# Explicitly generating 1 in an ideal without prime support

### The Question

Let $R$ be a unital commutative ring, and let $a,b_1,b_2\in R$. The following is a basic commutative algebra exercise.

Lemma. If $Ra+Rb_1=R$ and $Ra+Rb_2=R$, then $Ra+Rb_1b_2=R$.

Proof. Let $P$ be any prime ideal containing $Ra+Rb_1b_2$. Since $b_1b_2\in P$, then $b_1\in P$ or $b_2\in P$. In either case, $P=R$. Therefore, no prime ideal contains $Ra+Rb_1b_2$, so it is all of $R$.

Its a slick proof, but its also very nonconstructive. My question is; given $f_1,f_2,g_1,g_2\in R$ such that $$f_1a+g_1b_1=1= f_2a+g_2b_2$$ can you construct $f,g\in R$ such that $$fa+gb_1b_2=1?$$ I'm willing to be fairly lax in my standards for a 'construction', in that it doesn't have to be a closed formula. I just don't want it to use an embedding in a hypothetical prime ideal.

### My Motivation

My practical interest in this comes from a non-commutative analog of this problem. I am considering non-commutative $R$ and quasi-commuting elements $a,b_1,b_2$. That is, $ab_1=\lambda b_1a$ for some unit $\lambda$ (and likewise for other pairs).

I would like to deduce that $$Ra+Rb_1=R \text{ and } Ra+Rb_2=R \text{ implies } Ra+Rb_1b_2=R$$ Quasi-commuting elements are close enough to commutative that many constructions still work. However, one tool which does not generalize is primary decomposition. Therefore, I would like a more explicit commutative proof, in the hopes that it will work in the quasi-commuting case also.

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In the commutative case, you can construct $f$ and $g$ by $f=f_1af_2 + f_1g_2b_2 + g_1b_1f_2$ and $g = g_1g_2$ (to prove this, multiply $f_1a+g_1b_1=1$ with $f_2a+g_2b_2=1$). In the noncommutative case, I don't feel that it will work, but don't take my intuition too seriously. The problem is, I think, that you don't require $a$, $b_1$, $b_2$ to quasi-commute with $f_1,f_2,g_1,g_2$. – darij grinberg Feb 1 '12 at 23:03
PS: For my $f$ and $g$ to work, you need only three quasi-commuting pairs: $\left(b_1,g_2\right)$, $\left(a,g_2\right)$ and $\left(a,b_2\right)$. – darij grinberg Feb 1 '12 at 23:05
Alternatively, you can multiply $f_1a+g_1b_1=1$ by $b_2$ from the right, then plug that for $b_2$ in $f_2a+g_2b_2=1$. Then factor $a$. But you will need $a$ to commute with $b_2$. – Mahdi Majidi-Zolbanin Feb 2 '12 at 0:40
You will get $f=f_2+g_2f_1b_2$ and $g=g_2g_1$. – Mahdi Majidi-Zolbanin Feb 2 '12 at 0:45

Here is an example to show it can be done. Consider the q- polynomial ring $R=k_q[x,y]$, the free algebra on two variables over a field $k$ with defining ideal $qxy=yx$ where $q \in k^{\ast}$. (Usually I would be worried about whether $q$ is a root of unity or not but my example doesn't depend on this).
We have $R(1+xy)+Rx=R$ since $1 \cdot (1+xy)+(-q^{-1}y)\cdot x=1$. Similarly (and rather trivially) we have $R(1+xy)+Rxy=R$. In your notation I'm taking $a=(1+xy), b_1=x,b_2=xy$
The left ideal that you are interested in is $R(1+xy)+Rx^2y$, where the order I multiply $b_1$ and $b_2$ doesn't matter as I can cancel the $q$-powers if I want. I would like $f(x,y)$ and $g(x,y)$ such that $$1=f(x,y)(1+xy)+g(x,y)x^2y$$ I can take $f(x,y)=1-xy$ and $g(x,y)=q^{-1 } y$ and this will solve the problem. In this ring I think it may always be possible (provided that the elements you start with satisfy the hypotheses, which is quite restrictive), due to the fact you can write things nicely in a standard form as I have done above. Also the graded structure may may play a part in this positive example.
EDIT: I should point out in relation to darij's comment above that in this ring everything "nearly"quasi-commutes, and for homogeneous elements this is true; it is only when we consider non-homogeneous elements that we will get different powers of $q$ appearing to stop them quasi-commuting. EDIT 2: In fact I just noticed the second comment about only needing 3 quasi-commuting pairs, which may be what my example shows explicitly.