Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here is a question from an old exam:


1.

a) Let $R$ be a UFD and let $A=a_{0}x^{m}+...+a_{m}\ne 0 , B=b_{0}x^{n}+...+b_{n} \ne 0$ in $R[x]$ with $\gcd(a_{0},...,a_{m})\in R^{*}$, $\gcd(b_{0},...,b_{n})\in R^{*}$. For $C=AB=c_{0}x^{m+n}+...+c_{m+n}$. Show that $\gcd(c_{0},...,c_{m+n})\in R^{*}$.

b) With $K=\mathbf{C}(t)$, show that $A=x^{2011}+x+t$ in $K[x]$ is irreducible.


I didn't know how to begin, so I asked an older student and he told me to use $f:R\rightarrow S=R/pR$. In retrospect I still don't know how to begin despite the hint. (For b he told me to use $R=\mathbf{C[t]}$ together with a).). Help is greatly appreciated.

share|improve this question
    
Presumably in (b) you mean $\mathbb C(t)$? –  Thomas Andrews Nov 16 '11 at 20:41
    
Thomas Andrews, Yes :). Thanks Arturo Magidin for editing my question. –  PumaDAce Nov 16 '11 at 20:48

3 Answers 3

up vote 3 down vote accepted

For (b): it is even more general. Let $P(x)\in \mathbb C[x]$. Then $P(x)+t\in K[x]$ is irreducible.Suppose
$$ P(x)+t=f_1(x)f_2(x), \quad f_1(x), f_2(x)\in K[x].$$ Let $R_i(t)\in \mathbb C[t]$ be the lcm of the denominators (in $\mathbb C[t]$) of $f_i(x)$. Then $R_i(t)f_i(x)\in \mathbb C[t][x]$ and its coefficents are coprime. Applying (a) to the rhs of the equality
$$ R_1(t)R_2(t)(P(x)+t)=(R_1(t)f_1(x))\times (R_2(t)f_2(x))$$ we see the coefficients of the lfs are also coprome. So $R_1(t)R_2(t)$ is constant, so $R_1, R_2$ are constant and $f_1(x), f_2(x)\in \mathbb C[t][x]$. Now consider the first equality as an equality in $\mathbb C[x][t]$, it implies immediately that (for instance) $f_2(x)$ is constant. So $P(x)+t$ is irreducible.

For (a), Thomas already gave a proof. Let me give a different one. Suppose the $c_i$ are divisible by a prime element $p\in R$. Reducing $A,B$ mod $p$ we see that their product in $(R/pR)[x]$ is zero. But $(R/pR)[x]$ is an integral domain, so for instance $A=0$ mod $p$. But then $p$ divides $a_i$ for all $i\le m$.

share|improve this answer
    
Thank You, QiL. :) :) –  PumaDAce Nov 16 '11 at 22:11

Part (a) is easy.

Let $p$ be prime. Since the $a_i$ are relatively prime, let $i$ be the least value such that $p \not | a_i$. Similarly, let $j$ be the least value such that $p\not | b_j$. Then $$c_{i+j} = \sum_{m=0}^{i+j} a_mb_{i+j-m} = a_ib_j + \sum_{m=0}^{i-1} a_m b_{i+j-m} + \sum_{n=0}^{j-1} b_n a_{i+j-n}$$

But the two sums on the right are divisble by $p$, and $a_ib_j$ is not divisible by $p$, so $p\not | c_{i+j}$.

But that means that no prime divides all the $c_k$.

share|improve this answer
    
Thank You for letting me know Your proof, Thomas Andrews. :) :) –  PumaDAce Nov 16 '11 at 22:12

HINT $\ $ If not, then a prime $\rm\:p\:$ would divide the content of the product but neither factor. However, mod $\rm\:p\:,$ examining leading coefficients of $\rm\:A,B,AB\:$ yields $\rm\: p\nmid a,b\:$ but $\rm\:p\:|\:ab,\:$ contra $\rm\:p\:$ prime; i.e.

$\rm\quad in\ domain\ R/p: \ \ 0 \ne A = a\ x^j + \:\cdots,\ \ 0 \ne B = b\ x^k + \:\cdots\ \Rightarrow\ 0 \ne AB = ab\ x^{j+k} + \:\cdots$

In other words, primes $\rm\:p\in R\:$ remain prime in $\rm\:R[x]\:$ simply because the prime divisor property persists when multiplying leading coefficients. This is one form of Gauss's Lemma.

share|improve this answer
    
Thanks Bill Dubuque :) –  PumaDAce Nov 16 '11 at 22:15

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.