In this handout we cover how to factor polynomials—in particular, we give a polynomial time algorithm for factoring polynomials whose coefficients are in a finite field. That is, given an $\renewcommand{\F}{\mathbb{F}}f\in\F_q[x]$, write it as the decomposition

\begin{equation*} f = f_1\cdot f_2\dotsb f_n \end{equation*}

where $f_1$, $\dotsc$, $f_n$ are irreducible polynomials. An *irreducible* polynomial is one that cannot be written as a nontrivial product, i.e., $f$ is irreducible exactly when $f=gh$ implies at least one of $g$ and $h$ are a constant polynomial. Because the coefficients are from a field, all of the leading coefficients of the polynomials $f_i$ are invertible. Thus, one can write the decomposition as

\begin{equation*} f = c f_1\cdot f_2\dotsb f_n \tag{$*$} \end{equation*}

where $c\in\F_q$ and all the $f_i$ are *monic* (have a leading coefficient of 1).
We consider the factorization problem as finding the decomposition ($*$) given $f$.
By dividing both sides by $c$ we also make the leading coefficient on the left-hand side also 1. Thus, without loss of generality we will assume that the $f$ to factor is monic.

Throughout this handout the reader can think of $\F_q$ as being $\renewcommand{\Z}{\mathbb{Z}}\Z_p$ (i.e., the field of residues modulo a prime $p$), as $\Z_p$ is the simplest kind of finite field. However, the algorithms we discuss will also work in general finite fields $\F_q$.

We will break the problem of factoring $f\in\F_q[x]$ into various special cases. One important special case is to handle the case when $f$ is *squarefree*, meaning that in the decomposition $f=\prod_i f_i$ all of the $f_i$ are distinct. That is, a polynomial is squarefree when it is not divisible by the square any polynomial (except constant polynomials). For example, $f=(x+1)^2$ is not squarefree and $f=(x+1)(x-1)$ is squarefree over any field $\Z_p$ with $p>2$.

A second special case of the factoring problem is to $f$ into *distinct-degree factors*, i.e., write $f$ as
$f = g_1\dotsb g_n$ where $g_i$ is the product of all irreducible factors of degree $i$. For example, if $f=(x+1)^2(x-1)(x^2+3)(x^3+x+1)\in\F_5[x]$ then $g_1=(x+1)^2(x-1)$, $g_2=x^2+3$, $g_3=x^3+x+1$, and $g_4=\dotsb=g_8=1$.

The third special case is to factor the $g_i$ that appear in the distinct-degree factorization. That is, given a $g_i\in\F_q[x]$ of degree $d$ whose irreducible factors are all guaranteed to be of a known degree $i$ (and hence there must be $d/i$ of them), find all $d/i$ irreducible factors $g_i=h_1\dotsm h_{d/i}$. For example, given $g_1=x^3+x^2-x-1\in\F_5[x]$ (which is a product of only linear factors), find its decomposition into linear factors $g_1=(x+1)^2(x-1)$.

First, we consider the problem of taking a squarefree monic $f\in\F_q[x]$ of degree $n$ and writing it as the product $g_1\dotsm g_n$ where $g_i$ is the product of the irreducible polynomials of degree $i$ dividing $f$. Our algorithm for doing this will be based on a generalization of Fermat's little theorem in $\F_q[x]$.

Recall Fermat's little theorem says if $p$ is prime then $a^p\equiv a\pmod{p}$, i.e., all $a\in\Z_p$ are roots of the polynomial $x^p-x\in\Z_p[x]$. In fact, over general finite fields $\F_q$ one still has that all $a\in\F_q$ are roots of $x^q-x$. Thus, Fermat's little theorem can equivalently be written as

\begin{equation*} x^q-x=\prod_{a\in\F_q}(x-a). \end{equation*}

Thus, given $f$ one can compute $g_1$, the product of all linear factors of $f$, via $g_1:=\gcd(f,x^q-x)$.

Generalizing this, one can also show that

\begin{equation*} x^{q^2}-x = \prod_{\substack{\alpha\in\F_q[x],\text{ irred.}\\\deg(\alpha)\leq2}}\alpha(x) . \end{equation*}

In follows that once the linear factors of $f$ have been removed (by dividing by $g_1$) one can find $g_2$, all the quadratic factors of $f/g_1$, via $g_2:=\gcd(f/g_1,x^{q^2}-x)$.

