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Find a left ideal of $\mathbb{H}[X]$ that is a maximal, but not a ideal.

$\mathbb{H}[X]$ is just the polynomial extension.

$\mathbb{H}=\{a+bI+cJ+dK \mid a,b,c,d \in \mathbb{R} \} .$

$\mathbb{H}[X]=\{\alpha_0+\alpha_1X+\cdots+\alpha_nX^n: \alpha_0,\alpha_1,\dots,\alpha_n\in \mathbb{H}\}$

First thought is something like $\mathbb{H} \times (XI)$. However, sort of stuck. I know you have to use the non-commutativity of quaternions somewhere.

Second thought is that you have to use something about inverses. Well, it can't have $1$ in it. Since you would get itself. So, ... I'm pretty sure $\mathbb{H} \times (XI)$ is the left ideal. However, can't see how you would get not a right ideal.

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    @BrunoStonek Yeah,$I$can see what you are saying. However, to be honest I'm more worried about passing exams at the moment. That question came up with exam two years ago. The exam is tomorrow.2012-01-25

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New answer

Let $\mathbb H$ be the $\mathbb R$-algebra generated by the symbols $i_0$ and $j_0$ subject to the relations $ i_0^2=-1=j_0^2,\quad i_0\ j_0=-j_0\ i_0. $

Let $X$ be an indeterminate and define the $\mathbb R[X]$-algebra $\mathbb H[X]$ by $ \mathbb H[X]:=\mathbb R[X]\underset{\mathbb R}{\otimes}\mathbb H. $ We must find a maximal left ideal $\mathfrak a$ of $\mathbb H[X]$ which is not an ideal.

Put $ \mathbb C:=\frac{\mathbb R[X]}{(X^2+1)} $ and denote by $x\in\mathbb C$ the canonical image of $X$.

Note that the left ideal $\mathfrak b\subset\mathbb H[X]$ generated by $X^2+1$ is an ideal. In particular $ A:=\frac{\mathbb H[X]}{\mathfrak b} $ is a $\mathbb C$-algebra.

If $\pi:\mathbb H[X]\to A$ is the canonical projection, and if $\mathfrak c$ is a maximal left ideal of $\mathbb H[X]$ which is not an ideal, then $\mathfrak a:=\pi^{-1}(\mathfrak c)$ fits the bill.

Let $M$ be the $\mathbb C$-algebra of two by two matrices with entries in $\mathbb C$, and let $i,j\in A$ be the canonical images of $1\otimes i_0,1\otimes j_0\in\mathbb H[X]$.

One easily checks that there is a unique $\mathbb C$-algebra morphism $\phi:A\to M$ satisfying $ \phi(i)=\begin{pmatrix}x&0\\0&-x\end{pmatrix},\quad \phi(j)=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, $ and that $\phi$ is bijective.

It is also easy the verify that the left ideal of $M$ generated by $ e:=\begin{pmatrix}1&0\\0&0\end{pmatrix}=\phi\left(\frac{1-ix}{2}\right) $ is maximal, and that it is not an ideal.

This implies that we can put $ \mathfrak c:=A\ \frac{1-ix}{2}\quad, $ or, in other words, that $ \pi^{-1}\left(A\ \frac{1-ix}{2}\right) $ is a maximal left ideal of $\mathbb H[X]$ which is not an ideal.

Old answer

Let $\mathfrak a$ be the ideal of $\mathbb H[X]$ generated by $X^2+1$, let $A$ be the quotient $\mathbb H[X]/\mathfrak a$.

It suffices to find a maximal left ideal of $A$ which is not an ideal.

Write $x\in A$ for the canonical image of $X$. Let us identify $\mathbb R[x]$ to $\mathbb C$. Then $A$ is isomorphic, as a $\mathbb C$-algebra, to $M_2(\mathbb C)$.

(See the very beginning of the text of Gaëtan Chenevier named Lecture 6 on this html page. Here is a direct link to the pdf file.)

As a result, $A$ contains an idempotent $e$ with $1\neq e\neq0$, and $Ae$ is the sought-for maximal left ideal.

EDIT. One can put $ e:=\frac{1+ix}{2}\quad. $ (I use the traditional notation $i,j,k$ instead of the $I,J,K$ in the question.)

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I am assuming that in the polynomial ring $R=\mathbb{H}[X]$ the indeterminate $X$ commutes with all the constants, for otherwise what is $Xq$ for some $q\in\mathbb{H}$?

Unless I am very wrong, there is a one-sided division algorithm in $R$. So if we consider the left ideal $R(X-I)$ then the quotient space $R/R(X-I)$ is a 1-dimensional space over $\mathbb{H}$, because all the cosets of $R(X-I)$ contain a constant polynomial (divide any polynomial from the right by $X-I$ and take the remainder). Hence the left ideal $R(X-I)$ is maximal, because it is of 'codimension one'.

However, $R(X-I)$ is not an ideal of $R$. We have $ J(X-I)-(X-I)J=(JX-JI)-(JX-IJ)=-JI+IJ=2K. $ So any ideal containing $(X-I)$ must also contain the unit $2K$, and hence is equal to all of $R$.

We need to be careful here. Because $\mathbb{H}$ is not commutative, we don't get a ring homomorphism from $\mathbb{H}[X]$ to $\mathbb{H}$ by evaluating polynomials at a selected element $q\in\mathbb{H}$.