Question about some algebra theorem

Question

1.

order-presserving = monotonic

But we haven't define order structure on the ring.

2.

I try to prove x is a unit <=> (x)=A ,and fail.

That's what I think:

=>

we want to prove (x)=A.

(x)=A mean Ax=A, so we need to prove f:a->ax is bijective

bijective=injective+surjective

surjective: For every ax,there exist a such that f(a)=ax

injective: failed

presume I succeed in proving it

then |A|=|Ax|

hence A=Ax

<=

I have no idel to prove it.

:(

Answer 1

I suppose we talk about commutative rings.

1. The order is on the set of ideals of the ring (by inclusion) and not on the ring itself.

2. Very good thoughts. Well, actually, in this context $Ax$ denotes the set $\{ax\,\mid\,a\in A\}$, and once you showed surjectivity of $a\mapsto ax$, you are done, as it already proves that this set $Ax$ contains all elements of $A$ (and of course, it cannot contain more).
   However, if $x$ is a unit, it means it has a multiplicative inverse $x^{-1}$, and then $a\mapsto ax^{-1}$ is inverse of $a\mapsto ax$, so it is also injective: $$ax=bx \ \implies a=axx^{-1}=bxx^{-1}=b\,.$$

For the other direction, assume that $Ax=A=(1)$, in particular $1\in Ax$, i.e. $1=ax$ for some $x$, that is, $x\,|\,1$, $\ x\,$ is a unit.

Answer 2

1. The order in question is the partial order by containment on the ideals of the ring, but not on the elements of the ring. This partial order on the ideals always exists.

2. Now for the second question:

presume I succeed in proving it. then |A|=|Ax| hence $A=Ax$

Well, for finite rings that would be true, but there is no reason why $|A|=|Ax|$ should imply that $A=Ax$ in general. This is all a bit beside the point anyway. You don't need to worry about bijectivity of this map. It's simpler than that.

Let's start again. Of course, we always shave that $Ax\subseteq A$. The main problem is to show the other containment, that $A\subseteq Ax$. If $x$ is a unit, then look at $(Ax^{-1})x\subseteq Ax$. If instead you assume that $A\subseteq Ax$, then consider that $1\in Ax$, and go from there.

The thing to take away is that as soon as an ideal contains a unit, it becomes the whole ring. (The converse is obvious for rings with identity.)

Your original strategy of trying to show $a\mapsto ax$ is onto $A$ is actually a valid path, but I just wanted to mention this direct approach too.