Can the definition of algebraic Integer

The roots of polynomials, such as $x^3 + bx^2 + cx + d = 0$, with integer (or rational) coefficients.

be accurately paraphrased as

Any real or complex number $A + iB$ where $A$ and $B$ are integers.

If not, what belongs to my definition that does not belong to the usual one?

  • $\begingroup$ What about $\sqrt 2$? It is root of $x^2 - 2$. Your numbers are gaussian integers $\endgroup$ Jul 17, 2015 at 10:27
  • 2
    $\begingroup$ Your definition of algebraic integer isn't even correct. An algebraic integer is a complex number which is the root of a monic polynomial with integer coefficients. $\endgroup$
    – lokodiz
    Jul 17, 2015 at 10:34
  • $\begingroup$ You got it the other way around. The numbers you mention are all algebraic integers, but not all algebraic integers take that form. For example think about the roots of unity. $\endgroup$
    – A.P.
    Jul 17, 2015 at 12:05

2 Answers 2


If by "i" you mean $i = \sqrt{-1}$, then your "paraphrase" is lacking. It would be like saying only beagles are dogs, thus ignoring terriers, colliers, retrievers, etc. Your "such as" seemed to indicate knowledge of algebraic integers of degree $3$, such as $1 + \root 3 \of 2$, which is a little difficult to represent as $A + iB$.

I recommend that you play around with Wolfram Alpha, asking it questions like:

  • is 1/2 + sqrt(-7)/2 an algebraic integer?
  • is sqrt(-7)/2 an algebraic integer?
  • is 3^(1/5) an algebraic integer?
  • is pi an algebraic integer?
  • etc.

You're getting your Gaussian integers mixed up with your algebraic integers. All Gaussian integers are algebraic integers, but not all algebraic integers are Gaussian integers.

The polynomials are of crucial importance to determining what is or what isn't an algebraic integer. Take a look at this number: $$\frac{28 + \root 3 \of {10} + 19(\root 3 \of {10})^2}{3}$$ With an approximate decimal value of $39.4482$, this is clearly not your typical integer. But it is a solution to $x^3 - 28x^2 + 198x - 25626 = 0$ (so then $b = 28$, $c = 198$ and $d = -25626$).

Hey, where's $a$? It's there, it's equal to $a = 1$, so you can just leave it out. Which brings me to my next point: all algebraic integers are algebraic numbers, but not all algebraic numbers are algebraic integers. This is not an algebraic integer, but it is an algebraic number: $$\frac{\root 3 \of {11}}{3}$$ The relevant polynomial is $27x^3 - 11 = 0$. We have $b = 0$, $c = 0$, $d = -11$ but $a = 27$.

At the risk of sounding monotonous, the root of the polynomial is important, and trying to leave it out of a definition is just not worth it.

But the nice thing about Gaussian integers is that you can very easily guess the relevant polynomial. If $a + bi$ is a Gaussian integer, then the polynomial is probably is $x^2 - 2a + (a^2 + b^2)$.


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