Ideal in a polynomial ring quesiton

Question

Let $I$ be an ideal in $k[x_1,\ldots,x_n]$:

(a) Prove that $1 \in I$ only if $I = k[x_1,\ldots, x_n]$

(b) More generally, prove that $I$ contains a non zero constant if $I = k[x_1,\ldots,x_n]$.

Answer 1

Or more generally:

Let $R$ be a unitary ring and $I\subseteq R$ a (left) ideal. The following statements are equivalent

(1) $I=R$

(2) there exists an invertible element $r\in R$ such that $r\in I$

(3) $1\in I$.

(1) $\Rightarrow$ (2) Obvious, $1$ is such element.

(2) $\Rightarrow$ (3) Let $r\in I$ be invertible. Then $1=r^{-1}\cdot r\in I$ since $I$ is a (left) ideal.

(3) $\Rightarrow$ (1) Let $a\in R$. Then $a=a\cdot 1\in I$ again since $I$ is a (left) ideal.

Now all you have to do to complete your statement is to prove that polynomial $f\in k[x_1,\ldots,x_n]$ is invertible if and only if it is a nonzero constant. Can you complete the proof?

Answer 2

Let $I$ be an ideal in a ring $R$ with 1. Then we have that, trivially, that $I$ contains a unit (actually all units of $R$).

Conversely, if $I$ contains a unit $r \in R$. This implies that there is an $s \in R$ such that $r\cdot s = 1$. By definition of an ideal, this implies that $rs = 1 \in I$. However, this implies that if $x \in R$, then $x \cdot 1 = x \in I$. Therefore we have that $I = R$.

Relating this to your question: can you determine the units in the polynomial ring?

Answer 3

for $1)$ we have $<1>=\{f(x)\bullet 1; f(x)\in k[x_1,..,x_n]\} =\{f(x); f(x)\in k[x_1,..,x_n] \}=k[x_1,..,x_n] $ then if $1\in I \Rightarrow k[x_1,..,x_n]\subseteq I$, there for $k[x_1,..,x_n]= I$

for $2)$ let $a$ be any nonzero constant in $k$ and $a\in I\Rightarrow a^{-1}\in I$ then $1\in I$ , therefor from $(1)$ we have $I=k[x_1,..,x_n]$