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.
\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}.
\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


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