Ideal Arithmetic, and the Chinese Remainder Theorem.

When we first introduced ring ideals, we defined them as kernels of ring homomorphisms. This is a valid way to view them. However, they’re often used simply as sets, and the rest of the homomorphism structure is ignored. The reason they’re called “ideals” is they have some very nice properties which make them ideal to work with, in some situations.

We’ve already used some of these properties in the proof that a PID is a UFD.

Definition: Left and Right Ideals

Note: This expands on an earlier result about when a sub-ring is an Ideal.

We’re going to give an example of a ring with a left ideal that is not a right ideal. However, before we do that, we’re going to introduce a concept we should have introduced a while ago.

Definition: Category

By now, you’re already used to some categories. For example, the category of groups, from earlier entries. We finally gave the definition, because it allows us to define a useful type of structure.

Definition: Free Object

This is typically drawn out like this

Example: A Free Set

Let’s consider a fairly basic example. Suppose \( \mathcal{C} \) is the category of sets, and so \(F\) and \(G\) are sets. In the category of sets, morphisms are just functions (isomorphisms are bijections, etc…).

In order for \( \phi \), a function, from \(F\) to \(G\) to be determined by \(f\) and an arbitrary function, \(g : S \rightarrow G \), we have to have

\[ \phi = g \circ f^{-1} \]

which means that \( f \) has to be a bijection.

Example: A Free Ring

We introduced Free objects so that we could investigate a free Ring on a set of two elements. Let \(S = \{ X, Y \} \), and let \(F\) be a ring with only one unit, \(1\), which is free on \(S\). To simplify the notation, let’s denote \( f(X) \) as “\(X\)” and \(f(Y)\) as “\( Y \)” in \(F \).

Notice that if we let \( G = F \), then, by definition, for an arbitrary

\[ g : \{X, Y\} \rightarrow F \]

there is a unique endomorphism

\[ \phi : F \rightarrow F \]

where

$$ \begin{align} \phi(X) = g(X) \\ \phi(Y) = g(Y) \\ \end{align} $$

For example, if

$$ \begin{align} g(X) &= X^2 + 12 + 3YX - XY \\ g(Y) &= X^3 \\ \end{align} $$

then \( \text{Im}(\phi) \) would be isomorphic to

\[ F / \mathfrak{a} \]

where \( \mathfrak{a} \) is the ideal

\[ \mathfrak{a} = \langle X - (X^2 + 12 + 3YX - XY), Y - X^3 \rangle \]

More generally we have,

$$ \text{Im}(\phi) \cong F / \langle X - g(X), Y - g(Y) \rangle $$

Note: Just because we have defined something, that doesn’t mean it necessarily exists. For example, consider the category of groups with exactly one element (the identity). It is a trivial category, but is still a category. Now let \(S\) be any set with more than one element. It is easy to see that there is no free group from that category on \(S\).

---

Example: A left Ideal which is not a right Ideal

Now we’re going to apply some of what we’ve been discussing to provide an example of a left Ideal which is not a right ideal.

Let \(F\) be the ring with only one unit, \(1\), which is free on the set \( S = \{ X, Y \} \) (It’s the same ring from the previous example). Now, let

$$ \begin{align} \mathfrak{a} :=& \{ r \cdot X | r \in F \} \\ =& F \cdot X \\ \end{align} $$

Notice that \( \mathfrak{a} \) is a left idea, because for any \( r \in F \) and \( a \in \mathfrak{a} \), we have

$$ \begin{align} ra &= r(r_1X) \\ &= r_2X \in \mathfrak{a} \\ \end{align} $$

where \( r_2 = r \cdot r_1 \). However,

$$ \mathfrak{a} \cdot Y \nsubseteq \mathfrak{a} $$

because

$$ XY \neq rX $$

for any \( r \in F\). So \( \mathfrak{a} \) is not a right-ideal.

We’ll also point out that \( F \) is an example of a non-commutative integral domain, and a ring which is not a principal ideal domain, since the ideal \( \langle X, Y \rangle \) is not principal.

