Let $C[0,1]$ be the set of all continuous functions from $[0,1]$ to $\Bbb{R}$ which is a commutative ring with 1 with respect to the pointwise addition and multiplication: $$(f+g)(x) = f(x)+g(x)$$ and $$(fg)(x) = f(x)g(x).$$ Define $I = \{f \in C[0,1] : f(\frac{1}{3}) = f(\frac{1}{2}) = 0 \}$. (

Prove that $I$ is an ideal which is neither prime nor maximal.

Let $f,g \in I$, then $f(\frac{1}{3}) = f(\frac{1}{2}) = 0 = g(\frac{1}{3}) = g(\frac{1}{2}) $. Now, $$(f+g)(\frac{1}{3}) = f(\frac{1}{3}) + g(\frac{1}{3}) = 0 + 0 = 0$$ Similarly

$$(f+g)(\frac{1}{2}) = f(\frac{1}{2}) + g(\frac{1}{2}) = 0 + 0 = 0$$

Therefore $I$ is closed under addition.

Let $f \in C[0,1]$ and $g \in I$. $g \in I$ implies that $g(\frac{1}{3}) = g(\frac{1}{2}) = 0$. We claim that $fg \in I$.

$$(fg)(\frac{1}{3}) = f(\frac{1}{3}) \cdot g(\frac{1}{3}) = f(\frac{1}{3}) \cdot 0 = 0$$ Similarly

$$(fg)(\frac{1}{2}) = f(\frac{1}{2}) \cdot g(\frac{1}{2}) = f(\frac{1}{2}) \cdot 0 = 0.$$

Therefor $I$ is an ideal in $C[0,1]$.

We will show that this is not a prime ideal and hence it cannot be maximal also.

An ideal $I$ is said to be a prime ideal if $fg \in I$ implies either $f \in I$ or $g \in I$.

So to prove an ideal is not prime, we need to find $fg \in I$ such that $f \notin I$ and $g \notin I$.

Consider the polynomials $f = x - \frac{1}{3}$, $g = x - \frac{1}{2}$ and $fg = (x - \frac{1}{3})(x - \frac{1}{2})$. Polynomials are continuous and hence $f,g,fg \in C[0,1]$. Note that

$$(fg)(\frac{1}{3}) = 0 \text{ and } (fg)(\frac{1}{2})= 0.$$

Therefore $fg \in I$. But $f(\frac{1}{2}) \ne 0$ and $g(\frac{1}{3}) \ne 0$. Hence neither $f$ nor $g$ in $I$. This proves the claim.

**Note that $\frac{1}{3}$ and $\frac{1}{2}$ can be replaced with any two distinct numbers in $[0,1]$ and the result still holds true**)Prove that $I$ is an ideal which is neither prime nor maximal.

**Solution**:**Clam: $I$ is an ideal in $C[0,1]$**Let $f,g \in I$, then $f(\frac{1}{3}) = f(\frac{1}{2}) = 0 = g(\frac{1}{3}) = g(\frac{1}{2}) $. Now, $$(f+g)(\frac{1}{3}) = f(\frac{1}{3}) + g(\frac{1}{3}) = 0 + 0 = 0$$ Similarly

$$(f+g)(\frac{1}{2}) = f(\frac{1}{2}) + g(\frac{1}{2}) = 0 + 0 = 0$$

Therefore $I$ is closed under addition.

Let $f \in C[0,1]$ and $g \in I$. $g \in I$ implies that $g(\frac{1}{3}) = g(\frac{1}{2}) = 0$. We claim that $fg \in I$.

$$(fg)(\frac{1}{3}) = f(\frac{1}{3}) \cdot g(\frac{1}{3}) = f(\frac{1}{3}) \cdot 0 = 0$$ Similarly

$$(fg)(\frac{1}{2}) = f(\frac{1}{2}) \cdot g(\frac{1}{2}) = f(\frac{1}{2}) \cdot 0 = 0.$$

Therefor $I$ is an ideal in $C[0,1]$.

We will show that this is not a prime ideal and hence it cannot be maximal also.

**Clam: $I$ is not a prime ideal.**An ideal $I$ is said to be a prime ideal if $fg \in I$ implies either $f \in I$ or $g \in I$.

So to prove an ideal is not prime, we need to find $fg \in I$ such that $f \notin I$ and $g \notin I$.

Consider the polynomials $f = x - \frac{1}{3}$, $g = x - \frac{1}{2}$ and $fg = (x - \frac{1}{3})(x - \frac{1}{2})$. Polynomials are continuous and hence $f,g,fg \in C[0,1]$. Note that

$$(fg)(\frac{1}{3}) = 0 \text{ and } (fg)(\frac{1}{2})= 0.$$

Therefore $fg \in I$. But $f(\frac{1}{2}) \ne 0$ and $g(\frac{1}{3}) \ne 0$. Hence neither $f$ nor $g$ in $I$. This proves the claim.

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