My professor gave us the following definition:
Let $R$ be a commutative ring. $r \in R$ is prime if $r.R$ is a prime ideal. Then he concluded that this definition tells us $r \notin R^*$ but I do not understand why, could anyone explain this to me please?
Answer(based on comments given below the question):
A prime ideal is by definition a proper ideal, and if $r$ is a unit then $rR = R.$
The proof can be found in this link If an ideal contains the unit then it is the whole ring and here it is:
Proposition. Let $I$ be a left or right ideal of a ring $R$ with unity. Then $I = R$ if and only if $I$ contains a unit.
Proof: Let $I$ be a left ideal of $R$. If $I = R$ then $1 \in I$, so $I$ contains a unit. Conversely, if $u \in I$ is any unit, let $u^{-1}$ be its inverse. For any $r \in R$, we have $r = r (u^{-1} u) = (r u^{-1}) u \in I$, because $I$ is a left ideal. Thus $I = R$. This logic is easily adapted for right ideals to show the same result. Certainly it then is true for two-sided ideals and ideals of unital commutative rings.