We’ll start by defining Quotient Rings

Definition: Quotient Ring
Let \(R\) be a ring and \(I\) be an ideal in \(R\). Then let the quotient ring \(R/I\) $$ \begin{align*} R/I = \{a + I \ | \ a + R\} \end{align*} $$ be the set of all \(I\)-cosets in \(R\) with operations:
  1. Addition: \((a + I) + (b + I) = (a + b) + I\).
  2. Multiplication: \((a + I) + (b + I) = (ab) + I\).


Note that there are other notations used for \(I + a\). One notation is \([a]\) or \(\bar{a}\).

The claim is that these operations are well defined and will make the quotient set with the operations defined a ring. For addition, it’s already part of how we defined quotient groups? For multiplication, we need to show that if \(a + I = a' + I\) and \(b + I = b' + I\). Then

$$ \begin{align*} ab + I = a'b' + 1 \end{align*} $$

To see this, write \(a' = a + u\) and \(b' = b + v\) for some \(u, v \in I\). Then

$$ \begin{align*} a'b' &= (a + u)(b + v) \\ &= ab + av + ub + uv \end{align*} $$

\(av + ub + uv\) is in \(I\) since \(I\) is an ideal. Therefore, \(a'b' + (av+ub+uv)\) and \(ab + I\) represent the same ideal.

Next, we need to show that \(R/I\) is an abelian group with addition. And then we need to show that multiplication is associative and distributes over addition. (exercise)


Quotient Homomorphism

Definition: Quotient Homomorphism
Let \(R\) be a ring and \(I\) be an ideal in \(R\). Define $$ \begin{align*} \pi: R &\rightarrow R/I \\ \pi(a) &= a + I \end{align*} $$


So just send any element to its coset. We already know this makes a group homomorphism for quotient groups. It also makes a ring homomorphism. We just need to check. It is also surjective because every coset comes from some element in \(I\). The kernel of the homomorphism \(\ker(\pi) = I\).


Example 1

Take \(R = \mathbf{Z}\) and take \(I\) to be a principle ideal generated by some element \(d\) so \(I = (d) = \mathbf{Z}d\). Then

$$ \begin{align*} R/I = \mathbf{Z}/\mathbf{Z}d = \mathbf{Z}_d \end{align*} $$



Example 2

Take the ring of polynomials \(R[x]\). If we quotient this ring by the principle ideal generated by \((x^2 + 1)\), then we get the ring of complex numbers. \(R[x]/(x^2+1) \cong \mathbf{C}\)


Quotient Rings of Polynomial Rings

Let \(K\) be a field. Let \(R = K[x]\) be the ring of polynomials over the field \(K\) and let \(I \subseteq R\). From this, form the quotient ring \(S = K[x]/I\). Notice here that since the ideal is in a ring over a field, then we know every ideal is principle. So \(I\) is a principle ideal generated by some polynomial and we can write \(I = (f)\) for some \(f \in R\).

For example, \(R/(0) \cong R\). If \(I\) is not \(\{0\}\), then choose \(f\) to be monic of minimal degree. Now write

$$ \begin{align*} f = x^n + a_{n-1}x + ... + a_0, a_0,...,a_{n-1} \in K \end{align*} $$

Then \(S = K[x] /(f)\). We have the following claim based in this

Proposition
Every element in \(S\) has the form \(r + I\) for some unique \(r \in K[x]\) such that \(\deg(r) < n\).


Proof
Use division with remainder. If \(g + I \in S\) where \(g \in K[x]\), then there exists unique \(q, r \in K[x]\) such that \(g = qf + r\) where \(\deg(r) < n = \deg(f)\). This means we can write \(g - r = qf\). We know that \(qf \in I\) since it’s a multiple of \(f\). Since \(g - r = qf\), then \(g-r\) is also in \(I\). This implies that \(g\) and \(r\) are in the same \(I\)-coset. Therefore \(g+I = r+I\). \(\blacksquare\).

\(r + I\) is the canonical form of an element in \(S = K[x] / (f)\).


Example

Suppose \(K = \mathbf{Q}\) be the rationals ring, \(S = \mathbf{Q}[x]/(f)\) be the quotient ring and let \(f = x^2 - 2\) so \(I = (f)\) is the ideal generated by this polynomial \(f=x^2-2\).

