which of the following statements are true in ring theory?
(a) Let $R$ is a commutative ring with unity and $I$ be an ideal. Then $R/I$ is an integral domain.
(b) If $R$ is a commutative ring with unity the units of $R[x]$ are the units in $R$.
(a) I think it is not true as $R/I$ is an integral domain iff I is an prime ideal but I have no counter example.
(b) I have no idea.
can I get some help.Thanks for your cooperation.
(a) Which ideals are prime in $\mathbb Z$?
(b) Hint: $(1-aX)(1+aX)=1-a^2X^2$.
What happens if $R$ is a ring and $a \in R$ is such that $a^2=0$? Can you think of any such ring?
(a): Think of a ring $R$ that is not an integral domain, and let $I$ be the zero ideal, i.e. $I=\{0\}$.
(b): Remember, given a ring $S$, we say that an element $y\in S$ is a unit when there is some $z\in S$ such that $yz=1_S$ (the symbol $1_S$ denotes the multiplicative identity element of $S$).
The polynomial ring $R[x]$ consists of polynomials whose coefficients are elements of $R$: $$a_0+a_1x+a_2x^2+\cdots+a_nx^n,$$ where $n$ can be any integer greater than $0$, and each of the $a_i$'s come from $R$. They are added and multiplied like any other polynomials you have worked with before.
Clearly, if $a\in R$ is a unit (so that there is some $b\in R$ such that $ab=1_R$), then the polynomial $$a+0x+0x^2+\cdots$$ is a unit in $R[x]$, because $$\begin{align*} (a+0x+0x^2+\cdots)(b+0x+0x^2+\cdots)&=ab+(a0+0b)x+(a0+0+0b)x^2+\cdots\\ &=ab+0x+0x^2+\cdots\\ &=1_R+0x+0x^2+\cdots \end{align*}$$ and $1_R+0x+0x^2+\cdots$ is the mutliplicative identity of $R[x]$. (Usually, we don't bother writing all of these $0x^i$ terms, and just think of elements of $R$ as also being elements of $R[x]$. These elements of $R[x]$ are called constant polynomials.)
The question is whether there can be units of $R[x]$, other than the units of $R$ themselves. Thus, we're looking for non-constant polynomials (i.e., polynomials with some $x$'s in them) that can be multiplied by other polynomials to produce $1$. For simplicity, you might first take a look at linear polynomials, i.e. polynomials with an $x$ term, but no $x^2,x^3,\ldots$ terms. Of course, there's no guarantee that if there are non-constant polynomials that are units, that any of them will be linear; but this is a good place to start.
So, let's say we have our ring $R$, and elements $a,b\in R$ (with $b\neq 0$, so that we're looking at a non-constant polynomial) such that the polynomial $a+bx\in R[x]$ is a unit. In fact, let's make an even bigger simplifying step: assume that $a+bx$ is its own inverse, or in other words, that $$(a+bx)^2=a^2+2abx+b^2x^2=1.$$ (Again, we don't know that even if there are any polynomials that are units, and are linear, that they would do this; but again, it's a first thing to look at. Remember, mathematicians are lazy.)
This can only happen if $2ab=0$ and $b^2=0$ in $R$. Can you think of a ring $R$ and $a,b\in R$ satisfying these relations?
