What does it mean to say that an element 'satisfies' a polynomial?

In the context of finite fields, the definition of a primitive element $\alpha$ is given by: $\alpha$ is primitive if it generates all elements of $F_q - \{0\}$ when raised to powers up to $q-1$.

And for definition of a primitive polynomial: we say that an irreducible polynomial that $\alpha$ satisfies is also primitive.

To me it's unclear what is meant by $\alpha$ satisfying a polynomial, although my guess is that $\alpha$ is a zero of the polynomial. Also, I ask this question because I do not understand the definition of a primitive polynomial so if someone has a different/clearer definition(or explanation) then I would appreciate that.

• Purely as a matter of language, I would say that it’s ungrammatical or worse to say that an element satisfies a polynomial. Rather one should restrict the word “satisfy” to sentences with a free variable, like $f(x)=0$. That is, an element satisfies the sentence when the sentence becomes true when the variable is set equal to the named element. – Lubin May 14 '14 at 16:59
• I agree, this is what led to my confusion. – McT May 14 '14 at 17:34
• One should perhaps better say that $α$ satisfies a polynomial equation, which would be the verbal equivalent of the comment of Lubin. The short form is not always recognizable as a short form. – LutzL May 14 '14 at 17:46

That's exactly what it means. Ie, $a$ satisfies the polynomial $f(x)$ iff $f(a) = 0$.