Using the previous result, we know that Every element of \(S\) can be written as \(r + I\) for a unique \(r \in Q[x]\) such that \(\deg(r) < \deg(f)=2\). Therefore, we can write \(r + I\) as

$$ \begin{align*} (a + bx) + I = a + b\bar{x} \quad a,b \in \mathbf{Q} \end{align*} $$

This notation is a shorthand for: given \(a \in \mathbf{Q}\), then write \(a\) or \(\bar{a}\) instead of writing \(a + I \in S\) where \(a+I\) is the coset of constant polynomials in \(S\). Also write \(\bar{x}\) for the element \(x + I\). So, then

$$ \begin{align*} (a + b\bar{x}) + (a' + b'\bar{x}) &= (a+a') + (b+b')\bar{x} \\ (a + b\bar{x}) (a' + b'\bar{x}) &= aa' + ab'\bar{x} + ba'\bar{x} + bb'\bar{x}^2 \end{align*} $$

but now we have \(\bar{x}^2\). That’s because we still need to divide by \(f\). If we divide \(\bar{x}^2\) by \(x^2 - 2\), the remainder is 2. So \(\bar{x}^2\) is really 2 in the ring \(R/(f)\). Then we can simplify this to

$$ \begin{align*} (a + b\bar{x}) (a' + b'\bar{x}) &= aa' + ab'\bar{x} + ba'\bar{x} + 2bb' \\ &= (aa' + 2bb') +(ab'+ba')\bar{x} \end{align*} $$

Remark: This quotient ring \(S\) is isomorphic to \(\mathbf{Q}(\sqrt{2})\) which is a subring of \(\mathbf{R}\). \(\mathbf{Q}(\sqrt{2})\) is the ring

$$ \begin{align*} \{a+b\sqrt{2} \ | \ a,b \in \mathbf{Q}\} \end{align*} $$

The isomorphism between these rings is

$$ \begin{align*} a+b\bar{x} \rightarrow a + b\sqrt{2} \end{align*} $$



Homomorphism Theorem

We start by a recipe to form a new homomorphism from a quotient ring.

Homomorphism Theorem
Let \(\varphi: R \rightarrow S\) be a ring homomorphism and \(I \subseteq R\) be an ideal. If \(I \subseteq \ker(\varphi)\) (ie. \(\varphi(I) = \{0\}\)), then there exists a ring homomorphism \(\bar{\varphi}: R/I \rightarrow S\). Such that \(\bar{\varphi}(a + I) = \varphi(a)\)


To prove this we need to show that the new homomorphism \(\bar{\phi}\) is well defined and then we need to show that it’s a ring homomorphism. And then we have the isomorphism theorem as follows

Isomorphism Theorem
Let \(\varphi: R \rightarrow S\) be a surjective ring homomorphism and \(I \subseteq R\) be an ideal. If \(I = \ker(\varphi)\), then we have an isomorphism of rings $$ \begin{align*} \bar{\varphi}: R/I &\rightarrow S \\ \bar{\varphi}(a + I) &= \varphi(a) \end{align*} $$


Example 1

Let \(R = \mathbf{Q}[x]\) and \(S = \mathbf{R}\). Since we want to build a homomorphism from a polynomial ring, then we should always think of the substitution principle. Recall that the substitution principle says that given that we have a homomorphism (in this case let this homomorphism be the identity function), then we can create the following evaluation homomorphism

$$ \begin{align*} \varphi: \mathbf{Q}[x] &\rightarrow \mathbf{R} \\ \varphi(a) &= a \quad \text{if } a \in \mathbf{Q} \subseteq \mathbf{Q}[x] \\ \varphi(x) &= \sqrt{2} \\ \end{align*} $$

So in this case, it’s just an evaluation. So, given some polynomial \(f\), then

$$ \begin{align*} \varphi: \mathbf{Q}[x] &\rightarrow \mathbf{R} \\ \phi(f) &= f(\sqrt{2}) \end{align*} $$

What is the kernel of this homomorphism?

$$ \begin{align*} I = \ker(\varphi) &= \{ f \in \mathbf{Q}[x] \ | \ \varphi(f) = 0 \} \\ &= \{ f \in \mathbf{Q}[x] \ | \ f(\sqrt{2}) = 0 \} \end{align*} $$

So it’s any polynomial that has \(\sqrt{2}\) as a root. Since \(\mathbf{Q}\) is a field, then we know that the kernel is principle. So it’s generated by a single monic polynomial of minimal degree. First of all, this kernel isn’t trivial. For example \(x^2 - 2\) has a \(\sqrt{2}\) as a root of degree 2. If we know that we don’t have constant polynomials or polynomials of degree 1, then we know \(x^2 - 2\) is a generator for the group. Constant non-zero polynomials don’t have roots. Linear polynomials can’t have \(\sqrt{2}\) as a root. The roots will be rationals since the coefficients are rational.

