Let $R$ be a commutative ring with unit and let $a,b\in R$ be two elements which together generates the unit ideal. Show that $a^2$ and $b^2$ also generate the unit ideal together.
My Work:
Unit ideal is $R$. So, $R=\{ra+sb|r,s\in R\}$. Let $S=\{ra^2+sb^2|r,s\in R\}$. Then $S\subset R$. But, stuck in proving $R\subset S$. Can anyone please help me?
A useful characterization of the unit ideal is that an ideal is the unit ideal if and only if it contains $1$. e.g. your hypothesis is that there are elements $m$ and $n$ such that $am + bn = 1$.
Another hint, since I think I always have to spend a while to rediscover it every time I feel like I need to prove this, is as an intermediate step, try and show that $a^2$ and $b$ generate the unit ideal together. There may be other approaches that don't use this intermediate step, though.
${\bf Hint}\quad\ (a,b)\,(a^2,b^2)\, =\, (a,b)^3.\, $ Specializing $\,(a,b) = (1)\,$ immediately yields the result.
Remark $ $ To prove this ideal equality is very easy: $ $ expand both sides using the distributive law to prove that both sides $\,= (a^3, a^2b, ab^2,b^3),\,$ just like expanding polynomials. See here for more.
For related generalizations such as the Freshman's Dream for ideals and gcds see this answer.
Suppose you have a ring $R$, and $\mathfrak a,\mathfrak b$ are ideals with $\mathfrak a+\mathfrak b=1$, that is, they are comaximal. Then we can find $a,b\in\mathfrak a,\mathfrak b$ resp. with $a+b=1$. Then $a^2+2ab+b^2=1$, and $a^2\in\mathfrak a^2$, $2ab+b^2\in \mathfrak b$. This means $\mathfrak a^2+\mathfrak b=1$. Doing the same with this, we obtain $\mathfrak a^2+\mathfrak b^2=1$, (i.e. apply the last with $\mathfrak a'=\mathfrak a^2$). If $\mathfrak a=(a)$ and $\mathfrak b=(b)$, we get your result.
Of course, we can generalize this to any $m,n$ to get $\mathfrak a^m+\mathfrak b^n=1$. A related situation is true: if the radicals of $\mathfrak a,\mathfrak b$ generate the whole ring, then in fact $\mathfrak a,\mathfrak b$ generate the whole ring. If we have an equation $x+y=1$; and if $x^n\in \mathfrak a,y^n\in\mathfrak b$, show by means of the binomial theorem that $(x+y)^{n+m}=1$ can be written as a sum $A+B=1$ where $A\in\mathfrak a,B\in\mathfrak b$.
Since I'm not afraid to use the Axiom of Choice let me post a one liner proof:
$(a^2,b^2)\ne R\Rightarrow\exists\ \mathfrak m\in\operatorname{Max}(R)$ s.t. $(a^2,b^2)\subset \mathfrak m$. Then $a^2,b^2\in\mathfrak m$, so $a,b\in\mathfrak m$, a contradiction.
Yet one more argument: Our hypothesis is that there are integers $M$ and $N$ with $Ma+Nb=1$; our aim is to find integers $M'$ and $N'$ with $M'a^2+N'b^2=1$.
Well, take $Ma+Nb=1$ and cube: $$ 1=M^3a^3+3M^2Na^2b+3MN^2ab^2+N^3b^3=(M^3a+3M^2Nb)a^2\>+\>(3MN^2a+N^3b)b^2\,, $$ and there you are.
