# Simple Problem - Establishing an Isomorphism

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle \Phi (s \otimes r ) = sr$$

and

$$\displaystyle {\Phi}' \ : \ S \to S \otimes_R R$$

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 ...

Peter

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
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).

Last edited: