Is There a Typo in the Refinement of Schanuel's Lemma in Passman's Book?

[Sudharaka]

Hi everyone, :)

Here's a doubt that I came to my mind when reading A Course in Ring Theory by Passman.

On Chapter 8 (Projective Dimension) it states the Schanuel's Lemma;

And then it gives a refinement of Schanuel's Lemma as follows:

where the equivalence relation ~ is defined as follows:

​

I think that the proof of the refinement of Schanuel's lemma has a typo in it. I have drawn a red circle to highlight where the probable mistake is. I think instead of zero it should be $Q$ and $Q'$. That is the sequences should be,

\[0\rightarrow B\oplus Q\rightarrow P\oplus Q\rightarrow A\oplus Q\rightarrow 0\]

\[0\rightarrow B'\oplus Q'\rightarrow P'\oplus Q'\rightarrow A'\oplus Q'\rightarrow 0\]

To apply Schanuel's lemma we need to have short exact sequences. However for $B\oplus 0$ and $B'\oplus 0$ it is not guaranteed that the sequences will be exact.

Am I correct in my assumption? Is this really a typo. I find it hard to believe that a book like this one will have a typo. :)

 

[Deveno]

Presumably, the surjections:

$P \oplus Q \to A \oplus Q$
$P' \oplus Q' \to A \oplus Q'$ are given by:

$(p,q) \mapsto (\beta(p),q)$
$(p',q') \mapsto (\beta'(p'),q')$

For the sequences to be exact, we require that the kernel of these maps be the image of what we map into the middle direct sum.

Since these maps are injective on $0 \oplus Q$ and $0 \oplus Q'$, the image of whatever we map into the middle direct sum must be $0_Q$ and $0_{Q'}$ (on the "second summand side").

So it's not a typo, the lemma actually implies:

$(B \oplus 0) \oplus (P \oplus Q)\cong A \oplus Q$

(and similarly for the "primes")

but trivially, $B \oplus 0 \cong B$, so that:

$B \oplus (P \oplus Q) \cong (B \oplus 0) \oplus (P \oplus Q)\cong A \oplus Q$

Indeed, if one were to use your proposed "correction" one would have:

$(b,q) \mapsto (\iota(b),q) \mapsto (0,q) \neq (0,0)$ for $q \neq 0$,

violating exactness.

 

[Sudharaka]

Presumably, the surjections:

$P \oplus Q \to A \oplus Q$
$P' \oplus Q' \to A \oplus Q'$ are given by:

$(p,q) \mapsto (\beta(p),q)$
$(p',q') \mapsto (\beta'(p'),q')$

For the sequences to be exact, we require that the kernel of these maps be the image of what we map into the middle direct sum.

Since these maps are injective on $0 \oplus Q$ and $0 \oplus Q'$, the image of whatever we map into the middle direct sum must be $0_Q$ and $0_{Q'}$ (on the "second summand side").

So it's not a typo, the lemma actually implies:

$(B \oplus 0) \oplus (P \oplus Q)\cong A \oplus Q$

(and similarly for the "primes")

but trivially, $B \oplus 0 \cong B$, so that:

$B \oplus (P \oplus Q) \cong (B \oplus 0) \oplus (P \oplus Q)\cong A \oplus Q$

Indeed, if one were to use your proposed "correction" one would have:

$(b,q) \mapsto (\iota(b),q) \mapsto (0,q) \neq (0,0)$ for $q \neq 0$,

violating exactness.

Thanks for your reply. I think I am getting a hold of this. The confusion that was in my mind was how to define the first map (between $B\oplus 0$ to $P\oplus Q$ or between $B'\oplus 0$ to $P'\oplus Q'$). So for example if we denote the R-homomorphisms of the first sequence as \(\alpha\) and \(\beta\);

\[0\rightarrow B\oplus 0\overset{\alpha}{\longrightarrow} P\oplus Q\overset{\beta}{\longrightarrow} A\oplus Q\rightarrow 0\]

then \[(b,\,0)\overset{\alpha}{\longrightarrow} (\alpha(b),\,q)\overset{ \beta}{\longrightarrow}(\beta \alpha (b),\,q)\] and similarly for the second sequence, \[(b',\,0)\overset{\alpha '}{\longrightarrow} (\alpha '(b'),\,q')\overset{\beta '}{\longrightarrow}(\beta '\alpha '(b'),\,q')\] Am I correct?

 

[Deveno]

Since the "alpha" maps are injective, the image under them is:

$(\alpha(b),0)$

(I tend to use the letter "iota" to make it clear we are talking about an injection).

The subsequent surjection takes this to $(0,0)$ in $A \oplus Q$ since:

$\beta\alpha = 0$ by exactness.

 

[Sudharaka]

Since the "alpha" maps are injective, the image under them is:

$(\alpha(b),0)$

(I tend to use the letter "iota" to make it clear we are talking about an injection).

The subsequent surjection takes this to $(0,0)$ in $A \oplus Q$ since:

$\beta\alpha = 0$ by exactness.

Oh….. I should have seen that. (Headbang)

I think now I understand everything. Need to go through this lemma once again to get a clear picture of things. :) So basically each element $(b,\,0) \in B\oplus 0$ is mapped to $(\alpha(b),\,0)\in P\oplus Q$ under the injection, and then mapped to $(0,\,0)\in A\oplus Q$ under the surjection. When it comes to the surjective map between $P\oplus Q\mbox{ and }A\oplus Q$ for a general element $(p,\,q)\in P\oplus Q$ we should have,

\[(p,\,q)\overset{\alpha}{\longrightarrow}(\beta( p),\, q)\]

I think I am correct. :)