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Here is a question from an old exam:


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.

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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
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 $i\le m$.

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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 $c_k$.

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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. But mod $\rm\:p\:,$ examining leading coef's of $\rm\:A,B,AB\:$ $\Rightarrow$ $\rm\: p\nmid a,b\:$ but $\rm\:p\:|\:ab,\:$ contra $\rm\:p\:$ prime, i.e.

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

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

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Thanks Bill Dubuque :) – PumaDAce Nov 16 '11 at 22:15

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