Endomorphism ring of Q is a division algebra

We show that $Q$ is not semisimple nor simple, but $\text{End}_\mathbb{Z}(\mathbb{Q})$ is a division algebra.

Consider $A=\mathbb{Z}$ (as a $\mathbb{Z}$ -algebra). Consider $M=\mathbb{Q}$ as a right $\mathbb{Z}$ -module.
Lemma:
$\mathbb{Q}$ is not semisimple nor simple.

Suppose to the contrary $\mathbb{Q}=\bigoplus_{i\in I}N_i$ , where $N_i$ are simple $\mathbb{Z}$ -modules (i.e. $N_i\cong\mathbb{Z}/p_i\mathbb{Z}$ ). Then there exists nonzero $x\in\mathbb{Q}$ such that $x$ has finite order (product of primes). This is impossible in $\mathbb{Q}$ .
Lemma:
$\text{End}_\mathbb{Z}(\mathbb{Q})\cong\mathbb{Q}$ as $\mathbb{Z}$ -algebras.

Define $\Psi:\mathbb{Q}\to\text{End}_\mathbb{Z}(\mathbb{Q})$ where $q\in\mathbb{Q}$ is mapped to $\lambda_q\in\text{End}_\mathbb{Z}(\mathbb{Q})$ , where $\lambda_q(x)=qx$ . Let $k\in\mathbb{Z}$ , $q,q_1,q_2\in\mathbb{Q}$ .

We can check that $\Psi$ is a $\mathbb{Z}$ -algebra homomorphism.

Let $q\in\ker\Psi$ . Then $\Psi(q)=\lambda_q=0$ , $\lambda_q(x)=qx=0$ for all $x\in\mathbb{Q}$ . This implies $q=q\cdot 1=0$ . Hence $\Psi$ is injective.

Let $\phi\in\text{End}_\mathbb{Z}(\mathbb{Q})$ . Let $x=\frac{a}{b}\in\mathbb{Q}$ , where $a,b\in\mathbb{Z}$ . $\phi(x)=a\phi(\frac 1b)=\frac ab\cdot b\phi(\frac 1b)=\frac ab\cdot\phi(1)=\phi(1)\cdot x=\lambda_{\phi(1)}(x)$ . Hence $\Psi$ is surjective.

Thus $\text{End}_\mathbb{Z}(\mathbb{Q})\cong\mathbb{Q}$ is a division algebra, but $\mathbb{Q}$ is not simple.

Leave a Reply Cancel reply

Enter your comment here...

Fill in your details below or click an icon to log in:

Email (required) (Address never made public)

Name (required)

Website

You are commenting using your WordPress.com account. ( Log Out / Change )

You are commenting using your Google+ account. ( Log Out / Change )

You are commenting using your Twitter account. ( Log Out / Change )

You are commenting using your Facebook account. ( Log Out / Change )

Cancel

Connecting to %s