Let $A$ and $B$ be $F$-algebras, and let $φ_i : A → B$ be an $F$-algebra homomorphism, for $i$ ranging over some index set $I$. Define the equalizer $E$ to be the set of $a ∈ A$ such that for all $i,j ∈ I$, we have $φ_i(a) = φ_j(a)$.
I need to prove that E is a sub-$F$-algebra of $A$.
I am wondering what we need to check to show that $E$ is a sub-$F$- algebra. The $R$-algebra $A$ is defined as an ordered pair $(A,i)$ where $i: R\rightarrow A$ is a ring homomorphism. But we have not been given that what is a sub-$F$-algebra.
I guess this question is asking us to prove that regard $F$ algebra $A$ as a $F$ vector space, we need to prove that $E$ is a subspace.
So I am thinking about this:
check of closure under addition:
if $a,b\in E$, $\phi_i(a+b)=\phi_i(a)+\phi_j(b)=\phi_j(a)+\phi_j(b)=\phi_j(a+b)$.
check of closure under scaler multiplication:
if $a\in E,m\in F$, $\phi_i(ma)=\phi_i(m)\phi_i(a)=m\phi_i(a)=m\phi_j(a)=\phi_j(m)\phi_j(a)=\phi_j(ma)$.
check of $0$ and $1$: $\phi_i(0)=0=\phi_j(0),\phi_i(1)=1=\phi_j(1)$
So am I correct? Could someone please check if I get it? Thanks a lot!