p_i) =i$, but $deg(x_i) =1$. So they aren't isomorphic as graded rings.
$endgroup$
– Matt Samuel
Jan 14 at 3:43
$begingroup$
You are correct. However as you enter this area I urge you to think of the two rings as different. For example, $deg(p_i) =i$, but $deg(x_i) =1$. So they aren't isomorphic as graded rings.
$endgroup$
– Matt Samuel
Jan 14 at 3:43
|
show 3 more comments
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$begingroup$
Which deduction is that exactly?
$endgroup$
– Matt Samuel
Jan 14 at 3:32
$begingroup$
@MattSamuel The paragraph outside the box.
$endgroup$
– user549397
Jan 14 at 3:34
$begingroup$
When you say "the epimorphism," do you mean $x_imapsto p_i$? That's correct, but $x_imapsto p_{r+1-i}$ works too. That's what it means to be algebraically independent.
$endgroup$
– Matt Samuel
Jan 14 at 3:37
$begingroup$
@MattSamuel Yes, the epimorphism means $ Rmapsto R $ and $ [x_1, x_2, ..., x_r]mapsto [ p_{sigma (1)}, p_{sigma (2)},..., p_{sigma (r)} ] $ where $ sigmain S_r $.
$endgroup$
– user549397
Jan 14 at 3:40
$begingroup$
You are correct. However as you enter this area I urge you to think of the two rings as different. For example, $deg(p_i) =i$, but $deg(x_i) =1$. So they aren't isomorphic as graded rings.
$endgroup$
– Matt Samuel
Jan 14 at 3:43