Jtdylktuy

Doing some exercises on rings and ideals I found some hurdles on showing that some ideals on the ring $left(mathbb{R}^{mathbb{R}}, +, cdotright)$ are (or not) principal. Hints?

1. Let be $f,g in mathbb{R}^{mathbb{R}}$. Show that the ideal generated by $left(f,gright)$ in $mathbb{R}^{mathbb{R}}$ is a
   principal ideal.


2. Given $f in mathbb{R}^{mathbb{R}}$ and $Z(f):={x in
   mathbb{R},|,f(x)=0}$, then the set $J:={f inmathbb{R}^{mathbb{R}},:,|mathbb{R}setminus Z(f)|lt infty}$
   is an ideal of the ring $mathbb{R}^{mathbb{R}}$ but not a
   principal one.

Showing that $J$ is an ideal is just a matter of set and ideal properties, but the non-principality of that ideal gave me hard times.

Let $R = mathbb R^{mathbb R}$.

If $f,g in R$, then $(f,g) = {af+bg : a,b in R}$. We want to show that this is principal.

Fix $x in mathbb R$. Then, consider the ideal generated in $mathbb R$ by $f(x)$ and $g(x)$. Since $mathbb R$ is a principal ideal domain, there is a single element $y$ depending on $x$ which satisfies $(y) = (f(x),g(x))$. (In fact, since $mathbb R$ is a field, note that it has only two ideals, the whole ring and zero. So $y = 0$ or $y = 1$ holds!)

Define the function $h$ as follows : $h(x)=y$, where $y$ satisfies $(y) = (f(x),g(x))$ as ideals in $mathbb R$.

Of course, $h in R$. We claim that $(h) = (f,g)$. To see this, note that $h in (f,g)$, since for each point $x$ there exists $a_x,b_x in mathbb R$ such that $h(x) = a_xf(x)+b_xg(x)$. Define the functions $a,b in R$ by $a(x) = a_x$ and $b(x) = b_x$ to see that $h = af+bg in (f,g)$.

Next, $f in (h)$, since for each $x$ we have $f(x) in (h(x))$ , so there exists $c_x$ such that $f(x) = c_xh(x)$.You know the rest. Similarly for $g$.

Hence, $(h) = (f,g)$ is a principal ideal!

Note : When do you think $mathbb R^X$ satisfies such a property as the one above? Can you give some conditions on $X$?

A description in words of $J$ is given by those functions which vanish at all but finitely many points. We know it is an ideal, as you have shown that. We need to show that it is not principal.

Well, suppose it was principal. Let $h$ be a generator for the ideal $J$, so that $(h) = J$.

Then, $h$ vanishes only at finitely many points, say a finite set $F$. Consider a function $g$ which does not vanish at some $y notin F$,but is in $J$, then $g=hf$ implies $g(y) = h(y)f(y) = 0$, a contradiction. Hence, $g notin (f)$ but $g in J$, a contradiction.

Note : Do you think finiteness was important here? What about "vanishing at exactly two points", or "vanishing at infinitely many points"?

The point of the above exercise is that $R$ is special, in that every ideal is either principal, or infinitely generated.

$1)$ Any function $h$ such that $h(x) = 0 Leftrightarrow f(x) = 0$ and $g(x) = 0$ will generate $(f,g)$.

$2)$ Suppose $J = (f)$ for some $f$. Then $f$ is non-zero on some finite set $S$. Pick some $g in J$ such that $g(y) neq 0$ for some $y notin S$. Then $g neq hf$ for any function $h$ (since if it were, $g(y) = h(y)f(y) = 0$, a contradiction).

The ring $mathbb{R}^{mathbb{R}}$ is quite well behaved. For $finmathbb{R}^{mathbb{R}}$ set $Z(f)={xinmathbb{R}:f(x)=0}$.

Hint 1: $(f,g)=(f^2+g^2)$

Hint 2: $(f)={hinmathbb{R}^{mathbb{R}}:Z(h)supseteq Z(f)}$; if $J=(f)$, then $fin J$, so $Z(f)$ is cofinite (its complement is finite). Can you find a contradiction?