one-dimensional $\mathbb{R}$-Algebra $A$ I want to prove that every one-dimensional $\mathbb{R}$-Algebra $A$ is isomorphic if the multiplication is not always $0$.
My idea is the following: i do this by showing that there is a "one-element" in $A$.
I take the Element $a$ of a basis of $A$, so we have $a \neq 0$ 
and $A = \mathbb{R}a$.
Then there must be an element $r \in \mathbb{R}$ such that $ra = a^2$, and $e := r^{-1}a$ gives us the "$1$".
However, after thinking about it, i found a problem: how do i know that $a^2$ is not $0$ - it sounds easy, but i have not found an explanation yet.
 A: If $a^2$ were zero, then for all $s,r\in\mathbb{R}$ we'd have $(sa)(ra)=sra^2=0$, hence the multplication operation always yields $0$. By assumption we are not considering this case.
A: If by the phrase, "the multiplication is not always $0$" it is understood that the multiplication in $A$ is non-trivial, that is, there exist $c, d \in A$ with $cd \ne 0$, then for any basis element of $a \in A$ we may write $c = ra$, $d = sa$ for some $r, s \in \Bbb R$; so $0 \ne cd = (ra)(sa) = rsa^2$. Note that $r, s \ne 0$ since $cd \ne 0$; we may also conclude $a^2 \ne 0$ as well, since $a^2 = 0 \Rightarrow rsa^2 =0$.  
Note Added Sunday 14 June 2015 11:48 AM PST:  I thought it might be fun to flesh out an answer to the question tacitly asked in the first line of our OP supinf's post, viz., that any two $\Bbb R$-algebras $A_1$, $A_2$ with non-trivial multiplication are isomorphic.  I will do this by showing that any such $\Bbb R$-algebra $A$ is isomorphic to $\Bbb R$.  We have seen through supinf's efforts and remarks, and by my observations above, that for any basis element $a \in A$ we have $ 0 \ne a^2 = ra$ for some $0 \ne r \in \Bbb R$; following supinf, we set $e = r^{-1}a$; then
$e^2 = (r^{-1}a)^2 = r^{-2}a^2 = r^{-2}ra = r^{-1}a = e; \tag{1}$
that is, $e$ is an idempotent in $A$; and since $a$ is a basis element for $A$, $e$ is too:  for $b = sa \in A$,
$b = sa = sr(r^{-1}a) = sre = \beta e, \tag{2}$
with $\beta = sr$; from this, we see that multiplication in $A$ must commute; taking $\alpha e, \beta e \in A$ we have
$(\alpha e)(\beta e) = \alpha \beta e^2 = \beta \alpha e^2 = (\beta e)(\alpha e); \tag{3}$
next, we note that $e$ is in fact a multiplicative identity for $A$:
$ (\alpha e)e = \alpha e^2 = \alpha e, \tag{4}$
and
$e(\alpha e) = (\alpha e) e = \alpha e^2 = \alpha e. \tag{5}$
Since every $a \in A$ is of the form $a = \alpha e$, 
$1a = 1(\alpha e) = (1 \alpha) e = \alpha e = a; \tag{6}$
using this, we have next that if $\alpha e \ne 0$ we must have $\alpha \ne 0$, so $\alpha^{-1} \in \Bbb R$ is available to us, and
$(\alpha^{-1}e) (\alpha e) = \alpha^{-1} \alpha e^2 = 1e^2 = e^2 = e, \tag{7}$
showing every $0 \ne \alpha e \in A$ has a multiplicative inverse $\alpha^{-1} e \in A$; looks like $A$ is a field.
We now define a map $\theta: \Bbb R \to A$ by
$\theta(\alpha) = \alpha e \tag{8}$
for $\alpha \in \Bbb R$; then
$\theta(\alpha + \beta) = (\alpha + \beta)e = \alpha e + \beta e = \theta(\alpha) + \theta(\beta), \tag{9}$
$\theta(\alpha \beta) = (\alpha \beta)e = (\alpha \beta)e^2 = (\alpha e)(\beta e) = \theta(\alpha)\theta(\beta); \tag{10}$
$\theta: \Bbb R \to A$ is also clearly surjective, since for any $\alpha e \in A$ we have
$\theta(\alpha) = \alpha e; \tag{11}$
furthermore, $\theta$ is injective as well, since
$\theta(\alpha_1) = \theta(\alpha_2) \tag{12}$
implies
$\alpha_1 e = \alpha_2 e \tag{13}$
or
$(\alpha_1 - \alpha_2)e = 0, \tag{14}$
whence
$\alpha_1 - \alpha_2 = 0 \tag{15}$
or
$\alpha_1 = \alpha_2, \tag{16}$
since $e \ne 0$.
$\theta$ is thus an isomorphism 'twixt $\Bbb R$ and $A$, establishing the desired conclusion.  
It's probably worth observing that for $r, \alpha \in \Bbb R$
$\theta(r \alpha) = r\alpha e = r\theta(\alpha), \tag{17}$
so $\theta$ is an algebra isorphism as well.  End of Note.
