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Let $K$ be an algebraically closed field and consider the $K$-algebra $K[t]$ i.e the $K$-algebra of all polynomials in the indeterminate $t$ with coefficients in $K$.

Now consider an ideal $I$ of $K[t]$ then since $K[t]$ is PID we have that $I= \langle p(t) \rangle$ for some polynomial $p(t) \in K[t]$.

Suppose now that we have the situation that $K[t]/I \cong A$ where $A$ is any connected $K$-algebra of dimension $3$. (by connected algebra I mean that is $A$ is not the direct product of two algebras).Here $\cong$ means isomorphism as $K$-algebras.

Question: Can we always guarantee that $K[t]/I$ is isomorphic to $K[t]/(t^{3})$ as $K$-algebra? or that $K[t]/(t^{3})$ is isomorphic to $K^{3}?$ are these all the possible cases? or what can we say about the ideal $I$?

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@Arturo Magidin: apologies, I think it reads fine now. –  Learner Nov 7 '11 at 5:30
    
I'm still unclear on what $K^3$ is... –  Arturo Magidin Nov 7 '11 at 5:35
    
@ArturoMagidin - I believe that $K^3$ denotes the $K$-algebra $K\oplus K \oplus K$. –  user5137 Nov 7 '11 at 5:37
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@Jack: In which case it should be clear that $K[t]/(t^3)$ cannot be isomorphic to it, since $K[t]/(t^3)$ contains nontrivial nilpotents but $K\oplus K\oplus K$ does not... –  Arturo Magidin Nov 7 '11 at 5:38

1 Answer 1

up vote 1 down vote accepted

If we factor $p(t) = (t-a_1)^{r_1}\cdots (t-a_k)^{r_k}$, with $a_i$ pairwise distinct, $r_i\gt 0$ for all $i$, we have that $$K[t]/I \cong K[t]/(t-a_1)^{r_1} \times K[t]/(t-a_2)^{r_2}\times\cdots\times K[t]/(t-a_k)^{r_k}..$$

Moreover, for any $a_i\in K$, we have that $K[t]/(t-a_i)^r \cong K[t]/(t^r)$, with the isomorphism induced by mapping $t$ to $a_i+(t^r)$ in $K[t]/(t^r)$.

So if $p(t) = (t-a+1)^{r_1}\cdots (t-a_k)^{r_k}$, then $$K[t]/I \cong K[t]/(t^{r_1}) \times \cdots \times K[t]/(t^{r_k}).$$

Since you want $K[t]/I$ to not be a nontrivial direct product, that means that $p(t)$ is a power of an irreducible polynomial, so $K[t]/I \cong K[t]/(t^r)$, where $r=\deg(p)$.

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Very nice, thank you. And if $K[t]/I$ has dimension $3$ then $r=3$ no? –  Learner Nov 7 '11 at 5:44
    
The dimension of $K[t]/(t^n)$ as a $K$-vector space is $n$, yes. –  Arturo Magidin Nov 7 '11 at 5:59

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