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Let $k$ be a field and $k[x]$ polynomial ring, and take the module $k[x]/(x-a)^2$ for arbitrary $a\in k$. How to show that this module is not semisimple?

I was thinking the easiest way is to use this (notation from Lang, XVII. Chapter 2. Conditions Defining Semisimplicity) characterization:

SS 3. If $E$ is semisimple, then every submodule $F$ of $E$ is a direct sumand, i.e. there exists submodule $F'$ such that $E = F\bigoplus F'$.

So I was thinking of taking the submodule generated by $x-a$, but this got me nowhere. Any ideas on an elegant way to prove this?

toxic
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1 Answers1

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$\newcommand{\x}{\bar{x}}$For simplicity (no pun intended), I shall give a sketch for $a = 0$, you can do this for the general case easily.

Following your suggestion, we show that $k[x]/(x^2)$ cannot be written as $(\x) \oplus N$ for any $k[x]$-submodule $N$ of $k[x]/(x^2)$.

Suppose that such an $N$ existed. Note that $N$ is generated by some element of the form $a + b \x$ with $a, b \in k$. (Why?)
We must also have $a \neq 0$. (Since $N \cap (\x) = 0$.)

Thus, the element $u := 1 + \frac{b}{a}\x$ is in $N$. However, note that $\frac{b}{a}\x$ is a nilpotent and thus, $u$ is a unit. Conclude.

(Technically, it does not make sense to call an element of a module a unit. But in the above, I'm being a bit sloppy and using the equivalence of $N$ as a $k[x]$-submodule and $N$ as an ideal in $k[x]/(x^2)$. If this is not okay, simply note that $(1 - bx/a) \cdot u = 1$ is an element of $N$. But the submodule generated by $1$ is the whole module.)

Aryaman Maithani
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    Maybe we didn't even need to use that $N$ is prime, just that it is not $0$. So take an arbitrary element from $N$, divide it by $x^2$ and it is automatically of the form mentioned above (also $a\neq 0$). But why is $u$ a unit if the aforementioned element is nilpotent? What's the inverse? – toxic Jun 06 '22 at 13:40
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    @toxic: I didn't use that $N$ is prime. $N$ is a submodule of a quotient of $k[x]$ and $k[x]$ is a PID. That is the reason why $N$ is generated by a single element. I'm not sure what you mean by dividing an element of $N$ by $x^2$. $x^2$ is not a unit. To answer your last question: If $s$ is a nilpotent in any ring, then $1 - s$ has $1 + s + s^2 + \cdots$ as its inverse (the summation is finite). – Aryaman Maithani Jun 06 '22 at 15:25