This process can be generalized; the proper generalization of Fermat's last theorem that enables finding all irreducible factors of degree $d\geq1$ is

\begin{equation*} x^{q^d}-x = \prod_{\substack{\alpha\in\F_q[x],\text{ irred.}\\\deg(\alpha)\mid d}}\alpha(x) . \end{equation*}

Once all factors of $f$ less than degree $d$ have been removed from $f$ (via $f\cdot\prod_{i<d} g_i^{-1}$) all irreducible factors of degree exactly $d$ is found with $g_d:=\gcd(f\cdot\prod_{i<d} g_i^{-1},x^{q^d}-x)$.

How efficient would this be? If it was computed directly it would be an issue, since it would require $O(nq^d)$ field operations for computing $\gcd(f,x^{q^d}-x)$ where $\deg(f)=n$. Since $d=\Theta(n)$ in the worst case, computing this gcd would take exponential time in $n$. Thus, this proposed approach is seemingly totally infeasible.

A simple observation reduces the exponential running time to a polynomial one. Note that when $\gcd(f,x^{q^d}-x)$ the polynomial does not need to be constructed explicitly; it is enough to construct the polynomial $x^{q^d}-x \bmod f$ which has degree at most $n$. At first this might not seem useful, since computing $x^{q^d}-x \bmod f$ directly would also require exponential time.

However, we can use repeated squaring in order to compute the modular exponentiation $x^{q^d} \bmod f$ efficiently. Also, since

$$ x^{q^d} \bmod f = (x^{q^{d-1}} \bmod f)^q \bmod f $$

and $x^{q^{d-1}} \bmod f$ was used on the previous iteration (in order to extract the factors of $d-1$ from $f$). Thus, on the $i$th iteration one can save $x^{q^i} \bmod f$ and reuse it on iteration $i+1$ without needed to recompute it.

Input: Squarefree and monic $f\in\F_q[x]$ of degree $n$

$h:=x$, $f_\text{orig}:=f$

for $i$ from 1 to $n$:

$\hspace{1em} h := h^q \bmod f_\text{orig}$

$\hspace{1em} g_i := \gcd(h-x,f)$

$\hspace{1em} f := f / g_i$

return $g_1$, $\dotsc$, $g_n$

Assuming naive multiplication, the first operation in the loop uses $O(n^2\log q)$ field operations and the last two operations in the loop use $O(n^2)$ field operations. Thus, the total cost is $O(n^3\log q)$ field operations. With fast multiplication and fast gcd algorithms this cost can be brought down to $O^{\sim}(n^2\log q)$.

Next, we consider the problem of splitting a polynomial whose irreducible factors are all of the same degree (like $g_i$ from above). That is, given a squarefree monic $f\in\F_q[x]$ of degree $n$ and the knowledge that all of $f$'s irreducible factors have degree $d$ we want to decompose $f$ as $f=g_1\dotsm g_{k}$ where $k=n/d$. In fact, it will be sufficient to develop an algorithm that can find any nontrivial factorization $f=g_1\cdot g_2$ with neither $g_1$ or $g_2$ a constant polynomial, because then the problem is split into two smaller subproblems and we can run our splitting algorithm on both $g_1$ and $g_2$ separately.

In this section we assume that $q$ is odd, though a similar algorithm can be developed for the even case.

The Chinese Remainder Theorem says that solving a modular equation mod $n=p_1\dotsm p_k$ (where the $p_i$ are distinct primes) is equivalent to solving the equation mod each prime $p_i$ individually:

\begin{equation*} f(x)\equiv 0\pmod{n} \qquad\Longleftrightarrow\qquad \left\{\begin{aligned} f(x)&\equiv0\pmod{p_1}\\ &\;\vdots\\ f(x)&\equiv0\pmod{p_k} \end{aligned}\right\}. \end{equation*}

It also provides an efficiently-computable way to take a solution of the system of equations on the right-hand side and translate it into a solution on the left-hand side. For example, if $x\equiv a_i\pmod{p_i}$ for $1\leq i\leq k$, it tells you how to find an $x\in\Z_n$ so that $x\equiv a\pmod{n}$. In ring theoretic terms this means that the ring $\Z_n$ is equivalent to the "direct product" of the rings $\Z_{p_i}$:

\begin{align*} \Z_n &\cong \Z_{p_1}\times\Z_{p_2}\times\dotsb\times\Z_{p_k} \\ &\text{via the mapping} \\ x &\mapsto (x\bmod p_1,x\bmod p_2,\dotsc,x\bmod p_k) \end{align*}

