# Set of algebraic integer form a ring.

An algebraic integer is a complex number that is the root of monic polynomial with integer coefficients. Show that the set of algebraic integers is a subring of $C$. (Hint: Use symmetric function theorem).

I also know that $\alpha \in$ $C$ is an algebraic integer if and only if $m_\alpha,_Q \in Z[x].$

Thanks for the help.

• The most explicit proof that $\overline{\mathbb{Z}}$ is a ring is with the resultant. I'm not sure what means using the symmetric function theorem, can you elaborate on what you proved earlier ? Feb 7, 2017 at 11:48
• what is the meaning of $m_\alpha,_Q$? Feb 7, 2017 at 15:23
• $m_\alpha,_Q$ is minimal polynomial of $\alpha$ over field $Q$ Feb 7, 2017 at 15:50
– user321268
Feb 7, 2017 at 22:24
• See this thread for a couple standard ways. Oct 5, 2020 at 8:36

Let $$\alpha,\beta\in\mathbb{C}$$ be algebraic integers; e.g. $$\alpha^3+\alpha+1=0$$ and $$\beta^2+1=0$$.

Since the polynomials that $$\alpha$$ and $$\beta$$ satisfy are monic, we can, by division, write any element of $$\mathbb{Z}[\alpha,\beta]$$ as an integer combination of elements of the basis, $$B$$: $$\left\{1,\alpha,\alpha^2,\beta,\alpha\beta,\alpha^2\beta\right\}\tag1$$ That is, we can write any $$p\in\mathbb{Z}[\alpha,\beta]$$ as $$p=\overbrace{\begin{bmatrix}p_0&p_1&p_2&p_3&p_4&p_5\end{bmatrix}}^{\mathbb{Z}^n}\overbrace{\ \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}\ }^B\tag2$$ which means we can write \begin{align} p\ \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix} &=\overbrace{\begin{bmatrix} p_{0,0}&p_{0,1}&p_{0,2}&p_{0,3}&p_{0,4}&p_{0,5}\\ p_{1,0}&p_{1,1}&p_{1,2}&p_{1,3}&p_{1,4}&p_{1,5}\\ p_{2,0}&p_{2,1}&p_{2,2}&p_{2,3}&p_{2,4}&p_{2,5}\\ p_{3,0}&p_{3,1}&p_{3,2}&p_{3,3}&p_{3,4}&p_{3,5}\\ p_{4,0}&p_{4,1}&p_{4,2}&p_{4,3}&p_{4,4}&p_{4,5}\\ p_{5,0}&p_{5,1}&p_{5,2}&p_{5,3}&p_{5,4}&p_{5,5} \end{bmatrix}}^{\mathbb{Z}^{n\times n}} \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}\tag3\\[6pt] pB&=MB\tag4 \end{align} Since $$M$$ satisfies its characteristic polynomial $${\large\chi}_M(M)=0\tag5$$ we have that $$p$$ also satisfies $${\large\chi}_M$$: $${\large\chi}_M(p)=0\tag6$$ Since $${\large\chi}_M$$ is a monic integer polynomial, $$p$$ must be an algebraic integer.

Example 1

For example, $$(\alpha+\beta)\begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}=\begin{bmatrix} 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 \\ -1 & -1 & 0 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & -1 & -1 & 0 \end{bmatrix}\begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}$$ and $${\large\chi}_M(\lambda)=\det(M-I\lambda)=\lambda^6+5\lambda^4+2\lambda^3+4\lambda^2-4\lambda+1$$.

Thus, $$\alpha+\beta$$ is an algebraic integer.

Example 2

Furthermore, $$\alpha\beta\ \begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}=\begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -1 & -1 & 0 \\ 0 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 \end{bmatrix}\begin{bmatrix}1\\\alpha\\\alpha^2\\\beta\\\alpha\beta\\\alpha^2\beta\end{bmatrix}$$ and $${\large\chi}_M(\lambda)=\det(M-I\lambda)=\lambda^6-2\lambda^4+\lambda^2+1$$.

Thus, $$\alpha\beta$$ is an algebraic integer.

Conclusions

The key points that can be extrapolated from the examples above are that

(1) since $$\alpha$$ and $$\beta$$ both satisfy monic integer polynomials, $$\mathbb{Z}[\alpha,\beta]$$ can be written as a $$\mathbb{Z}$$-module generated by the Kronecker product (or flattened outer product) of $$\left\{1,\alpha,\alpha^2,\dots,\alpha^{k-1}\right\}$$ and $$\left\{1,\beta,\beta^2,\dots,\beta^{m-1}\right\}$$.

(2) the action of $$p\in\mathbb{Z}[\alpha,\beta]$$ on the basis of $$n=km$$ generators, from the Kronecker product above, can be represented by $$M\in\mathbb{Z}^{n\times n}$$.

(3) since $$M$$ satisfies its own characteristic polynomial, $${\large\chi}_M(M)=0$$, so does $$p$$, $${\large\chi}_M(p)=0$$

(4) the characteristic polynomial of an integer matrix is a monic integer polynomial.

Therefore, if $$\alpha$$ and $$\beta$$ are algebraic integers, then each $$p\in\mathbb{Z}[\alpha,\beta]$$ is an algebraic integer.