The question asks us to determine the properties of the set \( A = \{f \in C[0, 1] \mid f\left( \frac{1}{4}\right)=f\left( \frac{3}{4}\right) = 0\} \) within the ring \( C[0, 1] \), which is the ring of all real-valued continuous functions defined on the closed interval \( [0, 1] \). We need to check if A is an ideal, a prime ideal, or a maximal ideal in \( C[0, 1] \).

Ideal Property of A

To show that A is an ideal in \( C[0, 1] \), we need to verify two conditions:

1. A is a non-empty subset of \( C[0, 1] \) and is closed under subtraction.
2. For any element \( f \in A \) and any element \( h \in C[0, 1] \), the product \( hf \) is in A (closure under multiplication by arbitrary ring elements).

• A is clearly non-empty since the zero function \( z(x) = 0 \) for all \( x \in [0, 1] \) is continuous and satisfies \( z(1/4) = 0 \) and \( z(3/4) = 0 \), so \( z \in A \).
• Let \( f, g \in A \). By definition of A, \( f(1/4) = 0 \), \( f(3/4) = 0 \), \( g(1/4) = 0 \), and \( g(3/4) = 0 \). Consider the function \( f - g \). It is also continuous since f and g are continuous. Evaluate \( (f-g)(1/4) = f(1/4) - g(1/4) = 0 - 0 = 0 \). Evaluate \( (f-g)(3/4) = f(3/4) - g(3/4) = 0 - 0 = 0 \). Since \( (f-g)(1/4) = 0 \) and \( (f-g)(3/4) = 0 \), the function \( f - g \) belongs to A. Thus, A is closed under subtraction.
• Let \( f \in A \) and \( h \in C[0, 1] \). By definition of A, \( f(1/4) = 0 \) and \( f(3/4) = 0 \). The function \( h \) is any continuous function on \( [0, 1] \). Consider the product function \( hf \). It is continuous since h and f are continuous. Evaluate \( (hf)(1/4) = h(1/4)f(1/4) \). Since \( f(1/4) = 0 \), \( (hf)(1/4) = h(1/4) \cdot 0 = 0 \). Evaluate \( (hf)(3/4) = h(3/4)f(3/4) \). Since \( f(3/4) = 0 \), \( (hf)(3/4) = h(3/4) \cdot 0 = 0 \). Since \( (hf)(1/4) = 0 \) and \( (hf)(3/4) = 0 \), the function \( hf \) belongs to A. Thus, A is closed under multiplication by elements of \( C[0, 1] \).

Both conditions are met, so A is an ideal in \( C[0, 1] \).

Prime Ideal Analysis for A

An ideal P in a commutative ring R is called a prime ideal if P is a proper ideal and for any \( a, b \in R \), if \( ab \in P \), then either \( a \in P \) or \( b \in P \). In our case, R = \( C[0, 1] \) and P = A.

We need to check if for any \( f, g \in C[0, 1] \), if \( fg \in A \), does it imply \( f \in A \) or \( g \in A \)?

\( fg \in A \) means \( (fg)(1/4) = 0 \) and \( (fg)(3/4) = 0 \). This is equivalent to \( f(1/4)g(1/4) = 0 \) and \( f(3/4)g(3/4) = 0 \). For A to be prime, whenever \( f(1/4)g(1/4) = 0 \) and \( f(3/4)g(3/4) = 0 \), it must follow that either (\( f(1/4) = 0 \) and \( f(3/4) = 0 \)) or (\( g(1/4) = 0 \) and \( g(3/4) = 0 \)).

Let's try to find a counterexample. Can we find continuous functions f and g such that \( fg \in A \), but \( f \notin A \) and \( g \notin A \)?

• \( f \notin A \) means either \( f(1/4) \neq 0 \) or \( f(3/4) \neq 0 \).
• \( g \notin A \) means either \( g(1/4) \neq 0 \) or \( g(3/4) \neq 0 \).
• We need \( f(1/4)g(1/4) = 0 \) and \( f(3/4)g(3/4) = 0 \).

