In some notes on abstract algebra I found a number theory problem in which I was given to prove that the algebraic integers, $\overline{\mathbb{Z}}$ form a subring of $\mathbb{C}$ using the tensor product. The solution that I've found kinda generalizes some notions of field extensions. But DK if it's right.
Here's my work:
I took two arbitrary elements, $\lambda$ and $\lambda '$ in $\overline{\mathbb{Z}}$ and then, since $\mathbb{Z}[\lambda]$ and $\mathbb{Z}[\lambda ']$ are $\mathbb{Z}$ algebras, their tensor product, $\mathbb{Z}[\lambda] \otimes_{\mathbb{Z}} \mathbb{Z}[\lambda ']$ is also a $\mathbb{Z}$ algebra (the general result is for any ring $R$ and any algebras $A$ and $B$ over it).
Since $\lambda \in \overline{\mathbb{Z}}$, we could choose a monic, irreductible polynominal, $P_{\lambda}$ such that $\lambda$ is a root of this polynominal. If we were to choose another polynominal of smaller degree that's not monic in $\mathbb{Z}[T]$ for which $\lambda$ is not a root, its product with another polynominal won't be monic anymore. Since $P_{\lambda}$ is monic, by the division algorithm, $\forall f \in \mathbb{Z}[T]$, $\exists$ unique $q;g \in \mathbb{Z}[T]$ such that $f= qP_{\lambda}+g$. Take the evaluation homomorphism $\psi_{\lambda}:\mathbb{Z}[T] \longmapsto \mathbb{C}$ and hence, by the first ISO theorem, $$\frac{\mathbb{Z}[T]}{ker(\psi_{\lambda})} \cong Im( \psi_{\lambda})=\mathbb{Z}[\lambda]$$ and by the previous result, we get that $ker(\psi_{\lambda})=<P_{\lambda}>$. The same is true for $\lambda '$.
Hence, $$\mathbb{Z}[\lambda] \otimes_{\mathbb{Z}} \mathbb{Z}[\lambda '] \cong \frac{\mathbb{Z}[T]}{<P_{\lambda}>} \otimes_{\mathbb{Z}} \frac{\mathbb{Z}[T]}{<P_{\lambda '}>} $$.
Now we will take a bilinear map $\varphi :\mathbb{Z}[\lambda] \times \mathbb{Z}[\lambda '] \longmapsto \mathbb{Z}[\lambda;\lambda ']: (a;b) \longmapsto ab$ which is surjective. By the universal property of the tensor prodcut, we get a $\mathbb{Z}$-module homomorphism $\widetilde{\varphi}:\mathbb{Z}[\lambda] \otimes_{\mathbb{Z}} \mathbb{Z}[\lambda '] \longmapsto \mathbb{Z}[\lambda;\lambda ']: a \otimes_{\mathbb{Z}} b \longmapsto ab$ which is also a ring homomorphism but $\varphi$ and $\mu : \mathbb{Z}[\lambda] \times \mathbb{Z}[\lambda '] \longmapsto \mathbb{Z}[\lambda] \otimes_{\mathbb{Z}} \mathbb{Z}[\lambda '] : (a;b) \longmapsto a \otimes_{\mathbb{Z}} b$ are surjective and so is $\widetilde{\varphi}$, since $ \varphi= \widetilde{\varphi} \circ \mu$.
Since for any monic polynominal $f \in \mathbb{Z}[T],\, \mathbb{Z}[T]/<f> \cong \mathbb{Z}^{deg(f)}$, we have: $$ \mathbb{Z}[\lambda] \otimes_{\mathbb{Z}} \mathbb{Z}[\lambda '] \cong \mathbb{Z}^{n} \otimes_{\mathbb{Z}} \mathbb{Z}^{m} \cong \mathbb{Z}^{mn}$$ where $n:=deg(P_{\lambda})$; $m:=deg( P_{\lambda'}) $ and hence, $\exists$ $k(a) \in \mathbb{N^{*}}$ such that for every $a \in \mathbb{Z}[\lambda;\lambda '],$ $$\frac{\mathbb{Z}[T]}{ <P_{a}>} \cong \mathbb{Z}[a] \cong \mathbb{Z}^{k(a)}$$ and $a$ could take values like $\lambda + \lambda '$ or $\lambda \lambda'$ for which there is a monic polynominal $P_{a}$ and $a$ is a root of it.
Therefore, $<\overline{\mathbb{Z}};+;\times>$ is a ring.
