1
$\begingroup$

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.

enter image description here

  • 1
    That sort of argument is standard in the theory of CSAs. You use it to prove that the tensor product of CSAs is a CSAa, and what not. I'd suggest you keep reading on the subject and then the argument will grow to be natural to you. Asking for "intuition" is sometimes a misguided thing to do: this proof, as many others, are what you should *base* your intuition, rather. – 2017-01-01
  • 0
    You don't (I guess) ask for intuition on how to ride a bike, you do it. this is sort of the same thing. – 2017-01-01
  • 0
    @MarianoSuárez-Álvarez Can you describe the structure of the argument in your own words, to help me get a sense of the generalization? – 2017-01-01
  • 0
    Your image does not have a complete argument (you left out the part where the meaning of the notation is described, for example, so what I and J are, and what is the relation between them is anyone's guess...) so no. But the general idea is always the same: you want to show that something is zero in the tensor product, so you pick a possible counterexample of minimal rank. This minimality gives you in some way or another information about the coefficients, so that you can used your hypothesis: in this case, that those coefficients are central. – 2017-01-01
  • 0
    You do not need to get a sense of any generalization. Just make sure that you understand why this worka and go on. If this sort of argument is sufficiently useful, you'll re- counter it in other times, and in other contexts, and then by noticing that you''ll grow to see a pattern and *then* something that you can call "Intuition" if you really must will outgrow. – 2017-01-01
  • 0
    Also, do exercises in that section. Somewhere the book (you did not tell us which book, which is always bad practice!) proves or suggests as an exercise that the tensor product of a CAS and another algebra has center equal to the center of the second factor. Try to do that, or read the proof if it is there. – 2017-01-01
  • 1
    The preoccupation with acquiring intuition is counterproductive, mostly because intuition is the result of a long process. Expecting to get there the first time one encounters something (like an algebraic argument) is quite weird. What people call Intuition is in an immense majority of cases having get used to something — naturally, that does not occur on first sight! – 2017-01-01
  • 0
    @MarianoSuárez-Álvarez Maybe "intuition" wasn't the most precise word. I want to get the "story" of the proof, so I can remember it at all! As it was, it just seemed like a bunch of steps, pulled from nowhere. – 2017-01-02
  • 0
    @MarianoSuárez-Álvarez Below find the kind of explanation I was looking for. I continue to think it's valuable. – 2017-02-13

1 Answers 1

0

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