This problem is an exercise from Drozd-Kirichenko's book Finite Dimensional Algebras, page 29.

Let $k$ be a field. Let $A$ be a $k$-algebra not necessarily with identity.

Let $\overline A$ be the algebra obtained from $A$ by adjoining an identity : $\overline {A}=\{(a,\alpha) \mid a\in A,\alpha \in k\}.$

a) Prove that the elements of the form $(a, 0)$ generate an ideal of $\overline A$ which is isomorphic to $A$ (as an algebra without an identity).

b) If $A$ has an identity, show that $\overline A \simeq A\times k$.

c) Show that every homomorphism $f : A\rightarrow B$ between two algebras without an identity extends uniquely to a homomorphism $j : \overline A\rightarrow \overline B$ between the algebras with identity, and that every homomorphism $j : \overline A\rightarrow \overline B$ maps $A$ into $ B$. In particular, $A \simeq B$ if and only if $\overline A \simeq \overline B$.

There is one extra information added in the translated version (in Chinese): every proper ideal of $\overline A$ is contained in $A$ (the ideal constructed in a) which is isomorphic to $A$).

My problem is how to prove this and how to use this to prove c): $j$ maps $A$ into $B$.

I've tried to prove : if an ideal $I$ of $\overline A$ contains any $(a,1)$, then it must be $\overline A$ itself. I've tried so many times but failed: I could not use $(a,1)$ to get $(a,0)$.

  • $\begingroup$ The algebra $\overline A$ is defined with componentwise scalar multiplication and addition, and the multiplication is defined by: $(a,\alpha)(b,\beta)=(ab+\beta a+\alpha b,\alpha \beta)$ $\endgroup$ – user12580 May 25 '16 at 13:58

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