Subrings $A$ of $\mathbb{F}_p[x]$ such that $\dim_{\mathbb{F}_p}\mathbb{F}_p[x]/A=1$.

Question

Suppose we have the ring $\mathbb{F}_p[x]$ of polynomials with coefficients in the Galois field with $p$ prime number elements. I want to find all subrings $A$ containing the identity such that when considered as vector spaces over $\mathbb{F}_p$, the quotient $\mathbb{F}_p[x]/A$ is isomorphic to $\mathbb{F}_p$. Also want to find the isomorphism classes of these subrings.

So far I have only been able to find one subring: the ring generated by $1,x^2,x^3$. I know that if $x\in A$, then $A=\mathbb{F}_p[x]$, so I don't want $n\cdot x$ to be in $A$. After this, trying to find other subrings, it becomes really messy from my point of view. So I don't think this might be a good way to approach the problem.

It also seems to me that we can view this as an extension problem: $$0\rightarrow A\rightarrow \mathbb{F}_p[x]\rightarrow\mathbb{F}_p\rightarrow 0$$ And for groups, I know that using the second homology group would help if I were given $Q$ and $N$ but not $G$ in the following sequence: $$0\rightarrow N\rightarrow G\rightarrow Q\rightarrow 0$$ My questions are if there is a standard method for computing $N$ given $G$ and $Q$, and if these methods generalize to ring extensions? Any help would be appreciated, specially references to literature.

Answer 1

Building upon the ideas from Dirk's answer (+1) and Arthur's comment under the question.

The subring $A$ cannot contain a polynomial of degree $1$ because, as an $\Bbb{F}_p$-algebra, it would then contain $x$ and be equal to $\Bbb{F}_p[x]$. As Dirk pointed out, this implies that $\Bbb{F}_p[x]=A\oplus x\Bbb{F}_p$, and consequently there exist unique constants $a_2$ and $a_3$ such that $$ P=x^2-a_2x, Q=x^3-a_3x\in A. $$ It is straightforward to verify that these polynomials satisfy the relation $$ Q^2=P^3+3a_2PQ-2a_3P^2+(a_2^3-a_2a_3)Q+(a_3^2-a_2^2a_3) P. (*) $$ Such a relation has to exist because the space of polynomials of degree $\le6$ in $A$ and vanishing at $x=0$ has dimension five, so the set of six polynomials $P^3,Q^2,QP,P^2,Q,P$ must be linearly dependent. Finding the relation was then simple linear algebra.

The equation $(*)$ implies that the $\Bbb{F}_p[P]$-module $$ \tilde{A}=\Bbb{F}_p[P]+Q\Bbb{F}_p[P] $$ is actually a subring. The main point is that $(*)$ proves that $\tilde{A}$ is stable under multiplication by $Q$.

It follows easily that the set $$ \mathcal{B}=\{P^i\mid i\in\Bbb{N}\}\cup\{QP^i\mid i\in\Bbb{N}\} $$ is an $\Bbb{F}_p$-basis for $\tilde{A}$. Because the collection $\mathcal{B}$ has monic polynomials of degrees $0,2,3,4,\ldots,$ it follows that $\tilde{A}$ has codimension one in $\Bbb{F}_p[x]$.

Obviously $\tilde{A}\subseteq A$ so we can conclude that $A=\tilde{A}$. Equally obviously distinct choices of $a_2,a_3$ give rise to distinct subspaces $\tilde{A}$, so we arrive at the answer:

Every such subgring $A$ is determined by two parameters $a_2,a_3\in\Bbb{F}_p$ when the ring $A$ is generated by $x^2-a_2x$ and $x^3-a_3x$. Different choices for the parameters give rise to different subrings $A$.

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A few closing remarks:

• Dirk was definitely on the right track. He may have missed the relation $$P^2+2a_2Q-a_2^2P=x^4+(a_2^3-2a_2a_3)x,$$ that ties his $a_4$ to $a_2$ and $a_3$. Similarly, we can tie all $a_i, i>4$, to $a_2,a_3$.
• Arthur was also on the right track, but missed a few possibilities. Also, when $p=2$, all the monomials $(x-a)^2$ have linear coefficient zero, so we get more. Particularly as...
• ... the argument seems to generalize to fields other than prime fields of a positive characteristic.

Answer 2

As all powers of $x$ are linearly independent, the only chance to get what you want is to have all but one power of $x$ contained in your ring $A$, plus maybe lower order terms. However, if your ring $A$ contains $x$, then it also contains $x^2,x^3$, etc., so it is directly the whole of $\mathbb{F}_p[x]$. Therefore, $A$ can't contain $x$.

But that means that $A$ must contain elements $y_i = x^i + a_ix$ for all $i \geq 2$, where $a_i \in \mathbb{F}_p$. On the other hand, every choice of such elements $y_i$ will give you a ring $A$ with the desired properties. Now I think every choice of the sequence $(a_i)_{i \geq 2}$ will give a different, non-isomorphic (as rings) versions of $A$, but it might be better if you properly check that.