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Pick out the true statement(s):

(a) The set of all $2\times 2$ matrices with rational entries (with the usual operations of matrix addition and matrix multiplication) is a ring which has no nontrivial ideals.

(b) Let $R = C[0, 1]$ be considered as a ring with the usual operations of pointwise addition and pointwise multiplication. Let $I = \{f : [0, 1] → R \mid f(1/2) = 0\}$. Then $I$ is a maximal ideal.

(c) Let $R$ be a commutative ring and let $P$ be a prime ideal of $R$. Then $R/P$ is an integral domain.

obviously (c) is true but no idea about (a) & (b)

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I think you're using $\,R\,$ in (b) with two different meanings. Perhaps you meant $\,\Bbb R\,$ for the $\,R\,$ in the definition of $\,I\,$ ...? – DonAntonio Sep 17 '12 at 2:58

For (b): Define a ring homomorphism $\varphi : C[0,1]\to \Bbb{R}$, $\varphi$ sends a continuous function $f(x)$ to its real value at $1/2$. Now $\varphi$ is a surjective because to any real number $z$ we can associate to it the constant function $f(x) = z$, and then $f(1/2) = z$. The kernel of $\varphi$ is precisely your $I$ and by the first isomorphism theorem we conclude that

$$C[0,1]/I \cong \Bbb{R}$$

from which it follows that $I$ is maximal.

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(a) (Corrected.) This is false true. Recall that row reduction can be carried out by multiplication by matrices in the ring. Thus, if $I$ is an ideal, and $A\in I$ is not the zero matrix, $I$ must contain either the identity matrix (if $A$ is non-singular) or the matrix $\begin{bmatrix}1&0\\0&0\end{bmatrix}$ (if $A$ is singular). But in the latter case multiplication by the matrices corresponding to elementary row and column operations can then be used to show that the identity matrix is in $I$ anyway.

(b) Suppose that $f\in R\setminus I$ and $J$ is an ideal of $R$ such that $I\cup\{f\}\subseteq J$. Then $f^2\in J$ (why?), and $g(x)=\left|x-\frac12\right|\in J$ (why?), so $f^2+g\in J$. Show that $\left(f^2+g\right)^{-1}\in R$ and hence that the constant $1$ function is in $J$, and use that to show that $J=R$.

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Actually (a) is true. Assuming there is an ideal $I$ with a nonzero element $A$ you can apply row/column operations to $A$ and the fact that $I$ is closed under addition, to deduce that the identity is in $I$. – Mikko Korhonen Sep 17 '12 at 6:59
@m.k.: You’re right; I forgot all about the elementary matrices. – Brian M. Scott Sep 17 '12 at 7:00

Option $C$ is also not true. Because it is not mentioned that $R$ is ring with identity, If $R$ is a ring without identity then $R/P$ is also ring without identity and hence can't be integral domain (Integral domain is commutative ring with identity which has no zero divisor other then zero.

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Hints for (b):

1) Prove $\,f\,$ is onto

2) What is then $\,R/I\,$?

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