In fact, this idea applies to more than the integers modulo $n$; the exact same relationship holds for polynomials modulo $f=g_1\dotsm g_k$ where the $g_i$ are distinct irreducible polynomials. Just like $\Z_n$ is the set of integers with arithmetic performed modulo $n$ (this is also denoted by $\Z/n\Z$ or $\Z/(n)$), the set of polynomials $\F_q[x]$ with arithmetic performed modulo $f$ is denoted by $\F_q[x]/(f)$. Then the Chinese Remainder Theorem for polynomials says that

\begin{align*} \F_q[x]/(f) &\cong \F_q[x]/(g_1)\times\F_q[x]/(g_2)\times\dotsb\times\F_q[x]/(g_k) \\ &\text{via the mapping} \\ f &\mapsto (f\bmod g_1,f\bmod g_2,\dotsc,f\bmod g_k) . \end{align*}

What does the arithmetic of $\F_q[x]/(g)$ look like when $g$ is an irreducible polynomial? In fact, every nonzero polynomial in $h\in\F_q[x]/(g)$ has an inverse $h^{-1}$ which can be computed by solving $h\alpha=1$ for $\alpha$ in $\F_q[x]/(g)$, i.e., solving $h\alpha+g\beta=1$ for $\alpha$, $\beta\in\F_q[x]$. We know how to solve this using the extended Euclidean algorithm on $h$, $g\in\F_q[x]$, assuming that $h$ and $g$ are coprime. (Which they are, since $g$ is irreducible and $\deg(h)<\deg(g)$.)

Thus, $\F_q[x]/(g)$ is a field! When $g$ has degree $d$, the field has the $q^d$ elements $\bigl\{\sum_{i<d} c_ix^i:c_0,\dotsc,c_{d-1}\in\F_q\bigr\}$.

Fermat's Little Theorem also applies to any finite field $\F$; if $\F$ has $q^d$ elements and $a\in\F$ then $a^{q^d-1}=1$. The original version of Fermat's Little Theorem is recovered when $q$ is prime and $d=1$ (so $\F=\Z_p$).

Moreover, since the nonzero elements of any finite field form a cyclic group, what we called the "square root" of Fermat's Little Theorem also holds in any finite field. That is, if $\F$ has $q^d$ elements and $a\in\F$ then $a^{(q^d-1)/2}=\pm 1$. (This is where we assume that $q$ is odd, so that $(q^d-1)/2$ is an integer.)

Now let's go back to the problem of factoring $f\in\F_q[x]$ of degree $n$ where we know that all of its irreducible factors $g_i$ are of the same degree $d$. Suppose we choose a random $\alpha\in\F_q[x]$ of degree less than $n$. If we compute $\alpha^{q^d-1}\bmod f$ what will we get? Note that Fermat's Little Theorem doesn't apply directly here, since $f$ is not irreducible and hence $\F_q[x]/(f)$ is not a field. However, the Chinese Remainder Theorem allows us to write $\F_q[x]/(f)$ as a direct product of fields where Fermat's Little Theorem *does* apply.

By the Chinese Remainder Theorem, computing $\alpha^{q^d-1}\bmod f$ is essentially equivalent to computing

\begin{equation*} (\alpha^{q^d-1}\bmod g_1,\dotsc,\alpha^{q^d-1}\bmod g_k) , \end{equation*}

and because each $g_i$ is specifically known to be irreducible and of degree $d$, by Fermat's Little Theorem the above vector of residues is $(1,1,\dotsc,1)$, at least assuming that $\alpha$ is coprime to each $g_i$. Since $\alpha$ was chosen randomly, it is likely $\alpha$ is coprime to each $g_i$. However, if this is not the case then things are even easier, since a factor of $f$ can be recovered by $\gcd(\alpha,f)$. Thus, we can assume that $\alpha$ and each $g_i$ are coprime.

Thus, assuming $\alpha$ is coprime to $f$ we do in fact have $\alpha^{p^d-1}=1$ in $\F_q[x]/(f)$, because $\alpha^{p^d-1}=1$ in each of $\F_q[x]/(g_i)$ for $1\leq i\leq k$.

