Ideals in algebras I need some hints to solve the next problem:

Let $A,B$ be algebras over a field $K$. I want to show that $I \subseteq A$ is an ideal if there exists $\rho :A\longrightarrow B$ an algebra morphism such that $\ker(\rho) = I$.

Thanks.

The claim is false, as stated. The correct statement should be:

$I\subseteq A$ is an ideal in the $K$-algebra $A$ if and only if there exist a $K$-algebra $B$ and an algebra morphism $\rho\colon A\to B$ such that $I=\ker\rho$.

One direction is easy. For the other one, can you give an algebra structure to $A/I$ (quotient group with respect to the addition)?
The statement cannot hold for fixed $A$ and $B$ for the simple reason that there can be no algebra morphism from $A$ to $B$ (for example, take $A$ the $2\times 2$ matrices over $K$ and $B=K$), if algebra morphisms are supposed to preserve the multiplicative identity (there can be only the zero morphism in the other case).