Consider functions f and g constructed such that:

• \( f(1/4) \neq 0 \) but \( f(3/4) = 0 \).
• \( g(1/4) = 0 \) but \( g(3/4) \neq 0 \).

For example, we can construct continuous functions using piecewise linear definitions or smooth bump functions. Let's consider simple functions:

• Let f be a continuous function such that \( f(1/4) = 1 \) and \( f(3/4) = 0 \). (e.g., a function that peaks at 1/4 and is 0 at 3/4 and beyond, and 0 before 1/4).
• Let g be a continuous function such that \( g(1/4) = 0 \) and \( g(3/4) = 1 \). (e.g., a function that is 0 before 3/4 and at 1/4, and peaks at 3/4).

With such f and g:

• \( f \notin A \) because \( f(1/4) = 1 \neq 0 \).
• \( g \notin A \) because \( g(3/4) = 1 \neq 0 \).
• \( (fg)(1/4) = f(1/4)g(1/4) = 1 \cdot 0 = 0 \).
• \( (fg)(3/4) = f(3/4)g(3/4) = 0 \cdot 1 = 0 \).
• So, \( fg \in A \).

Since we found functions f and g such that \( fg \in A \) but \( f \notin A \) and \( g \notin A \), A is not a prime ideal.

Maximal Ideal Test for A

An ideal M in a commutative ring R is maximal if M is a proper ideal and there is no proper ideal P of R such that \( M \subset P \subset R \). In \( C[0, 1] \), the maximal ideals are precisely those of the form \( M_c = \{f \in C[0, 1] \mid f(c) = 0\} \) for some \( c \in [0, 1] \).

Our ideal A is defined by functions being zero at two distinct points, \( 1/4 \) and \( 3/4 \).

Let \( M_{1/4} = \{f \in C[0, 1] \mid f(1/4) = 0\} \). \( M_{1/4} \) is a maximal ideal in \( C[0, 1] \).

Let \( M_{3/4} = \{f \in C[0, 1] \mid f(3/4) = 0\} \). \( M_{3/4} \) is a maximal ideal in \( C[0, 1] \).

The ideal A can be expressed as the intersection of these two maximal ideals:

\( A = \{f \in C[0, 1] \mid f(1/4) = 0 \text{ and } f(3/4) = 0\} = \{f \in C[0, 1] \mid f(1/4) = 0\} \cap \{f \in C[0, 1] \mid f(3/4) = 0\} = M_{1/4} \cap M_{3/4} \).

Since \( 1/4 \neq 3/4 \), \( M_{1/4} \) and \( M_{3/4} \) are distinct maximal ideals. For example, a function that is 0 at 1/4 but 1 at 3/4 is in \( M_{1/4} \) but not in \( M_{3/4} \). Similarly, a function that is 1 at 1/4 but 0 at 3/4 is in \( M_{3/4} \) but not in \( M_{1/4} \).

The intersection of two distinct maximal ideals is an ideal that is properly contained in each of the maximal ideals. For instance, A is properly contained in \( M_{1/4} \) because any function in \( M_{1/4} \) that is non-zero at 3/4 belongs to \( M_{1/4} \) but not to A. Since A is properly contained in the proper ideal \( M_{1/4} \), A cannot be a maximal ideal itself.

Alternatively, an ideal P is maximal if and only if the quotient ring \( R/P \) is a field. An ideal P is prime if and only if the quotient ring \( R/P \) is an integral domain. Since A is not a prime ideal, \( C[0, 1]/A \) is not an integral domain, and therefore not a field. Thus, A is not a maximal ideal.

Conclusion

Based on our analysis, A is an ideal in \( C[0, 1] \), but it is neither a prime ideal nor a maximal ideal.

Therefore, the statement that A is an ideal in \( C[0, 1] \) but is not a prime ideal in \( C[0, 1] \) is true.