Can the direct sum of a real algebra be embedded into itself?

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Does there exist a real algebra $A$ which admits an injective $A$-module homomorphism $A^n\to A$, where the image has co-dimension $1$ (and $n$ is strictly greater than $1$)?

I am asking for $n>1$, as for $n=1$ the map $C^\infty(\Bbb R)\to C^\infty(\Bbb R)$, $f\mapsto x\cdot f$ gives us an example where it does work.

This obviously cannot happen with any finite-dimensional algebras. With dimension I refer to the dimension of the algebra as a real vector space.

I am led to this question in an exercise, in which I would like to show that the ideal of those functions that vanish on a sub-manifold is not locally free as a $C^\infty$ module (in the case that the sub-manifold has co-dimension $>1$). This gives the question a slight differential geometry flavour.

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If such a module-homomorphism exists for certain $n$, then $A$ has in particular an ideal isomorphic to $A^n$.

So a counterexample, for all $n$ simultaneously, will be given by a simple infinite-dimensional algebra. There are many of these. I'm no algebraist, so I don't have maybe easier examples at hand. But any simple unital Banach algebra is algebraically simple. In particular any unital simple C$^*$-algebra (like $\mathcal O_2$, or $C_r^*(\mathbb F_2)$, or UHF$(2^\infty)$, among many) are counterexamples.