# Factor Rings of Polynomial Rings.

Let $\mathbb F$ be a field and $\mathbb F[x]$ the ring of polynomials with coefficients in $\mathbb F$. Let $p(x)$ be an irreducible polynomial in $\mathbb F[x]$. Let $k$ be a positive integer and consider the vector space $V$, over the field $\frac{\mathbb F[x]}{(p(x))}$with basis

$$1, p(x), p(x)^2, \ldots, p(x)^{k-1}.$$

That is, $$V=\left\{q_0+q_1p(x)+\cdots +q_{k-1}p(x)^{k-1}\;\;:\;\;q_i\in\frac{\mathbb F[x]}{(p(x))}\right\}.$$

Define in $V$ a product modulo $p(x)^k$. That is, if

$$v=v_0+v_1p(x)+\cdots +v_{k-1}p(x)^{k-1}\;\;\text{and}\;\;u=u_0+u_1p(x)+\cdots +u_{k-1}p(x)^{k-1},$$ then we multiply $v$ by $u$ the obvious way, using the distributivity and assuming that $p(x)^k=0$. This makes $V$ an algebra. My question is:

Is this algebra $V$ isomorphic to $\frac{\mathbb F[x]}{(p(x)^k)}$?

• Why do you call the basis elements $p(x)^i$ ? They don't have anything to do with $p(x)$, right? Or is the product different from what I think? – darij grinberg Jun 7 '13 at 14:56
• No and also V is a field. – Thibaut Dumont Jun 7 '13 at 15:03
• It might help you think about it more cleanly if you write $\pi = p(x) \in F[x]$ and write your basis of $V$ as $\{1, \sigma, \dots, \sigma^{n-1}\}$. Can you write down your map from $F[x]/(\pi)^n \to V$ explicitly? It's trivially a ring homomorphism (you defined the multiplication on $V$ so that it was); can you see that it suffices to check that $F[x]/(\pi)^n$ is $(F[x]/(\pi))^n$ as a vector space in the obvious way? Can you check whether $F[x]/(\pi) \cong (\pi)/(\pi)^2$ as abelian groups? (Consider the map $F[x] \to F[x]$ given by $f \mapsto \pi f$.) You may like to look up regular rings. – Billy Jun 7 '13 at 16:20

## 3 Answers

If I understand well $V=L^k$, where $L=\mathbb F[X]/(p)$ is a field. On $V$ you define a multiplication by $$(a_0,a_1,\dots,a_{k-1})(b_0,b_1,\dots,b_{k-1})=(a_0b_0,a_0b_1+a_1b_0,\dots,a_0b_{k-1}+\cdots+a_{k-1}b_0).$$

It's easy to see that $V\simeq L[Y]/(Y^k)$ and thus your problem reduces to the following:

Let $\mathbb F$ be a field and $p\in\mathbb F[X]$ irreducible. Set $L=\mathbb F[X]/(p)$. Is it true that $L[Y]/(Y^k)$ is isomorphic to $\mathbb F[X]/(p^k)$?

Denote $L$ by $\mathbb F(t)$, where $t$ is the residue class of $X$ modulo $(p)$. (Obviously $p(t)=0$.) I'll prove that $$L[Y]/(Y^k)\text{ is isomorphic to } \mathbb F[X]/(p^k)$$ whenever $p'(t)\neq 0$. (This happens, for instance, if the characteristic of $\mathbb F$ is $0$.)

Define a ring homomorphism $\varphi:\mathbb F[X]\to L[Y]/(Y^k)$ by sending $X$ to $y+t$, where $y$ denotes the residue class of $Y$ modulo $(Y^k)$. We have $\varphi(p(X))=p(y+t)=p(t)+yh(y)$, $h$ being a polynomial with coefficients in $L$ and having the property that $h(0)=p'(t)$. Since $p(t)=0$ we get $\varphi(p(X))=yh(y)$ and using that $y^k=0$ we obtain $\varphi(p^k(X))=0$, that is, $p^k(X)\in\ker\varphi$. But $\ker\varphi$ is a principal ideal (of $\mathbb F[X]$), so $\ker\varphi=(f(X))$ for some $f\in\mathbb F[X]$. Since $f(X)\mid p^k(X)$ we must have $f(X)=p^i(X)$ for $1\le i\le k$. Suppose $i<k$. Then $\varphi(p^i(X))=0$ which is equivalent to $y^ih^i(y)=0$, that is, $Y^ih^i(Y)\in(Y^k)$. We thus get $Y\mid h(Y)$, a contradiction (with $h(0)\neq 0$). As a consequence we have proved that $\ker\varphi=(p^k(X))$. It remains to prove that $\varphi$ is surjective. But this follows easily observing that $\varphi$ is a homomorphism of $\mathbb F$-vector spaces of the same dimension (equal to $k\deg p$).

• Could you explain how you know $\varphi$ is a homomorphism? – Thoth Jun 9 '13 at 6:28
• $\phi$ is the unique homomorphism of $F$-algebras mapping $X$ to $y+t$. It exists because of the general universal property of the polynomial algebra $F[X]$. – Martin Brandenburg Jun 11 '13 at 13:12
• @YACP: What about the case that $p$ is not irreducible? One may still ask if $F[X,Y]/(p(X),Y^k)$ is isomorphic to $F[X]/(p(X)^k)$. – Martin Brandenburg Jun 11 '13 at 13:15

See this follow-up question for a discussion of coefficient fields.

By the proof of Cohen's Structure Theorem, there is a coefficient field inside $A = \mathbb{F}[X]/(p^k)$, hence an injective map $L \to A$, whose image is a complement to the maximal ideal $\bar{p}A$ of $A$. Extend it to an $L$-algebra homomorphism $\varphi : B = L[Y]/(Y^k) \to A$ by sending $\bar{Y}$ to $\bar{p}$. We can now view both $A$ and $B$ as $L$-vector spaces. Since $(p^i)/(p^{i+1})$ is isomorphic to $L$ for all $i \geq 0$, $\dim_L A = \dim_LB = k$. Therefore $\varphi$ is an isomorphism.

No, the algebra you've defined is isomorphic to $\mathbb F[x]/x^k$, where $x$ is represented by $p(x) \in V$. It has dimension $k$ whereas $\mathbb F[x]/p(x)^k$ has dimension $k\cdot\deg p$.

• It often helps to use different indeterminate variables; i.e. say that it is isomorphic to $F[y]/y^k$ by the correspondence $p(x) \leftrightarrow y$. – Hurkyl Jun 8 '13 at 20:58