Lemmas: Operations on Ideals

We’ve already been using 3. We’ll leave the proof to you, the reader (it’s very simple). We use it implicitly in our (so far implicit) definition of \( \langle S \rangle \), the ideal generated by the set \(S\). I.e.

$$ \langle S \rangle := \bigcap\limits_{I \in \mathcal{I}} I $$

where

$$ \mathcal{I} := \{ S \subseteq R | S \text{ is an ideal } \} $$

Proof

1. Suppose \( \mathfrak{a} \) and \( \mathfrak{b} \) are left-ideals of a ring, \(R\). Let \( x \) and \( y \) be elements of \( \mathfrak{a} \cdot \mathfrak{b} \), so that we have

$$ \begin{align} x &= \sum\limits_{i = 1}^n a_i \cdot b_i \\ y &= \sum\limits_{i = 1}^m c_i \cdot d_i \\ \end{align} $$

which means that

$$ x \cdot y = \sum\limits_{k = 2}^{m + n} \sum\limits_{i + j = k} (a_i \cdot b_i) \cdot (c_j \cdot d_j)
$$

and since \( c_j \in \mathfrak{a} \), a left-ideal of \(R\), we have, for all \( i \) and \( j \),

\[ a_i \cdot b_i \cdot c_j \in \mathfrak{a} \]

Thus, \( x \cdot y \in \mathfrak{a} \cdot \mathfrak{b} \).

It’s very easy to see that \( x + y \in \mathfrak{a} \cdot \mathfrak{b} \). So \( \mathfrak{a} \cdot \mathfrak{b} \) is a subring of \(R\).

Now, let \( r \in R \) be arbitrary. Then \( r \cdot x \), where \( x \in \mathfrak{a} \cdot \mathfrak{b} \) is arbitrary, is equal to

$$ \begin{align} r \cdot &\sum\limits_{i = 1}^n a_i \cdot b_i \\ = &\sum\limits_{i = 1}^n (r \cdot a_i) \cdot b_i \\ = &\sum\limits_{i = 1}^n {a’}_i \cdot b_i \\ \end{align} $$

where \( {a’}_i = r \cdot a_i \) is in \( \mathfrak{a} \), since \( \mathfrak{a} \) is a left-ideal. So part 1 is established for left-ideals.

2. Again, suppose \( \mathfrak{a} \) and \( \mathfrak{b} \) are left ideals, and let \( x \) and \( y \) be members of \( \mathfrak{a} + \mathfrak{b} \), so that

$$ \begin{align} x = a_1 &+ b_1 \\ y = a_2 &+ b_2 \\ a_1, a_2 &\in \mathfrak{a} \\ b_1, b_2 &\in \mathfrak{b} \\ \end{align} $$

Then,

$$ \begin{align} x \cdot y &= (a_1 + b_1) \cdot (a_2 + b_2) \\ &= a_1 \cdot a_2 + a_1 \cdot b_2 + b_1 \cdot a_2 + b_1 \cdot b_2 \\ &= (a_1 \cdot a_2 + b_1 \cdot a_2) + (a_1 \cdot b_2 + b_1 \cdot b_2) \\ \end{align} $$

Since ideals are subrings, \( a_1 \cdot a_2 \in \mathfrak{a} \) and \( b_1 \cdot b_2 \in \mathfrak{b} \). Now, because \( \mathfrak{a} \) and \( \mathfrak{b} \) are left ideals, \( b_1 \cdot a_2 \in \mathfrak{a} \) and \( a_1 \cdot b_2 \in \mathfrak{b} \). Putting this together, the last line of the previous set of equations lies in \( \mathfrak{a} + \mathfrak{b} \). Thus \( x \cdot y \in \mathfrak{a} + \mathfrak{b} \).

It is very easy to see that \( x + y \in \mathfrak{a} + \mathfrak{b} \), so \( \mathfrak{a} + \mathfrak{b} \) is a subring of \( R \).

Next we’ll show that it’s a left ideal. Let \( r \in R \) be arbitrary, and \( x = a + b \in \mathfrak{a} + \mathfrak{b} \) also be arbitrary. Then,

$$ \begin{align} r \cdot x &= r \cdot a + r \cdot b \\ &= a’ + b’ \\ \end{align} $$

where \( a’ = r \cdot a \in \mathfrak{a} \) and \( b’ = r \cdot b \in \mathfrak{b} \) since \( \mathfrak{a} \) and \( \mathfrak{b} \) are both left ideals of \(R\). Therefore we have shown that \( \mathfrak{a} + \mathfrak{b} \) is a left ideal of \(R\).

