Group isomorphism: quotient of R[x] ≅ direct sum of R
In what way can we construct an isomorphism using \( R[x] \) and \( R \)?
If we restrict our aim to just a group isomorphism then we can achieve the following:
Let \( R \) be a commutative ring, \( R[x] \) a polynomial ring, and \( f(x) \in R[x] \) a polynomial. Let \( \varphi: R[x] \to R^{\bigoplus d} \) be a set function defined by sending \( g(x) \in R[x] \) to the tuple representation of the remainder of \( g(x) \) when divided by \( f(x) \).
Then this function induces an isomorphism of abelian groups:
Can you construct the proof?
Extra:
- How to construct the complex ring \( \mathbb{C} \)?
- How to construct a ring of polynomial evaluations?
\( R[x]/(f(x)) \cong R^{\bigoplus d} \). Proof idea.
If \( f(x) \) is a monic polynomial of degree \( d \), then mapping \( g(x) \) to \( r(x) \) in \( g(x) = q(x)f(x) + r(x) \) is a is a function from elements of \( R[x] \) to elements of \( R^{\bigoplus d} \). We will call this mapping \( \varphi \). It turns out that this mapping meets the requirements to be a homomorphism of abelian groups (for the addition operation). A taster of this is that if \( g_3(x) = g_1(x) + g_2(x) \) then \( r_3(x) = r_1(x) + r_2(x) \), and so the mapping preserves addition. What is left is to show preservation of the identity and inverses.
We then turn to the homomorphism decomposition theorem: \( \varphi \) should be decomposable into a surjective homomorphism followed by an injective homomorphism to the image. The surjective part is from \( R[x] \) to \( R[x]/\operatorname{ker} \varphi \). The injective part is from this set to \( \operatorname{im} \varphi \). It turns out that \( \operatorname{ker} \varphi = (f(x)) \), and so we have \( R[x]/(f(x)) \cong \text{some subset of } R^{\bigoplus d} \). But \( \varphi \) is itself surjective so the codomain is all of \( R^{\bigoplus d} \). The function is surjective as every polynomial of degree less than \( d \) is its own remainder, and thus fills up the entire codomain of d-tuples. And so we end up with \( R[x]/(f(x)) \cong R^{\bigoplus d} \).
The actual proof is best broken down into some lemmas.
Lemma 1
\( \varphi \) is a surjective function.
Proof. \( \varphi \) is a well-defined function because we can always calculate a unique remainder of \( g(x) \) when divided by \( f(x) \) (shown previously). \( \varphi \) is surjective, as every polynomial of degree less than \( d \) is its own remainder when divided by \( f(x) \), and so for any element in \( R^{\bigoplus d} \) we can find a polynomial of degree less than \( d \) that maps to it. (Remember: \( f(x) \) itself would require \( d+1 \) coordinates to represent.)
Lemma 2
\( \varphi \) is a abelian group homomorphism.
Proof. Let \( g_1(x), g_2(x) \in R[x] \). We must show that:
- \( \varphi(g_1(x) + g_2(x)) = \varphi(g_1(x)) + \varphi(g_2(x)) \)
- \( \varphi(0) = 0 \)
- \( \varphi(-g(x)) = -\varphi(g(x)) \)
-
Let \( g_1(x), g_2(x) \in R[x] \). We can write:
\[ \begin{align*} g_1(x) &= q_1(x) f(x) + r_1(x) \\ g_2(x) &= q_2(x) f(x) + r_2(x) \end{align*} \]for some \( q_1(x), q_2(x), r_1(x), r_2(x) \in R[x] \), unique to \( g_1(x) \) and \( g_2(x) \). Then:
\[ g_1(x) + g_2(x) = (q_1(x) + q_2(x)) f(x) + r_1(x) + r_2(x) \]As \( r_1(x) = \varphi(g_1(x)) \) and \( r_2(x) = \varphi(g_2(x)) \), we have:
\[ \begin{align*} \varphi(g_1(x) + g_2(x)) &= r_1(x) + r_2(x) \\ &= \varphi(g_1(x)) + \varphi(g_2(x)) \end{align*} \]as desired.
- \( 0 = 0 f(x) + 0 \), and so \( \varphi(0) = 0 \).
- TODO!
Lemma 3
\( \operatorname{ker} \varphi = (f(x)) \).
Proof. Let \( g(x) \in \operatorname{ker} \varphi \). Then \( \varphi(g(x)) = 0 \). We know that:
\( R[x]/(f(x)) \cong R^{\bigoplus d} \). Proof.
\( \varphi \) is a surjective group homomorphism, by Lemma 1 and Lemma 2. By the first isomorphism theorem for groups, we have:
By Lemma 3, \( \operatorname{ker} \varphi = (f(x)) \), and so:
as desired.
As a recap, here is the first isomorphism theorem for groups:
First isomorphism theorem for groups
Suppose \( \varphi: G \to H \) is a surjective group homomorphism. Then:
\( d \) and not \( d+1 \)
Why is the dimension of the vector space \( R^{\bigoplus d} \) and not \( R^{\bigoplus d+1} \)? We need \( d \) coordinates to represent a polynomial of degree \( d-1 \) because we also need 1 coordinate for the 0th term.
\( R[x]/(f(x)) \cong R^{\bigoplus d} \). Motivation and proof idea.
One way to construct an isomorphism between \( R[x] \) and \( R^{\bigoplus d} \) is to restrict the polynomial ring to only contain polynomials of degree less than \( d \) and then only consider the addition operation. This is a bit pointless, as the now reduced \( R[x] \) group is practically identical to \( R^{\bigoplus d)} \)—it just uses \( x, x^1, x^2 \) etc. as notation instead of commas that might be used for writing down elements of \( R^{\bigoplus d)} \).
Another way to restrict ourselves to polynomials of \( \operatorname{deg} \leq d \) is to "project" all polynomials onto the smaller set of polynomials of degree less than or equal to \( d \). One way to project \( g(x) \in R[x] \) onto the set of polynomials of degree less than or equal to \( d \) is first fix a polynomial \( f(x) \) of degree \( d \) and then map \( g(x) \) to the remainder of \( g(x) \) when divided by \( f(x) \). The uniqueness of the remainder given \( g(x) \) and \( f(x) \) insures that this projection is a set function. What is left to do is show that the function maps the \( + \) operation in \( R[x] \) to the \( + \) operation in \( R^{\bigoplus d} \). This is analogous to showing that for integers \( (a+b) \operatorname{mod} c \) is the same as \( (a \operatorname{mod} c) + (b \operatorname{mod} c) \).
Example
TODO: add the complex number example