We saw above that $\alpha^{p^d-1}=1$ in $\F_q[x]/(f)$. What about the square root $\alpha^{(p^d-1)/2}$ in $\F_q[x]/(f)$? Note that this is **not** necessarily $\pm1$, since recall that $\F_q[x]/(f)$ is **not** a field. In general rings (that are not fields), the identity 1 may have more square roots than just $1$ and $-1$.

Again, we can use the Chinese Remainder Theorem to evaluate what $\alpha^{(p^d-1)/2}\bmod f$ is. This is essentially equivalent to computing

\begin{equation*} (\alpha^{(q^d-1)/2}\bmod g_1,\dotsc,\alpha^{(q^d-1)/2}\bmod g_k) , \end{equation*}

which by the square root of Fermat's Little Theorem is a vector whose entries are all $\pm1$. If by chance this vector is $(-1,-1,\dotsc,-1)$ then it **will** be the case that $\alpha^{(q^d-1)/2}=-1$ in $\F_q[x]/(f)$.
However, this is quite unlikely, especially if $k$ is large. In fact, since $\alpha$ was chosen randomly, we expect that 50% of the entries of the vector will be $1$ and 50% of the entries of the vector will be $-1$. If there is at least one $1$ and $-1$ entry in the vector, then $\alpha^{(q^d-1)/2}\neq\pm1$ in $\F_q[x]/(f)$.

The scenario when $\alpha^{(q^d-1)/2}\bmod f\neq\pm1$ is the one that is beneficial, because that means that $\alpha^{(q^d-1)/2}\bmod g_i=1$ for some $i$ and $\alpha^{(q^d-1)/2}\bmod g_j=-1$ for some $j$. In other words, we have that $g_j$ divides $\alpha^{(q^d-1)/2}-1$ and $g_i$ does not divide $\alpha^{(q^d-1)/2}-1$. Thus, $\gcd(\alpha^{(q^d-1)/2}-1,f)$ reveals a nontrivial factor of $f$, since it definitely includes $g_j$ but not $g_i$.

Input: Squarefree and monic $f\in\F_q[x]$ of degree $n$ and $d$, the degree of all irreducible factors of $f$

Choose $\alpha\in\F_q[x]$ randomly of degree less than $n$.

If $g:=\gcd(\alpha,f)$ is nontrivial, then $f$ is split by $f=g\cdot(f/g)$.

Compute $A:=\alpha^{(q^d-1)/2}\bmod f$.

If $g:=\gcd(A-1,f)$ is nontrivial, then $f$ is split by $f=g\cdot(f/g)$.

If $g$ is trivial, then repeat the algorithm with another random $\alpha$.

The running time of the algorithm is dominated by the computation of $A$, which using repeated squaring requires $O(d(\log q)n^2)$ operations in $\F_q$. How many times do we expect to get unlucky, though? In the worst case $f$ has exactly two irreducible factors, i.e., $d=n/2$. In this case we expect 50% of the time $A$ will be $1$ or $-1$ which gives a trivial gcd. However, 50% of the time $A$ will not be $\pm1$ and as we saw above this results in a nontrivial $g$ being found. Thus, even in the worst case we do not expect to have to try too many random $\alpha$ before a factor is found.

In order to completely factor $f$, the above algorithm must be called recursively at most $n/d$ times, one for each factor of $f$. In the worst case, every time a factor $g$ is recovered it would be of degree exactly $d$, meaning that $g$ itself is irreducible and only a single recursive call needs to be made on $f/g$, a polynomial of degree $n-d$. In such a case the total cost of splitting $f$ completely would be $n/d$ times $O(dn^2\log q)$, which is $O(n^3\log q)$.

However, since $\alpha$ was chosen randomly, it is expected that $g$ will contain about half of the irreducible factors of $f$ and $f/g$ will contain the other half of the irreducible factors. In this case, there will be two recursive calls and each will be of size roughly $n/2$. In such a case the depth of the recursion is expected to be *logarithmic* in $n/d$, not linear in $n/d$. Thus, the expected running time of the algorithm is $O(dn^2\log(q)\log(n/d))$ field operations.

So far, we've assumed that the input polynomial $f\in\F_q[x]$ to factor is squarefree. In fact, everything in the algorithm we described works if $f$ is "squarefull" (meaning that it is divisible by an irreducible polynomial more than once) but we need to be a bit careful.