3. The proof is very easy, and left to the reader.

The proofs for the three cases where the ideals are right ideals are nearly identical. Combining the results for left and right ideals, we obtain the cases where the ideals are two-sided ideals of \(R\).

4. Finally, we’ll prove that if \( R \) is a commutative ring with units and \(\mathfrak{a} + \mathfrak{b} = R \), then \( \mathfrak{a} \cdot \mathfrak{b} = \mathfrak{a} \cap \mathfrak{b} \).

Suppose \( x \in \mathfrak{a} \cdot \mathfrak{b} \). So we have

$$ x = \sum\limits_{i = 1}^n a_i \cdot b_i $$

where \( a_i \in \mathfrak{a} \) and \( b_i \in \mathfrak{b} \). Notice that since \( R \) is commutative, \( \mathfrak{a} \) and \( \mathfrak{b} \) are two-sided ideals. This means that each term of the sum is in both \( \mathfrak{a} \) and \( \mathfrak{b} \). Thus, \( x \in \mathfrak{a} \cap \mathfrak{b} \), so we have

\[ \mathfrak{a} \cdot \mathfrak{b} \subseteq \mathfrak{a} \cap \mathfrak{b} \]

Finally, suppose that \( x \in \mathfrak{a} \cap \mathfrak{b} \).

Notice that since \( R \) is a commutative ring with units, that \( \mathfrak{a} + \mathfrak{b} = R \) implies that \( 1 \in \mathfrak{a} + \mathfrak{b} \). This means that for some \( a_1 \in \mathfrak{a} \) and \( b_1 \in \mathfrak{b} \) we have

$$ 1 = a_1 + b_1 $$

Now consider

$$ \begin{align} x &= 1 \cdot x \\ &= (a_1 + b_1) \cdot x \\ &= a_1 \cdot x + b_1 \cdot x \\ \end{align} $$

since \( x \in \mathfrak{b} \), \( a_1 \cdot x \in \mathfrak{a} \cdot \mathfrak{b} \). Similarly, since \( x \in \mathfrak{a} \), we see that \( b_1 \cdot x \in \mathfrak{a} \cdot \mathfrak{b} \). Thus their sum is also in \( \mathfrak{a} \cdot \mathfrak{b} \). This gives us

$$ \mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{a} \cdot \mathfrak{b} $$

QED

Theorem: The Chinese Remainder Theorem

Proof

The case where \( N = 1 \) is trivially satisfied with \( r_1 = r \).

Now, suppose \( N \geq 2 \). Let’s consider a fixed \( i \). Since \( \mathfrak{a}_i + \mathfrak{a}_j = R\), when \( j \neq i \), we have

$$ x_{i,j} + y_{i,j} = 1 $$

where \( x_{i,j} \in \mathfrak{a}_i \) and \( y_{i,j} \in \mathfrak{a}_j \). So let’s consider the product

$$ \prod\limits_{j = 1, j \neq i}^N (x_{i,j} + y_{i,j}) = 1 $$

Notice that when you expand it out, the only term without an \( x_{i,j} \) is the right-most term,

$$ y_{1,2} \cdot y_{1,3} \cdot … \cdot y_{N-1, N} $$

this means that we can express \(1\) as a sum of elements of \( \mathfrak{a}_i \) and an element of \( \displaystyle \prod\limits_{j = 1, j \neq i}^N \mathfrak{a}_j \), so we have

$$ \mathfrak{a}_i + \prod\limits_{j = 1, j \neq i}^N \mathfrak{a}_j = R $$

which means that there are \( u_i \) and \( v_i \in R \), such that

$$ u_i + v_i = 1 $$

where

$$ \begin{align} u_i &\in \mathfrak{a}_i \\ v_i &\in \prod\limits_{j = 1, j \neq i}^N \mathfrak{a}_j \\ \end{align} $$

Then if we define

$$ z_i := r_i \cdot v_i $$

we see that

$$ z_i \equiv \begin{cases} r_i, &\mod \mathfrak{a}_i \\ 0, &\mod \mathfrak{a}_j \\ \end{cases} $$

and so the sum,

$$ r = \sum\limits_{i = 1}^N z_i $$

satisfies

$$ r \equiv r_i \mod \mathfrak{a}_i $$

QED

That’s all for this entry. Thanks for reading!!