Essential ideals I am trying to get my head around essential ideals. In literature I found 2 definitions:
An ideal $I$ in a C*-algebra $A$ is essential in $A$
(i) if $aI = 0$ implies $a=0$, $a\in A$; or
(ii) if every other non-zero ideal in $A$ has a non-zero intersection with $I$.
I'm trying to show that they are equivalent. While I was able to show (i) implies (ii) by proving the contrapositive, I'm struggling with showing (ii) implies (i). 
Also, is it in general true that if $I$ is an essential ideal in a C*-algebra $A$ and $J$ is an essential ideal in a C*-algebra $B$ then $I \otimes J$ is an essential ideal in $A \otimes B$, where $\otimes$ denotes the minimal tensor product, and how can I prove this?
I am very grateful to anyone who can help with these questions. 
 A: (disclaimer: "ideal", as is usual in C$^*$-algebra theory, means "closed, two-sided, ideal")
The key observation is that, for ideals $I,J$ in a C$^*$-algebra, we have $ I\cap J=IJ$ (see the proof at the end).
Now, assuming (ii), let $a\in A$ with $aI=0$. Let $J=\overline{AaA}$, the ideal generated by $a$. Then $J\cap I=JI=0$, so $J=0$. Then $a=0$.

Claim 1:

If $I\subset A$ is an ideal, then $I^2=I$.

Proof. Let $a\in I^+$. Then $a=(a^{1/2})^2\in I^2$. As $I$ and $I^2$ are C$^*$-algebras, they are spanned by their positive elements; so $$ I^2\subset I\subset I^2.$$
Claim 2:

If $I,J\subset A$ are ideals, then $IJ=I\cap J$.

Proof. We have $$I\cap J=(I\cap J)^2\subset IJ\subset I\cap J.$$
A: You can solve this in a purely algebraic way, but using slightly different definitions, and then showing that for $C^*$-algebras, these definitions agree with those in the OP.

Let $A$ be an algebra, and let $I$ be a left ideal.  Consider the conditions:

*

*for $x\in A$, we have $xI=\{0\} \implies x=0$.  (That is, $xa=0$ for all $a\in I$ means $x=0$).

*whenever $J$ is a right ideal in $A$ with $I\cap J=\{0\}$, then $J=\{0\}$.

*whenever $J$ is a two-sided ideal in $A$ with $I\cap J=\{0\}$, then $J=\{0\}$.

Always (1)$\implies$(2).  Indeed, if (1) holds, and $J$ is a right ideal with $I\cap J=\{0\}$, then given $b\in J$ and $a\in I$, we have $ba\in J$ as $J$ is a right ideal, and $ba\in I$ as $I$ is a left ideal, so $ba\in I\cap J$, so $ba=0$.   By (1), necessarily $b=0$.  So $J=\{0\}$, showing (2).
Clearly (2)$\implies$(3).
Now suppose that $I$ has a non-degenerate product meaning that for $a\in I$, if $a a'=0$ for all $a'\in I$, then $a=0$.
We claim that under this assumption, (3)$\implies$(1).  Let (3) hold, and let $J = \{ x\in A : xI=\{0\}\}$.  For $x\in J$ and $y,z\in A$, we have that $yxzI \subseteq yx I \subseteq y\{0\}=\{0\}$, as $I$ is a left ideal.  So $J$ is a two-sided ideal.  Let $x\in I\cap J$, so $xI=\{0\}$ so $xa'=0$ for all $a'\in I$.  As $I$ has non-degenerate product, $x=0$.  So $I\cap J=\{0\}$ so by (3), $J=\{0\}$, so (1) holds.

Now suppose $A$ is a $C^*$-algebra, and $I$ is an ideal (so two-sided, and closed).  Then (1) is condition (i) in the OP, and (3) is condition (ii) in the OP.  As $I$ is a $C^*$-algebra in its own right, it has an approximate identity, and so certainly has non-degenerate product, so we have shown that (i) and (ii) are equivalent.
In fact, even if $I$ is not closed, its closure still has an approximate identity, $(e_\alpha)$ say.  If $aa'=0$ for all $a'\in I$, then by continuity, also $aa'=0$ for all $a'$ in the closure of $I$, so $a=0$.  Thus $I$ has non-degenerate product.
