Problem 33 - A Computational Introduction to Number Theory and Algebra

Question

Let \(R\) be a ring, and fix \(x_{1}, \ldots, x_{n} \in R\). Let $$ I:=\left\\{g \in R\left[X_{1}, \ldots, X_{n}\right]: g\left(x_{1}, \ldots, x_{n}\right)=0\right\\} $$ Show that \(I\) is an ideal of \(R\left[X_{1}, \ldots, X_{n}\right],\) and that \(I=\left(X_{1}-x_{1}, \ldots, X_{n}-x_{n}\right)\)

Step-by-Step Solution

Answer

Question: Show that the set \(I = \{f \in R[X_1, \ldots, X_n] \mid f(x_1, \ldots, x_n) = 0\}\) is an ideal of the polynomial ring \(R[X_1, \ldots, X_n]\), and that \(I\) is equal to the set generated by the polynomials \((X_1 - x_1, \ldots, X_n - x_n)\). Solution: We showed that \(I\) is nonempty, closed under addition, and closed under multiplication, thus proving that \(I\) is an ideal of the polynomial ring \(R[X_1, \ldots, X_n]\). By demonstrating that \(I\) contains all polynomials \((X_i - x_i)\) and that any polynomial in \(I\) is generated by \((X_1 - x_1, \ldots, X_n - x_n)\), we confirmed the equality of the two sets.

Step 1: Show that \(I\) is nonempty

To show that \(I\) is nonempty, we must provide atleast one element that belongs to \(I\). The zero polynomial, denoted by \(0\), is always an element of any polynomial ring. By substituting the values \(x_1, \ldots, x_n\) into the zero polynomial, we get \(0(x_1, \ldots, x_n) = 0\). Therefore, \(0 \in I\) and \(I\) is nonempty.

Step 2: Closed under addition

To show that \(I\) is closed under addition, let \(f, g \in I\). This means that \(f(x_1, \ldots, x_n) = 0\) and \(g(x_1, \ldots, x_n) = 0\). We must show that the sum \(f + g\) also belongs to \(I\). When we substitute the values \(x_1, \ldots, x_n\) into the sum, we get \((f + g)(x_1, \ldots, x_n) = f(x_1, \ldots, x_n) + g(x_1, \ldots, x_n) = 0 + 0 = 0\). Thus, \((f + g) \in I\) and \(I\) is closed under addition.

Step 3: Closed under multiplication

To show that \(I\) is closed under multiplication by any element of the polynomial ring, let \(f \in I\) and \(h \in R[X_1, \ldots, X_n]\). We must show that \(hf \in I\). When we substitute the values \(x_1, \ldots, x_n\) into the product, we get \((hf)(x_1, \ldots, x_n) = h(x_1, \ldots, x_n) \cdot f(x_1, \ldots, x_n) = h(x_1, \ldots, x_n) \cdot 0 = 0\). Thus, \((hf) \in I\) and \(I\) is closed under multiplication by any element of the given ring.

Step 4: \(I\) contains all \((X_i - x_i)\)

To show that \(I\) contains all polynomials \((X_i - x_i)\), we will substitute the values \(x_1, \ldots, x_n\) into each polynomial \((X_i - x_i)\), for \(1 \leq i \leq n\). For each polynomial, we get \((X_i - x_i)(x_1, \ldots, x_n) = x_i - x_i = 0\). Therefore, each \((X_i - x_i) \in I\).

Step 5: Any polynomial in \(I\) is generated by \((X_1 - x_1, \ldots, X_n - x_n)\)

To show that any polynomial in \(I\) is generated by \((X_1 - x_1, \ldots, X_n - x_n)\), let \(f \in I\). Since \(f(x_1, \ldots, x_n) = 0\), using polynomial division, we can write \(f\) as: $$ f(X_1, \ldots, X_n) = h(X_1, \ldots, X_n) (X_1 - x_1) + r_1(X_1, \ldots, X_n) $$ where \(h, r_1\) are polynomials and \(r_1\) doesn't have \(X_1\) as a term. Now, for \(r_1 \not\equiv 0\), we can repeat the process using \(X_2 - x_2\), ..., \(X_n - x_n\), we get: $$ r_{n-1}(X_1, \ldots, X_n) = h_n(X_1, \ldots, X_n) (X_n - x_n) + r_n(X_1, \ldots, X_n) $$ Hence, every \(f \in I\) can be expressed as a combination of elements \((X_i - x_i)\). Thus, \(I\) is generated by the set \((X_1 - x_1, \ldots, X_n - x_n)\). In conclusion, we have shown that \(I\) is an ideal of \(R[X_1, \ldots, X_n]\), and that it is equal to the set generated by the polynomials \((X_1 - x_1, \ldots, X_n - x_n)\).

Key Concepts

Ring Theory

Ring theory is a branch of abstract algebra that studies rings—algebraic structures equipped with two binary operations resembling addition and multiplication. A ring \( R \) is defined as a set along with two operations: addition (\( + \)) and multiplication (\( \cdot \)), which satisfy the axiomatic properties associated with commutativity, associativity, distributive law, and the presence of additive identity (0) and multiplicative identity (1, in a commutative ring).

Understanding ring theory is fundamental in deciphering the behaviors of polynomials, integers, and various algebraic systems when operations are applied to them. For example, when examining the set \( I \) in the exercise, proving that it is an ideal involves checking if it adheres to the axioms that define ring structure—namely, being closed under addition and multiplication.

Polynomial Ring

A polynomial ring is a collection of polynomials formed over a certain ring \( R \) with coefficients from that ring. It is denoted as \( R[X_1, \ldots, X_n] \) for polynomials in \( n \) indeterminates. This set becomes a ring under the usual polynomial addition and multiplication.

In the context of our exercise, \( R[X_1, \ldots, X_n] \) represents a polynomial ring where the operations of addition and multiplication are consistent with the operations of the coefficient ring \( R \). Each \( X_i \) acts as a placeholder for variables, and polynomials in this ring can be manipulated using algebraic rules to simplify expressions and solve equations.

Zero Polynomial

The zero polynomial is a crucial concept in both ring theory and polynomial rings. It plays the role of the additive identity, which means its addition to any polynomial in a given ring does not change the value of that polynomial. The zero polynomial is universally denoted by \( 0 \) and has the unique property that its value is zero regardless of the substitution for the variables.

As seen in Step 1 of our solution, the inclusion of the zero polynomial in set \( I \) confirms its non-emptiness and validates the structure of \( I \) as consistent with the requisite properties of a polynomial ring.

Closed Under Addition

For a set within a ring to be an ideal, it must be 'closed under addition.' This means that for any two elements \( f, g \) in the set, the sum \( f + g \) is also within the set. This closure property ensures that the set maintains a consistent internal structure under the addition operation.

In the exercise, we showed that taking any two polynomials in \( I \) and adding them yields another polynomial in \( I \) (as detailed in Step 2), thereby satisfying the 'closed under addition' criterion necessary for \( I \) to be an ideal of the ring.

Closed Under Multiplication

Similar to closure under addition, a set being 'closed under multiplication' is another requirement for it to be an ideal of a ring. This means that multiplying any element of the set by any element of the ring results in an expression that still belongs to the set. It is what makes an ideal an integral part of ring theory since it ties the set closely to the ring's structure.

The exercise demonstrates this property in Step 3 by showing that the product of any polynomial \( h \) from \( R[X_1, \ldots, X_n] \) and any polynomial \( f \) in the set \( I \) results in a polynomial that also belongs to \( I \). This adherence to ring multiplication rules is part of what defines \( I \) as an ideal.