Algorithmic decomposition of commutative semisimple algebra I have a question concerning this survey, page 6, statements (a) and (b). It is not necessary to look into that paper for understanding my question.
Let $k$ be a field (possibly of characteristic 0) and $A$ be a commutative semisimple $k$-algebra. We want to find ideals $I_1,\dotsc,I_n$ such that $A=\bigoplus_iI_i$. Let $u\in A$ be nonzero and $I=(u)$. Since $A$ is semisimple, $I$ must be a summand, so it has an identity element $e$.  The procedure outlined in the article is as follows:


*

*Let $k_0=k$, $i=0$.

*Let $v_i\in I$ be not a $k_i$-multiple of $e$ and compute the minimal polynomial $m$ of $v_i$ over $k_i$.

*If $m$ is irreducible, then $k_i[v_i]\subseteq I$ is a field. If $k_i[v_i]=I$ then we are done.


This is true because if $m$ is irreducible, then $k_i(m)=k_i[X]/(m)$ is a field. By the very definition of the minimal polynomial, $k_i[X]\to k_i[v_i]\subseteq I, X\mapsto v_i$ has kernel $(m)$, so $k_i[v_i]\cong k_i(m)$ is a field.


*Repeat from 1. for $k_{i+1}=k[v_i]$ and with the new $v_{i+1}$ linearly independent from the previous ones.

*The minimal polynomial $m$ over $k_i$ might not be irreducible over $k_i$; if $m=fg$ for $f,g\in k_i[X]$ nontrivial, then $f$ and $g$ are coprime, so $(f)+(g)=k_i[X]$ and $(f)\cap (g)=(m)$. It follows that $(f(v_i))+(g(v_i))=I$ are also coprime, and $(f(v_i))\cap (g(v_i))=(m(v_i))=0$, so $I=(f(v_i))\oplus (g(v_i))$.


Questions: 


*

*Why are $f, g$ coprime?

*If $f,g\in k_i[X]$ are coprime, why does not follow that $(f(v_i))$ an $(g(v_i))$ are coprime ideals?
 A: Question 1 is answered using the following simple fact:

Proposition 1. Let $k$ be a field. Let $A$ be a commutative semisimple $k$-algebra that is finite-dimensional as a $k$-vector space. Let $v \in A$. Then, the minimal polynomial of $v$ over $k$ is squarefree.

To prove this, we will need a simple lemma:

Lemma 2. Let $A$ be a commutative semisimple ring. Let $a \in A$ satisfy $a^2 = 0$. Then, $a = 0$.

Here and in the following, a commutative ring is said to be semisimple if its Jacobson radical is $0$. (When the ring is a $k$-algebra over a field $k$, and is finite-dimensional as a $k$-vector space, this definition agrees with all other known definitions of "semisimple".)
Proof of Lemma 2. For each $x \in A$, we have $\left(1-ax\right) \left(1+ax\right) = 1 - \left(ax\right)^2 = 1 - \underbrace{a^2}_{=0} x^2 = 1$. Thus, for each $x \in A$, the element $1 - ax$ of $A$ is invertible (with inverse $1 + ax$).
Recall that the Jacobson radical of $A$ is the set of all $b \in A$ such that for each $x \in A$, the element $1 - bx$ of $A$ is invertible. (This is one of the definitions of the Jacobson radical.) This shows that $a$ belongs to the Jacobson radical of $A$ (because for each $x \in A$, the element $1 - ax$ of $A$ is invertible). But the Jacobson radical of $A$ is $0$ (since $A$ is semisimple). Hence, $a = 0$ (since $a$ belongs to this Jacobson radical). This proves Lemma 2. $\blacksquare$
Proof of Proposition 1. Assume the contrary. Thus, the minimal polynomial of $v$ over $k$ is not squarefree. Denote this minimal polynomial by $m$. Hence, $m$ is not squarefree; thus, there exists a non-constant polynomial $p \in k\left[x\right]$ such that $p^2 \mid m$. Consider this $p$. There exists a polynomial $q \in k\left[x\right]$ such that $m = p^2 q$ (since $p^2 \mid m$). Consider this $q$. Now, $m\left(v\right) = 0$ (since $m$ is the minimal polynomial of $v$). But from $m = p^2 q$, we obtain $m\left(v\right) = \left(p\left(v\right)\right)^2 q\left(v\right)$. Hence, $\left(p\left(v\right)\right)^2 q\left(v\right) = m\left(v\right) = 0$. Note that $p \cdot pq = p^2 q = m \neq 0$, so that $pq \neq 0$.
Define $a \in A$ by $a = p\left(v\right) q\left(v\right)$. Then, $a^2 = \left(p\left(v\right) q\left(v\right)\right)^2 = \underbrace{\left(p\left(v\right)\right)^2 q\left(v\right)}_{=0} q\left(v\right) = 0$. Thus, Lemma 2 shows that $a = 0$.
Now, $\left(pq\right)\left(v\right) = p\left(v\right) q\left(v\right) = a = 0$. Hence, $m \mid pq$ (since $m$ is the minimal polynomial of $v$). Thus, $\deg \left(pq\right) \geq \deg m$ (since $pq \neq 0$ and $m \neq 0$). But from $m = p^2 q = p \cdot pq$, we obtain $\deg m = \deg\left(p \cdot pq\right) = \deg p + \deg\left(pq\right)$. Hence, $\deg \left(pq\right) \geq \deg m = \deg p + \deg\left(pq\right)$. Cancelling $\deg\left(pq\right)$ from this inequality, we obtain $0 \geq \deg p$ (since $pq \neq 0$). Thus, the polynomial $p$ is constant. This contradicts the fact that $p$ is not constant. This contradiction shows that our assumption was wrong, and Proposition 1 is proven. $\blacksquare$
Proposition 1 answers your Question 1 (about why $f$ and $g$ are coprime) once you apply it to $k = k_i$. Note that commutativity of $A$ is crucial; otherwise, e.g., $A$ could be a matrix ring and $v$ could be a nonzero nilpotent matrix therein.

