Proof that the ideals of $A \otimes B$ are ideals of $B$ when $A$ is a CSA I'm looking for your favorite treatment of the proof that if $A$ is a central simple algebra over $F$, then the two-sided ideals of $A \mathop{\otimes}\limits_F B$ are all just $A \mathop{\otimes}\limits_F I$ for $I$ a two-sided ideal of $B$.
Some of the proofs I've been reading have seemed like they could be stated more clearly.
Proofs that explain what they're doing and why are appreciated. In the proof below, I can follow the steps, but I have absolutely no intuition for it.

 A: I find the treatment in these notes by Artin to be much more clear. Let me provide in this answer the kind of intuition I was looking for:
Theorem: When $A$ is central simple, any $J \triangleleft A \mathop{\otimes}\limits_k B$ is $A \mathop{\otimes}\limits_k (B \cap J)$.
Proof: We know that $A \mathop{\otimes}\limits_k (B \cap J) \triangleleft J$; we show $J \triangleleft A \mathop{\otimes}\limits_k (B \cap J)$. Suppose for a second that we had a simple tensor $a \otimes b \in J$. If $1 \otimes b$ happens to be in $J$, then $a \otimes b \in A \mathop{\otimes}\limits_k (B \cap J)$ by definition. It turns out that this is true, by using simplicity of $A$.
Lemma: When $A$ is simple, any pure tensor in $J \triangleleft A \mathop{\otimes}\limits_k B$ is in $A \mathop{\otimes}\limits_k (B \cap J)$. 
Proof: If $a \otimes b$ is in $J$, with $a$ nonzero, then since $A$ is simple, there is some combination $\sum_i \alpha_i a \alpha'_i = 1$. Apply this sum product to $a \otimes b$ and get $1 \otimes b \in J$. QED
So we've proven the theorem when $j \in J$ is a pure tensor. Suppose now that $j$ has an expression
$$j = a_1 \otimes b_1 + a_2 \otimes b_2$$
and take $b_1$ and $b_2$ to be linearly independent and $a_1$ and $a_2$ nonzero. (I'm not really sure we need $b_1$ and $b_2$ linearly independent here, but it's good hygiene.)
There are two elements $j'$ and $j''$ that will be convenient for us to form. The element $j''$ won't always be defined; $j'$ is like a first try, and $j''$ is the second try, that exists if the first try fails. For $j'$, I take the combination
$$\sum \alpha_i a_1 \alpha_i' = 1,$$
and just apply it to all of $j$, and the other term $a_2 \otimes b_2$ just comes along for the ride.
$$j' = 1 \otimes b_1 + a'_2 \otimes b_2$$
I can form $j''$ if $a'_2$ doesn't commute with some element of $A$ (that is, if $a'_2 \not\in k)$, by taking $a \in A$ such that $[a_2', \alpha] \neq 0$ and then taking $[1 \otimes \alpha, j']$. This has one term fewer, and then I do my coefficient magic from the first part.
$$j'' = 1 \otimes b_2$$
Lemma: If either $j'$ or $j''$ is in $A \mathop{\otimes}\limits_k (B \cap J)$, then $j$ is. 
Proof: If $j' \in A \mathop{\otimes}\limits_k (B \cap J)$, then $j - a_1j'$ is a pure tensor in $J$, so it's in $A \mathop{\otimes}\limits_k (J \cap B)$, so $j= (j-a_1j') + a_1j'$ is an $A$ linear combination of things in $A \mathop{\otimes}\limits_k (B \cap J)$. If instead $j'' \in A \mathop{\otimes}\limits_k (B \cap J)$, then just replace $a_1j'$ with $a_2j''$ in the above sentence.
Lemma: Either $j'$ or $j''$ is in $A \mathop{\otimes}\limits_k (B \cap J)$. 
Proof: If $j''$ is defined, it is pure, and so we're done. If $j''$ isn't defined, it's because $a'_2$ is in $k$, and then $j'$ is in $J \cap B$ and so in $A \mathop{\otimes}\limits_k (J \cap B)$. QED
Now just to hammer it home, let's see how I'd do it with $j$ a sum of three terms, and we can see from there the method of induction.
$$j = a_1 \otimes b_1 + a_2 \otimes b_2 + a_3 \otimes b_3 
\\
j' = 1 \otimes b_1 + a_2' \otimes b_2 + a_3' \otimes b_3
\\
j'' = 1 \otimes b_2 + a_3'' \otimes b_3
$$
Lemma: If either $j'$ or $j''$ is in $A \mathop{\otimes}\limits_k (B \cap J)$, then $j$ is. 
Proof: If $j' \in A \mathop{\otimes}\limits_k (B \cap J)$, then $j - a_1j'$ is a rank two tensor in $J$, so it's in $A \mathop{\otimes}\limits_k (J \cap B)$, so $j= (j-a_1j') + a_1j'$ is an $A$ linear combination of things in $A \mathop{\otimes}\limits_k (B \cap J)$. If instead $j'' \in A \mathop{\otimes}\limits_k (B \cap J)$, then just replace $a_1j'$ with $a_2j''$ in the above sentence.
Lemma: Either $j'$ or $j''$ is in $A \mathop{\otimes}\limits_k (B \cap J)$. 
Proof: If $j''$ is defined, it is rank two, and so in $A \mathop{\otimes}\limits_k (B \cap J)$ by induction. If $j''$ isn't defined, it's because $a'_2$ is in $k$, and then $j'$ is rank two, and so in $A \mathop{\otimes}\limits_k (J \cap B)$. QED
