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Simple Problem - Establishing an Isomorphism


Well-known member
MHB Site Helper
Jun 22, 2012
In Example 7 in Dummit and Foote, Section 10.4. pages 369-370 (see attachment) D&F are seeking to establish an isomorphism:

\(\displaystyle S \otimes_R R \cong S \)

They establish the existence of two S-module homomorphisms:

\(\displaystyle \Phi \ : \ S \otimes_R R \to S \)

defined by \(\displaystyle \Phi (s \otimes r ) = sr \)


\(\displaystyle {\Phi}' \ : \ S \to S \otimes_R R \)

defined by \(\displaystyle {\Phi}' (s) = s \otimes 1 \)

D&F then show that \(\displaystyle \Phi {\Phi}' = I \) where I is the identity function on simple tensors ...

How does this establish that \(\displaystyle S \otimes_R R \cong S \) ... presumably this establishes \(\displaystyle \Phi \) as a bijective homomorphism ... but how exactly ...

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Well-known member
MHB Math Scholar
Feb 15, 2012
A bijective homomorphism is also called an isomorphism...

To be a bit more clear, let us suppose we have two $R$-module homomorphisms:

$f:M \to N$
$g: N \to M$

such that $g \circ f = 1_M$, the identity map on $M$.

Claim 1: $g$ is surjective.

Let $m \in M$. Then $m = g(f(m))$ so that $m$ has the pre-image under $g$ of $f(m)$.

Claim 2: $f$ is injective.

Suppose $f(m) = f(m')$. Then $m - m' = g(f(m)) - g(f(m')) = g(f(m) - f(m')) = g(0) = 0$, so $m = m'$.

Now, if we already know $f$ is surjective, this means $f$ is bijective, hence an isomorphism (and $g$ is also likewise an isomorphism).

Since any $s \in S$ has the pre-image (under $\Phi$) $s \otimes 1$, $\Phi$ is clearly surjective.

By showing $\Phi'\Phi$ is the identity on simple tensors, it follows from the bilinearity of $\otimes$ that it is the identity on any SUM of simple tensors, that is, on any tensor.

(it should be clear that the identity map $I$ on $S\otimes_R R$ is indeed an $S$-linear map).

(EDIT: you have the composition order of $\Phi$ and $\Phi'$ reversed)

(EDIT #2: D&F also show that $\Phi\Phi' = 1_S$ which means these morphisms are two-sided inverses of each other, hence each is a bijection).
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