# Isomorphic rings or not?

Prove that the ring $\mathbb F_2(T^2)[X]/((X^2+T^2)^2)$ is (or is not) isomorphic to $\mathbb F_2(T)[Y]/(Y^2)$.

Remark. The above question is related to this topic, where it's proved that for $p(X)=X^2+T^2$ such an isomorphism holds provided $p'\neq 0$, i.e., over fields of characteristic $\neq 2$.

• @MartinBrandenburg, the claim that «it is a wrong notation» is a common mistake... It can be traced back to the rather surprising idea that there is a correct notation. – Mariano Suárez-Álvarez Jun 19 '13 at 7:37

They are isomorphic.

Let $k := \mathbb{F}_2(T)$ and let $k' := \mathbb{F}_2(T')$ be its subfield generated by the element $T' := T^2$.

With this notation, $A = k'[X] / ((X^2 + T')^2)$ and $B = k[Y]/(Y^2)$, and we will construct an isomorphism between $A$ and $B$. Let $\epsilon$ be the image of $X^2 + T'$ in $A$; then $\epsilon^2 = 0$ and $A / A\epsilon \cong k'[X] / (X^2 + T')$ is isomorphic to $k$, and is therefore a field.

Hence $A$ is a local ring with non-zero principal maximal ideal $\epsilon A$ of square zero, and the residue field of $A$ is isomorphic to $k$. Note that $ann(\epsilon) = \epsilon A$ since $\epsilon \neq 0$ and $\epsilon A$ is a maximal ideal.

It will therefore be enough to find a subfield $F$ of $A$ isomorphic to $k$ such that $F \oplus \epsilon A = A$, because we can then write down an isomorphism from $B$ onto $A$ which sends $k$ onto $F$ and $Y$ to $\epsilon$.

Such a subfield is called a coefficient field in the literature; see for example the discussion before Theorem 28.3 in Matsumura's book "Commutative Ring Theory". Part (ii) of this theorem also implies that such a field $F$ exists in this situation. Incidentally, the existence of coefficient fields in equi-characteristic complete local rings is a key step in the proof of the important Cohen Structure Theorem.

Here is an explicit (far from canonical) example of such a coefficient field $F$ inside $A$. Let $R$ be the subring of $A$ generated by $X$. Then the image of $R$ in $A/\epsilon A \cong k$ is generated by $T$, so this image is the polynomial ring $\mathbb{F}_2[T]$. Hence $R$ is also isomorphic to $\mathbb{F}_2[T]$. Furthermore, if $0 \neq r \in R$ and $r a = 0$ in $A$ for some element $a \in A$, then $a = \epsilon b$ for some $b \in A$ since the image of $r$ in the field $A / \epsilon A$ is non-zero, but then $\epsilon rb = 0$ forces $rb \in ann(\epsilon) = \epsilon A$ so again $b \in \epsilon A$ and hence $a = \epsilon b \in \epsilon^2A = 0$. Thus no non-zero element of $R$ is a zero-divisor in $A$. Since $A$ is Artinian, every non-zero element of $R$ is actually a unit in $A$, so the embedding $\mathbb{F}_2[T] \hookrightarrow A$ extends to an embedding $\varphi : k = \mathbb{F}_2(T) \hookrightarrow A$. Its image $F$ is the required coefficient field.

• Good answer! I wonder if your proof can be extended somehow to the general case (which suggested in fact this problem as the OP mentioned): "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 $\mathbb F[X]/(p^k)$?" – user89712 Jan 5 '14 at 22:20
• Yes, the same proof will work. – Konstantin Ardakov Jan 5 '14 at 22:45
• @user: Thanks. I've also added an answer to original question on $F[X]/(p^k)$. – Konstantin Ardakov Jan 5 '14 at 23:21