So now we have a homomorphism \(\varphi\) from \(\mathbf{Q}[x]\) to \(\mathbf{R}\). Therefore, we can apply the homomorphism theorem to get a new homomorphism (quotient homomorphism) from \(\mathbf{Q}[x]\) to \(\mathbf{Q}[x]/(x^2-2)\).

$$ \begin{align*} \pi: \mathbf{Q}[x] &\rightarrow \mathbf{Q}[x]/(x^2-2) \\ \phi(f) &= f + (x^2 - 2) \end{align*} $$

The homomorphism takes a polynomial \(f\) in \(\mathbf{Q}[x]\) to its coset.

Notice now that we have a homomorphism \(\varphi: \mathbf{Q}[x] \rightarrow \mathbf{R}\) and we have the canonical homomorphism \(\pi: \mathbf{Q}[x] \rightarrow \mathbf{Q}[x]/(x^2-2)\). Before we can apply the isomorphism theorem, we have one issue which is that \(\varphi\) is not surjective. We can fix this by restricting the target to only the image of \(\varphi\) so

$$ \begin{align*} \varphi': \mathbf{Q}[x] &\rightarrow \varphi(\mathbf{Q}[x]) \\ f(x) &\rightarrow f(\sqrt{2}) \\ \end{align*} $$

So now we have a surjective homomorphism where the kernel is \(I = (x^2 - 2)\) and we formed the quotient ring \(\mathbf{Q}[x]/(x^2-2)\). Therefore, we can now apply the isomorphism to conclude that \(\mathbf{Q}[x]/(x^2-2)\) is isomorphic to \(\varphi(\mathbf{Q}[x])\). Moreover, we have an isomorphism defined by

$$ \begin{align*} \bar{\varphi}: \mathbf{Q}[x]/(x^2-2) &\rightarrow \varphi(\mathbf{Q}[x]) \\ f + (x^2 - 2) &\rightarrow \varphi(f) = f(\sqrt{2}) \end{align*} $$

Note that \(\varphi(\mathbf{Q}[x])\) are all the real numbers we can get by plugging in \(\sqrt{2}\) into polynomials with rational coefficients. So

$$ \begin{align*} f(x) &= a_0 + a_1x + ... + a_nx^n, \quad a_i \in \mathbf{Q} \\ f(\sqrt{2}) &= a_0 + a_1(\sqrt{2}) + ... + a_n(\sqrt{2})^n, \quad a_i \in \mathbf{Q} \end{align*} $$

But \((\sqrt{2}) = 2, (\sqrt{2})^3 = 2\sqrt{2}\) and in general \((\sqrt{2})^2k = 2^k\) while \((\sqrt{2})^{2k+1} = 2^k\sqrt{2}\). So any \(f(\sqrt{2})\) simplifies to \(a + b\sqrt{2}\). Therefore

$$ \begin{align*} \varphi(\mathbf{Q}) = \{ a + b\sqrt{2} \ | \ a,b \in \mathbf{Q} \} \end{align*} $$



Example 2

Suppose we have a homomorphism which is just the identity function \(id\) from \(\mathbf{R} \rightarrow \mathbf{R}\). Based on this create the evaluation homomorphism

$$ \begin{align*} \varphi: \mathbf{R}[x] &\rightarrow \mathbf{C} \\ \varphi(a) &\rightarrow a \\ \varphi(x) &\rightarrow i \end{align*} $$

This homomorphism is in fact surjective. The kernel of this homomorphism is generated by \((x^2+1)\). Therefore,

$$ \begin{align*} \mathbf{R}[x]/(x^2+1) \cong \mathbf{C} \end{align*} $$



Example 3

This take, let’s take the rational numbers instead so

$$ \begin{align*} \varphi: \mathbf{Q}[x] &\rightarrow \mathbf{R} \\ \varphi(a) &\rightarrow a \\ \varphi(x) &\rightarrow \sqrt[3]{2} \end{align*} $$

The kernel of this homomorphism is generated by \((x^3-2)\). The image of this homomorphism is a subset of \(\mathbf{R}\).

$$ \begin{align*} \mathbf{Q}[x]/(x^3-2) \cong \varphi(\mathbf{Q}[x]) \subseteq \mathbf{R} \end{align*} $$



References