Algebraic integer definition (basic question)

Question

The notes of a module I'm doing have the following definition:

Let $\alpha \in \mathbb{C}$. We say that $\alpha$ is an algebraic integer if $\alpha$ is the root of a monic polynomial with integer coefficients, that is, $p(\alpha)=0$ for some $p \in \mathbb{Z}[t]$.

My problem here is with what goes after the "that is" above. Is it not possible for $p(\alpha)=0$ for some $p \in \mathbb{Z}[t]$, where $p$ is not a monic polynomial? I thought $\mathbb{Z}[t] = a + bt$ with $a, b \in \mathbb{Z}$, so $b$ does not need to be 1 for $p \in \mathbb{Z}[t]$ to hold, or did I misunderstand something?

Answer 1

Your doubt is natural. The part that comes after the “that is” doesn't make sense. It should be “that is, $p(\alpha)=0$ for some monic polynomial $p(x)\in\Bbb Z[x]$”.

Answer 2

The key word(s) is(are) not "that is", but the following word "some". The idea is: $\alpha$ is an algebraic integer if there exists a monic polynomial $p\in\mathbb Z[x]$ such that $p(\alpha)=0$.

It doesn't mean that, if $\alpha$ is already an algebraic integer, $p(\alpha)$ cannot not be additionally true for some other polynomials, including those in $\mathbb{Z}[x]$ that are not monic, or even some that are not in $\mathbb Z[x]$... However, at least one of polynomials $p$ satisfying $p(\alpha)=0$ must be monic, with integer coefficients.

Example: $\alpha=i$ is an algebraic integer because we can take the monic polynomial $p(x)=x^2+1$ such that $p(i)=0$.

Example 2: $\alpha=\frac{1}{2}$ is not an algebraic integer, because no polynomial $p(x)=a_0+a_1x+\ldots+a_{n-1}x^{n-1}+x^n$ (with $a_i\in\mathbb Z$) satisfies $p(\frac{1}{2})=0$. Namely, if it was satisfied for some polynomial $p$, then the multiple: $2^{n-1}p(\frac{1}{2})=a_02^{n-1}+a_12^{n-2}+\ldots+a_{n-1}+\frac{1}{2}$ would also be zero, which it isn't, as it isn't even an integer.

(In fact, it can be shown that a number from $\mathbb Q$ is an algebraic integer if and only if it is an integer in the ordinary sense, i.e. if it is in $\mathbb Z$. This justifies the terminology "algebraic integer" - it is a generalisation of the notion of an "integer".)

Also, a correction in your question: $\mathbb Z[x]=\{a_0+a_1x+\ldots+a_nx^n\mid n\in\mathbb N, a_0,a_1,\ldots,a_n\in\mathbb Z\}$ i.e. it contains polynomials of arbitrary degrees.