I'm reading the book "noncommutative ring" writed by Herstein. In the page 15, the author says that
Let $F$ be a field and $A$ is an algebra over $F$.
Let $\rho$ be a maximal regular right ideal of $A$ considered as a ring. We claim that $\rho$ is automatically a subspace of $A$ over $F$. If not, $F\rho\nsubseteq F$; however, from the defining properties of an algebra $F\rho$ is a right ideal of $R$. By the maximality of $\rho$ we conclude that $A=F\rho+\rho$. Thus $A^2=(F\rho+\rho)A\subset(F\rho)A+\rho A\subset\rho(FA)+\rho A\subset \rho$. Since $\rho$ is regular there is an $a\in A$ such that $x-ax\in \rho$ for all $x\in A$; but $ax\in A^2\subset \rho$ leaving us with the contradiction that $\rho=A$. Having now established that every maximal regular right ideal of $A$ as a ring is a maximal regular right ideal of $A$ as an algebra we conclude, on invoking Theorem 1.2.2, that the algebra radical of $A$ coincides with the ring radical of $A$.
There are two typing errors.
"$F\rho\nsubseteq F$" should be "$F\rho \nsubseteq \rho$"
"$F\rho$ is a right ideal of $R$" should be "$F\rho$ is a right ideal of $A$"
My question is the sentence "$F\rho$ is a right ideal of $A$." The author assumes that $F\rho\nsubseteq \rho$. Hence there exists $f_0\in F$ and $x_0\in \rho$ such that $f_0 x_0\notin\rho$.
If $F\rho$ is a right ideal of $A$, then for all $f_1, f_2\in F$ and $x_1, x_2\in \rho$, $f_1 x_1+f_2 x_2\in \rho$. ($F\rho$ is an additive abelian group, so the additive closure holds.)
Let $f_1=f_0, x_1=x_0, f_2=0$. Then we get a contradiction.
Is it right? (I mean my argument.)