# Show that $P[X]$ is a prime ideal of $R[X]$

i) Let $P$ be a prime ideal of $R$. Show that $P[X]$ is a prime ideal of $R[X]$.
where $P[X]$ is the set of polynomials in $X$ with coefficients in $P$

ii) If $P$ is maximal in $R$, is $P[X]$ necessarily maximal in $R[X]$? Prove or give a counterexample.

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What have you tried? –  Tobias Kildetoft Feb 7 '13 at 15:22
You got few good answers. Could you bother to accept one of them? –  user26857 Feb 8 '13 at 15:16

## 4 Answers

Hint for both: Show $R[X]/P[X] \cong (R/P)[X]$.

For (i), use that an ideal $I$ of a commutative ring $A$ is prime if and only if $A/I$ is an integral domain.

For (ii), use that an ideal $I$ of $A$ is maximal if and only if $A/I$ is a field.

It is often useful to first look at the simple case of $R=\mathbb Z$ for examples.

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## First question

This can be done by Proof by Contradiction.

Let $f(x) = \sum\limits_{i=0}^n \alpha_i x^i$; and $g(x) = \sum\limits_{i=0}^m \beta_i x^i$ (of course, we also assume $\alpha_n \neq 0$, and $\beta_m \neq 0$), now assume that $f(x).g(x) \in P[x]$, but both $f(x); g(x) \notin P[x]$. So there exists the smallest $n_0$, and $m_0$, where $a_{n_0} \notin P$; and $b_{m_0} \notin P$. Can you find a contradiction from here?

## Second question

No, that's not true, consider $R = \mathbb{Z}$; and $P = 3\mathbb{Z}$.

We'll have $P[x] \subsetneq P[x] + <2x> \subsetneq R[x]$; since $1 \notin P[x] + <2x>$

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Answer to i)

I first checked the definition here: http://en.wikipedia.org/wiki/Prime_ideal, since I haven't touched prime ideals in a long time. I will use that the second condition for $R$ to be prime in $P$ is clearly equivalent to: $a\neq 0$ and $b\neq 0$ implies $ab\neq 0$ in $P/R$.

So there are two things to verify.

1) Since $P\subsetneq R$, it is clear that $P[x]\subsetneq R[X]$.

2) Let $f(X)=a_nX^n+\ldots+a_0$ ($a_n\neq 0$) and $g(x)=b_mX^m+\ldots+b_0$ ($b_m\neq 0$) be two nozero polynomials in $P[X]/R[X]=(P/R)[X]$.

Then $f(X)g(X)=a_nb_mX^{n+m}+\ldots +a_ob_0$. Since $P$ is prime and both $a_n$ and $b_m$ are nonzero, we have $a_nb_m\neq 0$, so $f(X)g(X)\neq 0$.

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Prime ideals can have zero divisors, in fact in a commutative ring $R$ any prime ideal necessarily contains every zero divisor. –  JSchlather Feb 7 '13 at 15:35
@JacobSchlather Thank you Jacob. I forgot to mention: in the quotient ring! –  1015 Feb 7 '13 at 15:50

While one can prove in a structural way that a prime ideal $\rm\:P\subset R\:$ extends to a prime ideal $\rm\:P[x]\subset R[x],\:$ via $\rm\,R[x]/P[x] \cong (R/P)[x],\:$ it is instructive to examine it at the element-level.

$$\begin{eqnarray} \rm f\not\in P[x]\:\Rightarrow\: &\rm mod&\rm\ P\!:\ \ f &\equiv&\rm\ \ f_j\ \ x^j\,\ \ \ +\,\cdots\, +\,f_0,\ &\rm\ \color{#C00}{f_j} &\rm\color{#C00}{ \not\equiv 0}\\ \rm g\not\in P[x]\:\Rightarrow\: &\rm mod&\rm\ P\!:\ \ g &\equiv&\rm\ \ g_k\ x^k\,\ \ +\,\cdots\,+\, g_0,\ &\rm\ \color{#C00}{ g_k} &\rm \color{#C00}{\not\equiv 0}\\ \rm \Rightarrow\: &\rm mod&\rm\ P\!:\ fg &\equiv&\rm f_j g_k x^{j+k} +\,\cdots\, + f_0 g_0,\ &\rm \color{#C00}{f_j g_k }&\rm \color{#C00}{\not\equiv 0}\ \ by\ P\ prime\\ \rm \Rightarrow\: &\rm fg \not\in&\rm\ P[x]\\ \end{eqnarray}$$

Thus primes $\rm\:P\subset R\:$ extend to primes $\rm\:P[x]\subset R[x]\:$ simply because the characteristic property of primes: $\rm\ \color{#C00}{a,b\not\in P\:\Rightarrow\:ab\not\in P}\:$ persists mod $\rm\,P\,$ for the $\,\ell$eading coefficients, i.e. $$\rm mod\ P\!:\ \ \ell(f),\ell(g)\not\equiv 0\:\Rightarrow\:\ell(fg)\not\equiv 0\qquad\quad$$ which immediately yields $\rm\ mod\ P\!:\ deg\,fg\, =\, deg\,f + deg\, g,\:$ an equivalent propert of polynomials over domains, which is more familiar.

The above is at the heart of some simple forms of Gauss's Lemma.

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