After the distinct-degree factorization step, we will have computed $g_1$, $\dotsc$, $g_n$ which are each squarefree polynomials because $x^{q^d}-x$ is a squarefree polynomial. However, the product of the $g_i$ will not equal $f$ exactly when $f$ is squarefull. Instead, we will have $f=S\cdot\prod_i g_i$ where $S$ is the duplicated factors of $f$.

The input to the equal-degree factorization step will be the $g_i$, so nothing changes in the equal-degree step which will still factor the $g_i$ into their irreducible components.

At the end, we take each irreducible factor of the $g_i$ and see if divides $S$ (and if so, how many times). This can be done with repeated quotient and remainder. Since the polynomials involved all have degree at most $n$ this will be less than the cost of the other parts of the algorithm.

Input: Monic $f\in\F_q[x]$ of degree $n$

$h:=x$, $f_\text{orig}:=f$

for $i$ from 1 to $n$:

$\hspace{1em} h := h^q \bmod f_\text{orig}$

$\hspace{1em} g := \gcd(h-x,f)$

$\hspace{1em} f := f / g$

$\hspace{1em}$Apply equal-degree factorization on $g$ to write $g=g_1\dotsm g_k$

$\hspace{1em}$Add $g_1$, $\dotsc$, $g_k$ to the list of irreducible factors

$\hspace{1em}$for $j$ from 1 to $k$:

$\hspace{2em}$while $g_j$ divides $f$:

$\hspace{3em} f := f / g_j$

$\hspace{3em}$Add another copy of $g_j$ to the list of irreducible factors

Ouput the list of irreducible factors of $f$

The bottleneck of the outer loop in the complete factoring algorithm is the cost of computing the equal-degree factorization $g=g_1\dotsm g_k$. During step $i$ of the outer loop say that the degree of $g$ is $m_i$. Then the equal-degree factorization on step $i$ will produce the $m_i/i$ factors $g_1$, $\dotsc$, $g_{m_i/i}$ and cost of finding these will be an expected $O(in^2\log(q)\log(m_i/i))$ field operations.

Note that we have

\begin{equation*} i\log(m_i/i) = m_i\frac{\log(m_i/i)}{m_i/i} \leq m_i, \qquad\text{since $\frac{\log x}{x}\leq 1$}. \end{equation*}

Thus iteration $i$ of the loop takes an expected $O(m_i n^2\log(q))$ field operations. Because $\sum_{i=1}^n m_i\leq n$ the total expected running time of the entire algorithm is $\sum_{i=1}^n O(m_i n^2\log(q))=O(n^3\log(q))$ field operations.

Lastly, we will see how factoring polynomials in $\F_q[x]$ can also be used as a basis for factoring polynomials in $\Z[x]$. The algorithm we present will have exponential running time, but with some additional cleverness can be made to run in polynomial time.

First, note that to factor polynomials in $\Z[x]$ actually requires the ability to factor integers. For example, suppose all coefficients of your input polynomial $f\in\Z[x]$ are divisible by the same number $N$. Then in order to write $f$ as a product of factors where each factor cannot be factored any further requires $N$ to also be factored. One workaround to this is to consider the factorization problem over $\mathbb{Q}[x]$ instead of $\Z[x]$, since over $\mathbb{Q}$ every nonzero constant is invertible and cannot be factored further. If we ignore the issue of factoring integer constants, then the factoring problem in $\mathbb{Q}[x]$ is equivalent to the factoring problem in $\Z[x]$. We will sidestep the issue by just assuming that $f\in\Z[x]$ is monic.

The coefficients of the polynomial $f\in\Z[x]$ can be reduced modulo $p$ to form a polynomial $\bar f\in\Z_p[x]$. Our idea will be to compute $\bar f$ for large enough $p$, then factor $\bar f$ over $\Z_p[x]$. This will provide a factorization

\begin{equation*} \bar f = g_1\dotsm g_k \tag{$**$} \end{equation*}

for irreducible polynomials $\bar g_i\in\Z_p[x]$. If a polynomial is irreducible in $\Z_p[x]$ this definitely implies it is irreducible in $\Z[x]$ (because equality in $\Z$ implies equality in $\Z_p$). However, the converse does not hold: a polynomial might factor *farther* over $\Z_p$ than it does over $\Z$.