Now, let us answer Question 2. Again, I shall state it in greater generality:

Proposition 3. Let $k$ be a field. Let $A$ be a commutative semisimple $k$-algebra. Let $v \in A$. Let $f, g \in k\left[x\right]$ be coprime polynomials such that $\left(fg\right)\left(v\right) = 0$. Then, $f\left(v\right) A + g\left(v\right) A = A$ and $f\left(v\right) A \cap g\left(v\right) A = 0$.

(Here, of course, "$f\left(v\right) A \cap g\left(v\right) A$" should be understood as $\left(f\left(v\right) A\right) \cap \left(g\left(v\right) A\right)$.)
Proof of Proposition 3. The polynomials $f, g$ are coprime elements of the principal ideal domain $k\left[x\right]$. Hence, Bezout's theorem says that there exist polynomials $a, b \in k\left[x\right]$ such that $af + bg = 1$. Consider these $a$ and $b$. Now, evaluating both sides of the polynomial identity $af + bg = 1$ at $v$, we obtain
\begin{align}
a\left(v\right) f\left(v\right) + b\left(v\right) g\left(v\right) = 1 .
\end{align}
Hence, each $u \in A$ satisfies
\begin{align}
\underbrace{\left(a\left(v\right) f\left(v\right) + b\left(v\right) g\left(v\right)\right)}_{= 1} u = 1 u = u.
\end{align}
Thus, each $u \in A$ satisfies
\begin{align}
u &= \left( a\left(v\right) f\left(v\right) + b\left(v\right) g\left(v\right)\right) u \\
&= f\left(v\right) \underbrace{a\left(v\right) u}_{\in A} + g\left(v\right) \underbrace{b\left(v\right) u}_{\in A} \\
&\in f\left(v\right) A + g\left(v\right) A .
\end{align}
In other words, $A \subseteq f\left(v\right) A + g\left(v\right) A$. Combining this with the obvious inclusion $f\left(v\right) A + g\left(v\right) A \subseteq A$, we see that $f\left(v\right) A + g\left(v\right) A = A$.
It remains to show that $f\left(v\right) A \cap g\left(v\right) A = 0$. It is here that we will use the semisimplicity of $A$. Indeed, let $c \in f\left(v\right) A \cap g\left(v\right) A$. Thus, $c \in f\left(v\right) A$ and $c\in g\left(v\right) A$. Hence,
\begin{align}
c^2 &= \underbrace{c}_{\in f\left(v\right) A} \underbrace{c}_{\in g\left(v\right) A}
\in f\left(v\right) A g\left(v\right) A
= \underbrace{f\left(v\right) g\left(v\right)}_{=\left(fg\right)\left(v\right) = 0} A = 0A = 0 .
\end{align}
In other words, $c^2 = 0$. Hence, Lemma 2 (applied to $c$ instead of $a$) yields $c = 0$.
Now, forget that we fixed $c$. We thus have shown that $c = 0$ for each $c \in f\left(v\right) A \cap g\left(v\right) A$. In other words, $f\left(v\right) A \cap g\left(v\right) A = 0$. Thus, the proof of Proposition 3 is complete. $\blacksquare$
