Quotient Modules of a polynomial ring Let $R = K[x], K$ a field. Define for $a \space \epsilon \space K$ the ideal $ I_a := (x-a)$ in $K[x]$ and see $I_a$ as an $R$ module. Using that for $ a \space \epsilon \space K$ and  $ b \space \epsilon \space K$, $I_a$ and $I_b$ are isomorphic, prove that the quotient modules $R/I_a$ and $R/I_b$ are not isomorphic as $R$ -modules.
How should I prove this? I have tried doing it with the first isomorphism theorem for modules, constructing a function $ \theta $  with $ ker(\theta) = I_a$ but I don't seem to get anywhere. Does anyone know a proper solution to this question? Thanks.

Suppose you have a isomorphism $f:R/I_b \to R/I_a$ as $R$ modules. Observe that $x=b$ in $R/I_b$ and $x=a$ in $R/I_a$. Then you have 
$$ bf(1)=f(b)=f(x)=xf(1)=af(1) $$
Now recall that the module structure in $R/I_a$ comes from the projection map $\pi$. Now by assumption $a\neq b$. And this means $(a-b)f(1)=0$ implies that $f(1)\in I_a$ since $a-b \notin (x-a)=ker(\pi)$ but this means $f(1)=0 $ in $R/I_a$ i.e. $f(1)\in ker(f)$ contradicting $f$ being an isomorphism. 