Note that if $p$ is chosen large enough, one can recover a polynomial $\alpha$ from its reduction $\bar\alpha$ modulo $p$. For example, suppose that you know the coefficients of $\alpha$ are all at most 5 in absolute value and the bar denotes reduction modulo $p=11$. If $\bar\alpha = x^3 - 5x^2 + 5x - 2$ then the coefficients of $\alpha$ and $\bar\alpha$ must be the same, because any other way of "lifting" the coefficients of $\Z_p$ to $\Z$ would introduce a coefficient $c$ with $|c|\geq6$. If the polynomial $\alpha$ we want to recover has a maximum coefficient of absolute value $N$, then we choose $p>2N$. Using the "symmetric range" $\{-(p-1)/2,\dotsc,(p-1)/2\}$ of residues mod $p$, we can capture all of $\alpha$'s coefficients exactly mod $p$, and therefore will be able to recover the $\alpha$ from $\bar\alpha$.

So by taking $p$ large enough we will be able to recover the coefficients of the factors of $f$ from their modular reductions—if we can compute their modular reductions. Say $f_1$ is an irreducible factor of $f$. Since $f\bmod p$ can only factor *farther* than $f$, it must be the case that some product of the $g_i$s in ($**$) must combine in order to give $f_1$, i.e., there is a set $S\subseteq\{1,\dotsc,k\}$ such that

\begin{equation*} f_1 = \prod_{i\in S} g_i . \end{equation*}

If we can find the set $S$ then we would be able to compute the product $f_1$ and we can easily test that $f_1$ is indeed a true factor of $f$ by checking that $f\bmod f_1=0$. The problem with this approach is that there seems no easy way to find the set $S$. Of course, we can try all possible subsets $S\subseteq\{1,\dotsc,k\}$ and figure out which ones yield true factors in $\Z[x]$, not $\Z_p[x]$. Of course, this requires exponential time in the number of factors.

Incidentally, it is easy to find the squarefree part of a polynomial in $\Z[x]$ or $\mathbb{Q}[x]$ (or more generally any field $\F$ where $1+1+\dotsb+1\neq0$ for arbitrary many additions). This is because in $\F[x]$ a factor divides $f=\sum_{i\geq0}a_ix^i\in\F[x]$ more than once if and only if it divides the derivative of $f$, defined by $f':=\sum_{i\geq1}ia_ix^{i-1}$.

Thus, the squarefree part of $f$ can be computed by $f/\gcd(f,f')$. You have to be careful over a finite field, as the precondition on the field isn't met (in that case $1+1+\dotsb+1=0$ when there are $p$ ones) and it is possible that $f'=0$ even when $f\neq0$. Though even in a finite field it still is the case that $\gcd(f,f')=1$ does imply that $f$ is squarefree.

Input: A squarefree and monic $f\in\Z[x]$ of degree $n$ and maximum coefficient in absolute value of $A$

Let $p\in[2B,4B)$ be a random prime where $B := 2^nA\sqrt{n+1}$

Factor $\bar f\in\Z_p[x]$ as $g_1\dotsm g_k$ for irreducible $g_i$ (mod $p$) and write the $g_i$ as polynomials with coefficients absolutely bounded by $p/2$

$T := \{1,\dotsc,k\}$

for all $S\subseteq T$, starting with the smallest $S$:

$\hspace{1em} g:=\prod_{i\in S} g_i$

$\hspace{1em}$if $f\bmod g=0$ then

$\hspace{2em} f:=f/g$

$\hspace{2em} T:=T\setminus S$

$\hspace{2em}$add $g$ to the list of irreducible factors

Output the list of irreducible factors of $f$

Unfortunately, the loop may run exponentially many times, since there are $2^k$ subsets of $T$.
There is a better method for determining which $g_i$ combine together to form actual irreducible factors of $f$, but it involves more mathematical machinery. In particular, an algorithm of Lenstra, Lenstra, and Lovász from 1982 is able to solve the factoring problem in $\mathbb{Q}[x]$ in polynomial time in $\deg(f)=n$ and in the size of the coefficients of $f$. At the time this was somewhat surprising, even to the discoverers. Their method is even totally deterministic, which at first seems paradoxical since it relies on the $\Z_p[x]$ factoring method that uses randomness. This is possible because they are able to show that they can find a prime $p$ in polynomial time (without relying on randomness) for which the $\Z_p[x]$ factoring algorithm is *guaranteed* to find the factorization in $\Z_p[x]$.