### $I = \{f \in C[0,1] : f(\alpha) = 0\}$ is a maximal ideal in $C[0,1]$ (NBHM) (CSIR)

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 $$(fg)(x) = f(x)g(x)..$$ Fix $\alpha \in [0,1]$ and define $I = \{f \in C[0,1] : f(\alpha) = 0 \}$.
Prove that $I$ is a maximal ideal in $C[0,1]$.
Solution:
Clam: $I$ is an ideal in $C[0,1]$
Let $f,g \in I$, then $f(\alpha) = 0 = g(\alpha)$. Now, $$(f+g)(\alpha) = f(\alpha) + g(\alpha) = 0 + 0 = 0$$
Therefore $I$ is closed under addition.
Let $f \in C[0,1]$ and $g \in I$. Now, $g \in I$ implies that $g(\alpha) = 0$.  We claim that $fg \in I$.
$$(fg)(\alpha) = f(\alpha) \cdot g(\alpha) = f(\alpha) \cdot 0 = 0$$
Therefor $I$ is an ideal in $C[0,1]$.

Clam: $I$ is a maximal ideal.
An ideal $I$ is said to be a maximal if whenever there exists an ideal $J$ such that $I \subseteq J \subseteq C[0,1]$ we have either $J = I$ or $J = C[0,1]$.
There is a characterization for the maximal ideals which we will use here.
Result: Let $R$ be a commutative ring with 1 and $I$ be an ideal in $R$ then $I$ is maximal in $R$ if and only if the quotient $\frac{R}{I}$ is a field.
We prove that $\frac{C[0,1]}{I} \cong \Bbb{R}$ which is a field and proof follows.

We will construct an onto ring homomorphism from $C[0,1]$ to $\Bbb{R}$ whose kernel $I$, then by the fundamental theorem of a ring homomorphism, we have $\frac{C[0,1]}{I} \cong \Bbb R$

Define a ring homomorphism $\phi: C[0,1] \to \Bbb{R}$ by $\phi(f) = f(\alpha)$ (the evaluation map). Now, $$\phi(f+g) = (f+g)(\alpha) = f(\alpha) + g(\alpha) = \phi(f) + \phi(g)$$ Similarly
$$\phi(fg) = (fg)(\alpha) = f(\alpha) \cdot g(\alpha) = \phi(f) \cdot \phi(g)$$
Therefore $\phi$ is a ring homomorphism.

We will calculate the Kernal of $\phi$:
Ker $\phi = \{f \in C[0,1]: \phi(f) = 0 \} = \{f \in C[0,1]: f(\alpha) = 0 \} = I$.

We left with verifying onto for $\phi$. Let $a$ be a real number. We have to find $f \in C[0,1]$ such that $\phi(f) = a$. But $\phi(f) = f(\alpha)$. So we can take $f$ to be the linear polynomial $f = x + (a - \alpha)$ (which has constant term $a - \alpha$). Now, for this $f$, $\phi(f) = f(\alpha) = \alpha + (a - \alpha) = a$ and the proof is complete.

### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft...