the unitization of $A$

Question

Let $A$ be an $F$-algebra. Then the set $F \times A$ becomes an $F$-algebra, which we denote by $ A^{*}$, if we define addition, scalar multiplication and product as follows: $$(\lambda, x) + (\mu, y) := (\lambda + \mu, x + y),$$ $$\mu(\lambda, x) := (\mu\lambda,\mu x),$$ $$(\lambda, x)(\mu, y) := (\lambda \mu,\mu x + \lambda y + xy).$$ We consider $A$ as a subalgebra of $A^*$ via the embedding $x \mapsto (0, x)$. Note that $A$ is actually an ideal of $A^*$. A crucial observation for us is that $A^*$ is a unital algebra.Indeed, $(1, 0)$ is its unity.The algebra $A^* $ is called the unitization of $A$.

My questions:

If $A$ is simple, is $A^* $ simple?

If a unital F-algebra $A$ is prime, is $A^* $ prime?

Answer 1

If $A$ is simple, is $A^\ast$ simple?

For a nonzero algebra $A$, $A^\ast$ is never simple, because, as you already said in your post, $\{0\}\times A$ is an ideal.

If a unital $F$-algebra is prime, is $A^\ast$ prime?

Not necessarily. When $A$ is commutative, $A^\ast$ is also commutative, so for example, $(1,-1)(0,1)=(0,0)$, showing $A^\ast$ is not a domain even if $A$ is.

Answer 2

You can tweak a bit your construction (called the Dorroh extension, or unitization of the algebra), to get a better behaved overalgebra $R$, the reduced algebra with $1$ of $A$. Concretely, $$R:=A^*/\text{Ann}_{A^*}(A).$$

If $A$ has zero annihilator (Ann$_A(A)=0$, e.g. if $A$ is semiprime) then $A$ is injected in $R$, Ann$_R(A)=0$, and if $A$ has $1$ then $A=R$. Moreover, if $A$ is prime (resp. primitive, resp. semiprime) then $R$ is prime (resp. primitive, resp. semiprime), $Z(A)\subseteq Z(R)$ and $Z(R)$ is the reduced algebra with $1$ of $Z(A)$.

See Polynomial identities in ring theory by Louis Rowen, Section 1.11 and its associated examples section, to get more details.