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 $x^2 - 2a + (a^2 + b^2)$.