# 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.

## Author: mathtuition88

https://mathtuition88.com/

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