# Prime ideals in polynomials rings

Let $A$ be a commutative ring, $\mathbb{q}\subset A$ an ideal of $A$, and $\mathbb{q}A[x]$ the ideal of $A[x]$ generated by $\mathbb{q}$ (consists of the polynomials with coefficients in $\mathbb{q}$). Show $\mathbb{q}$ is prime in $A \Leftrightarrow \mathbb{q}A[x]$ is prime in $A[x]$.

I've attempted to construct an isomorphism from $(A/\mathbb{q})[x]\to A[x]/\mathbb{q}A[x]$, but I'm not really sure how to go about doing so other than some appeal to the first isomorphism theorem for rings.

-

## 4 Answers

Just consider the natural map $A[x]\rightarrow A/q[x]$ and note that its kernel is exactly $qA[x]$.

-

It has been a long night; I can't believe I didn't realize that the kernel is $\mathfrak{q}A[x]$... – chris Feb 13 '12 at 18:47

To prove that $qA[x]$ is again prime you have to prove that for $P\cdot Q \in qA[x]$ either $P$ or $Q$ is in $qA[x]$. Induction on the sum of the degrees of $n=deg(P)$ and $m=deg(Q)$. If $n=0$ it follows from the fact that $q$ is prime. If $n>0$ $$PQ = a_{n}b_{m}X^{n+m}+\ldots$$ so $a_{n}b_{m}\in q$ and wlog $a_{n}\in q$. Now $P=P' +a_{n}X^{n}$ and $$PQ = P'Q+a_{n}X^nQ\in qA[x]\Rightarrow P'Q\in qA[x]$$ Now use the induction hypothesis!

-

We can assume $\mathfrak q=0$ (replace $A$ by $A/\mathfrak q$). Then the statement reduces to:

$A$ is a domain if and only if $A[x]$ is a domain,

which is clear.

-

+1 This is actually a very nice way to look at the problem. – Adrián Barquero Feb 13 '12 at 18:54

Hint: $q$ prime $\:\Rightarrow\:qA[x]$ prime follows from examining leading coefficients mod $q$, i.e.

${\rm mod}\ q:\: f,g \not\equiv 0\:\Rightarrow f \equiv a\:x^n+\cdots,\ \ g \equiv b\:x^m+\:\cdots,\ a,b\not\in q,\:$ where $\cdots$ means lower degree terms. Hence we deduce $\:fg \equiv ab\:x^{n+m}+\cdots,\ ab\not\in q\:\Rightarrow\: fg \not\equiv 0$.

Note: for $q = 0$ the proof becomes: domain $A\:\Rightarrow\:$ domain $A[x]$. The proof works by exploiting the multiplicativity of the leading coefficient map combined with the nonexistence of zero-divisors in a domain, to show that the product of nonzero polynomials remains nonzero because the same holds true for their leading coefficients.

This is a prototypical example of using a multiplicative map to transfer multiplicative properties from one monoid to another (here the nonexistence of zero-divisors). For some further examples, see this post on using norms to transfer multiplicative properties between $\mathbb Z$ and rings of algebraic integers, and also this post where one deduces the existence of inverses of domain elements $\ne 0$ that are algebraic over a subfield (generalizing rationalizing denominators).

Note: one can reduce to the case $A$ is a domain by factoring out by the prime $q$. However, when learning these techniques, it is instructive to compare the structural vs. element-wise proofs.

Note also that this may be viewed as a generalization of one form of Gauss's Lemma.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 45, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9367977380752563, "perplexity_flag": "head"}