$R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective, then is $R$ Noetherian? Let $R$ be a commutative ring with unity such that every surjective ring homomorphism $f:R\to R$ is injective, then is $R$ a Noetherian ring ? 
 A: Here's a counterexample.  Let $R=\mathbb{Z}[2^{1/3},2^{1/9},2^{1/27},\dots]\subset\mathbb{R}$.  Then $R$ is not Noetherian, since the ideal $(2^{1/3},2^{1/9},2^{1/27},\dots)$ is not finitely generated.  But the only ring-homomorphism $f:R\to R$ is the identity.  Indeed, any such $f$ must be the identity on $\mathbb{Z}$, and then $f\left(2^{1/3^n}\right)$ must be $2^{1/3^n}$ for each $n$ since $2^{1/3^n}$ is the only $3^n$th root of $2$ in $R$.

More generally, any non-Noetherian subring $R$ of $\overline{\mathbb{Q}}$ works.  Indeed, if $a\in R$, then it is a root of a nonzero polynomial $p(x)$ with integer coefficients.  If $a\neq 0$, $p(x)$ can be taken to have nonzero constant term.  The constant term of $p(x)$ is then in the ideal generated by $a$, since $p(a)=0$ and every other term of $p(a)$ is in the ideal generated by $a$.  Thus every nontrivial ideal in $R$ contains a nonzero integer.  It follows that any homomorphism $f:R\to R$ must be injective.
(In fact, any homomorphism $f:R\to R$ must also be surjective: it must send roots of any $p(x)\in\mathbb{Z}[x]$ to roots of $p(x)$, and there are only finitely many such roots.  An injection from a finite set to itself is surjective, so the image of $f$ must contain all the roots of $p(x)$ in $R$.)
