Question

Let \(\mathcal{A}\) be an associative algebra, and \(\mathbf{x} \in \mathcal{A}\). Show that \(\mathcal{A} \mathbf{x}\) is a left ideal. \(\mathbf{x} \cdot \mathcal{A}\) is a right ideal, and \(\mathcal{A} \mathbf{x} \cdot \mathcal{A}\) is a two-sided ideal.

Short answer

According to above steps, for any elements in \(\mathcal{A}\), \(ab\mathbf{x}, \mathbf{x}ab, a\mathbf{x}b\) are always in \(\mathcal{A} \mathbf{x}\), \(\mathbf{x} \cdot \mathcal{A}\) and \(\mathcal{A} \mathbf{x} \cdot \mathcal{A}\), respectively. So \(\mathcal{A} \mathbf{x}\) is a left ideal, \(\mathbf{x} \cdot \mathcal{A}\) is a right ideal and \(\mathcal{A} \mathbf{x} \cdot \mathcal{A}\) is a two-sided ideal in \(\mathcal{A}\).

Step-by-step solution

Check that \(\mathcal{A} \mathbf{x}\) is a left ideal

To show that \(\mathcal{A} \mathbf{x}\) is a left ideal, we need to show that for any elements \(a, b \in \mathcal{A}, \) when we perform the multiplication \((ab)\mathbf{x}\) it also results in an element of \(\mathcal{A} \mathbf{x}\). Then \(c = ab \in \mathcal{A}\), and \(c\mathbf{x}=(ab)\mathbf{x} \in \mathcal{A} \mathbf{x}\), which attests to the property of left-ideal. Hence, \(\mathcal{A} \mathbf{x}\) is a left ideal.

Evidence that \(\mathbf{x} \cdot \mathcal{A}\) is a right ideal

We follow a similar procedure for a right ideal. Here we need to show that for any \(a, b \in \mathcal{A}\), \(\mathbf{x}(ab) \in \mathbf{x} \cdot \mathcal{A}\). Let \(c = ab \in \mathcal{A}\), then \(\mathbf{x}c=\mathbf{x}(ab) \in \mathbf{x} \cdot \mathcal{A}\). This verifies that \(\mathbf{x} \cdot \mathcal{A}\) is a right ideal.

Demonstrate that \(\mathcal{A} \mathbf{x} \cdot \mathcal{A}\) is a two-sided ideal

In this case, we need to show that for any \(a,b,c \in \mathcal{A}\) and \(\mathbf{x}\), \(a\mathbf{x}cb \in \mathcal{A} \mathbf{x} \cdot \mathcal{A}\). Let \(d = acb \in \mathcal{A}\), then \(d\mathbf{x}d = a\mathbf{x}cb \in \mathcal{A} \mathbf{x} \cdot \mathcal{A}\). This confirms that \(\mathcal{A} \mathbf{x} \cdot \mathcal{A}\) is a two-sided ideal.

Key concepts

Left Ideal

In associative algebra, a left ideal is a subset of an algebra that absorbs multiplication from the left by elements of the algebra. This means that when we multiply any element of the algebra with a member of this subset from the left, the result stays within the subset. To think about this in simple terms, if we have an element

• p in our algebra
• and a subset L,

multiplying from the left, pL, results in something still within L.
Here, we want to show that \( \mathcal{A} \mathbf{x} \) forms a left ideal. This subset is made by multiplying each element of the algebra \( \mathcal{A} \) with a fixed element, \( \mathbf{x} \). The property holds because when you take any two elements \( a \) and \( b \) in \( \mathcal{A} \) and multiply them to get \( ab \), their product acting on \( \mathbf{x} \) gives \( (ab)\mathbf{x} \). This result is still within the set \( \mathcal{A} \mathbf{x} \), proving it's a left ideal.

Right Ideal

A right ideal in associative algebra is the counterpart to a left ideal, focused on multiplication from the right. If you have a subset R of an algebra, and it is a right ideal, multiplying any member of the algebra with a member of this subset from the right will keep you inside the subset.
Consider this as saying that if you take an element

• p in your algebra
• and a subset R,

combining them as Rp leaves you with something that remains in R.
For our case, \( \mathbf{x} \cdot \mathcal{A} \) is proposed to be a right ideal. Here, multiplication is done by first taking your fixed element \( \mathbf{x} \) and then multiplying everything in \( \mathcal{A} \).
For any elements \( a \) and \( b \) in \( \mathcal{A} \), the multiplication \( \mathbf{x}(ab) \) lies in \( \mathbf{x} \cdot \mathcal{A} \), thereby showing it fulfills the properties of a right ideal.

Two-Sided Ideal

A two-sided ideal is like having the best of both left and right ideals. It is a subset that remains closed under multiplication from both sides by any element of the algebra. This is central when you're dealing with quotients of algebra since a two-sided ideal lets you factor without losing structure.
To understand this easily, imagine a subset I of an algebra. For I to be a two-sided ideal, any combination with an element p from either side will still be in I. So,

• ap
• or pa where a is in the algebra

leaves the result in I.
In the given exercise, \( \mathcal{A} \mathbf{x} \cdot \mathcal{A} \) qualifies as a two-sided ideal. You take any element \( a \), \( b \), and \( c \) from \( \mathcal{A} \), and then form \( a\mathbf{x}cb \). This expression still resides within \( \mathcal{A} \mathbf{x} \cdot \mathcal{A} \), showing it's closed under multiplication from both sides. This property makes it a two-sided ideal, perfectly fitting well with the entire algebra.