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What are the unit elements of $R/I$ where $I$ = {all cont. functions on $[0,1]$ |$ f(0) = f(1) = 0$} where $I$ is an Ideal w.r.t pointwise addition and multiplication.enter image description here I am little bit confused about it. Can you please help me out?

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    hii suomynonA thank you for editing. Do you know the answer? – 2017-01-19
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    No I is not Maximal ideal. Because it is not prime ideal. Take a cont. function on [0,1] which is not 0 except the point 0 and take another same on [0,1] which is not 0 except the point 1. – 2017-01-19
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    Hiii Gerry Myerson $R = C[0,1]$ all cont. function over $[0,1]$ – 2017-01-19

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Here's another way to picture things.

$A=\{f\in R\mid f(0)=0\}$ and $B=\{f\in R\mid f(1)=0\}$ are both maximal ideals of $R$. For example, the first one is the kernel of the epimorphism $\phi:R\to \mathbb R$ which maps $f\mapsto f(0)$. Your ideal $I$ is equal to $A\cap B$.

By the Chinese remainder theorem, $\frac{R}{I}\cong R/A\times R/B\cong \mathbb R\times \mathbb R$. It is easy to see that the only units of the rightmost ring are ordered pairs with both entries nonzero, and this corresponds to elements of the quotient which are nonzero at both $0$ and $1$ on $[0,1]$. This takes care of the title question.

Now for an $h$ which is nonzero on $0$ and $1$, if $h(c)=0$ for some $c\in (0,1)$, it is still a unit in the quotient, even though $h$ itself is not a unit in $R$. I think the method you're pursuing in the notes is to attempt to show that $h\equiv l \pmod I$ where $l$ is another thing in $R$ which you're sure is a unit (i.e. something which is nonzero everywhere on $[0,1]$) since you know that the homomorphic image of a unit is also a unit.

So the only real mistake above is saying that $h(x)+I$ does not have a multiplicative inverse in $R/I$. It does have a multiplicative inverse: $l(x)+I$'s inverse. Of course $h$ doesn't have an inverse in $R$, but that is a separate matter.

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There is no problem, and this is a simple illustration of the situation you describe: the element $2\in\Bbb{Z}$ doesn't have a multiplicative inverse, but in the quotient $\Bbb{Z}/3\Bbb{Z}$ the coset $2+(3)$ is a unit.

Another example is the polynomial ring $\Bbb{R}[x]$ and the quotient $A=\Bbb{R}[x]/(x^2-1)$; the polynomial $x^2$ is clearly not invertible, but in $A$, $x^2$ is an element of the same coset as $1$.

Speaking about this more generally, we require some non-units to be units when mapped into a quotient ring, especially a quotient by a maximal ideal.