If $K[X]/I \cong K[X]/J$ with $I \subset J$ then $I=J$? Let $K$ be a field and $I,J$ are ideals of $K[X].$ Then if $K[X]/I \cong K[X]/J$     with $I \subset J$ then can one say that $I=J$ ?
 A: Yes.  
First, note that $K[X]/I$ has infinite $K$-dimension if and only if $I=(0)$.  So the result is true if either of $I$ or $J$ is $(0)$.
Assume now that $I$ and $J$ are both non-zero.  Recall that $K[X]$ is a principal ideal domain.  Hence there are polynomials $P$ and $Q$ such that $I=(P)$ and $J=(Q)$.  Then the $K$-dimension of $K[X]/I$ (or $K[X]/J$) is the degree of $P$ (or $Q$, respectively).  Thus $P$ and $Q$ have the same degree.  Finally, $I\subset J$ implies that $P$ is divisible by $Q$.  Thus $P$ and $Q$ are equal up to a non-zero scalar factor, and so $I=J$.
A: Let $R$ be any Noetherian ring and let $I\subseteq J\subseteq R$ be ideals such that $R/I$ and $R/J$ are isomorphic as rings.  Then $I=J$.
To prove this, we may first replace $R$ by $R/I$, and so we may assume $I=0$.  We then have a ring-isomorphism $R/J\to R$ for some ideal $J$ and want to show $J=0$.  Equivalently, we have a surjective ring-homomorphism $f:R\to R$ with kernel $J$, and want to show $J=0$.
So we are reduced to showing that if $R$ is a Noetherian ring and $f:R\to R$ is a surjective ring-homomorphism, then $f$ is injective.  To prove this, note that if $\ker f\neq 0$, then the inclusions $$0\subset\ker f\subset \ker f^2\subset\ker f^3\subset\dots$$ are all strict.  Indeed, each of these inclusions is obtained by applying $f^{-1}$ to the previous inclusion, and $f^{-1}$ preserves strict inclusions since $f$ is surjective.  So this is an infinite ascending chain of ideals in $R$, which contradicts the assumption that $R$ is Noetherian.

Alternatively, in the case $R=K[X]$, you can give a proof in the spirit of the other answers which used $K$-dimension (assuming the isomorphism was an isomorphism of $K$-algebras) which works assuming just a ring-isomorphism.  The trick is to use length of modules as a substitute for $K$-dimension.
First, note that the ring $K[X]/I$ is artinian iff $I\neq 0$, so if $K[X]/I\cong K[X]/J$ then $I=0$ iff $J=0$.  So let us assume $I,J\neq 0$.
In this case, $K[X]/J$ is a quotient of $K[X]/I$, and both are artinian rings.  Note that the length of $K[X]/J$ as a module over itself is the same as its length as a $K[X]/I$-module, since $K[X]/I$-submodules of $K[X]/J$ are the same as $K[X]/J$-submodules.  If $J\neq I$ so $K[X]/J$ is a proper quotient of $K[X]/I$, then $K[X]/J$ has smaller length than $K[X]/I$ as a $K[X]/I$-module, and hence the length of $K[X]/J$ as a module over itself is less than the length of $K[X]/I$ as a module over itself.  Thus the rings $K[X]/I$ and $K[X]/J$ cannot be isomorphic.
A: Hint: As Eric Wofsey mentioned, this proof only works if $\cong$ means "isomorphic as $K$-Algebras".
If $I\subset J$ we have 
$$(K[X]/I)/(J/I)\cong K[X]/J$$
So we get in the given situation
$$(K[X]/I)/(J/I)\cong K[X]/I$$
Since $K[X]/I$ is a finite dimensional vector space, this is only possible for $J/I=\{0\}$ i.e. $I=J$.
