Abstract Nonsense

Some Polynomial Irreducibility Criteria

Point of Post: In this post we discuss some (very few, about three) of the classic irreducibility criteria for polynomials, including Eisenstein’s criterion in its general form.

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Motivation

So, up until this point we’ve discussed Euclidean Domains, PIDs, and UFDs and (in our last set of posts) showed that certain polynomial rings are also UFDs and Euclidean domains. A central role in all three of these classes of rings (although particularly in UFDs) is the notion of irreducibility. Consequently, if we hope to put our theory to good use it would be helpful to have a list of easily checkable criteria for when a polynomial over some given ring is irreducible. In this post we create such a list (very short, about three), mostly for reference at a later date.

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Irreducibility Criteria

There are some basic one’s we’d like to begin with. Namely, in our last post we saw that a polynomial $p(x)\in k[x]$ (for some field $k$) is divisible by $x-a$ (with $a\in k$) if and only if $p(a)=0$. Now, note that if $p(x)$ is degree two or three, then being reducible is equivalent to having a root since any factorization of a degree one or two or three polynomial will involve a factor of degree one. From this we get the following theorem:

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Theorem: Let $k$ be a field and $p(x)\in k[x]$ with $\deg(p(x))\in\{2,3\}$. Then, $p(x)$ is reducible in $k[x]$ if and only if $p(x)$ has a root in $k$.

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This theorem implies that $x^2+1$ is irreducible in $\mathbb{R}[x]$ and perhaps slightly less obviously that $1+x+x^2$ is irreducible in $\mathbb{Z}_2[x]$  (since as a function $\mathbb{Z}_2\to\mathbb{Z}_2$ it’s identically $1$).

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Often times we want to prove something is irreducible in $\mathbb{Q}[x]$ but it is more tenable (whether we want to check roots or something) to check it in some $\mathbb{F}_p[x]$. Luckily we have the following:

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Theorem: Let $q(x)\in\mathbb{Q}[x]$ with leading coefficient not divisible by $p$, a prime. If $q(x)$ is irreducible in $\mathbb{F}_p[x]$ then $q(x)$ is irreducible in $\mathbb{Q}[x]$.

Proof: Suppose that $q(x)$ was reducible in $\mathbb{Q}[x]$, then Gauss’s lemma would tell us that $q(x)$ is reducible in $\mathbb{Z}[x]$, say $q(x)=f(x)g(x)$ where $\deg(f),\deg(g)<\deg(q)$, and since there is a natural homomorphism $\mathbb{Z}[x]\to\mathbb{F}_p[x]$ one would have that $q(x)$ is reducible in $\mathbb{F}_p[x]$. But, since the leading coefficient of  $q(x)$ is nonzero in $\mathbb{F}_p$ we know that $\deg(f),\deg(g)$ are still less than $\deg(q)$ even after coefficients have reduced. Thus, we have reduced $q(x)$ in $\mathbb{F}_p[x]$ which is impossible. $\blacksquare$

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This in conjunction with the previous theorem allows us to conclude that $1+x+x^2$ and $1+x+x^2+x^3$ are irreducible in $\mathbb{Q}[x]$ since they have no roots in $\mathbb{F}_2$ and $\mathbb{F}_3$ respectively.

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We now come to the classic irreducibility, Eisenstein’s irreducibility criterion. It easily shows things such as $x^4-4x^3+6$ or $x^4+10x^2+1$ are irreducible in $\mathbb{Q}[x]$. Roughly, what the theorem says is that if you have a polynomial whose leading coefficient is not divisible by a prime, every other coefficient is divisible by this (same) prime, and the constant term is not divisible by this (same) prime then the polynomial is irreducible in $\mathbb{Z}[x]$, and so by Gauss’s lemma, irreducible in $\mathbb{Q}[x]$. Indeed:

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Theorem(Eisenstein’s Irreducibility Criterion): Let $a_0+\cdots+a_nx^n=q(x)\in\mathbb{Z}[x]$ be such that there exists some prime $p$ such that $p\mid a_0,\cdots,a_{n-1}$, $p\nmid a_n$, and $p^2\nmid a_0$ then $q(x)$ is irreducible in $\mathbb{Q}[x]$.

Proof: By Gauss’s lemma it suffices to prove that $q(x)$ is irreducible in $\mathbb{Z}[x]$. So, suppose that $q(x)=(b_0+\cdots+b_mx^m)(c_0+\cdots+c_kx^k)$. Now, consider the reduction morphism $\varphi:\mathbb{Z}[x]\to\mathbb{F}_p[x]$. We have then that

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$\varphi(q(x))=\varphi(b_0+\cdots+b_mx^m)\varphi(c_0+\cdots+c_kx^k)$

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but note that $\varphi(q(x))=ux^n$ for some $u\in\mathbb{F}_p^\times$ (since each coefficient other than the leading one is divisible by $p$). But, since $\mathbb{F}_p$ is a field we know that $\mathbb{F}_p$ is a UFD and so $\varphi(b_0+\cdots+b_mx^m)=vx^m$ and $\varphi(c_0+\cdots+c_kx^k)=wx^k$ for $v,w\in\mathbb{F}_p^\times$. But, note that this implies that $p\mid b_0,c_0$ and so $p^2\mid b_0c_0=a_0$ which is a contradiction. $\blacksquare$

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This is the Eisenstein’s criterion that is usually encountered in a beginning undergraduate course, but thankfully having done material above this level should make clear that the above easily and fruitfully generalizes. Indeed, what was really used (secretly) was just that $\mathbb{Z}/(p)$ was a field, or more pertinently, an integral domain to get a contradiction to the fact that $a_0\notin (p)^2$. Clearly we should be able to generalize this to the more general case. Indeed:

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Theorem: Let $R$ be an integral domain and $a_0+\cdots+a_nx^n=q(x)\in R[x]$. Suppose that there exists a prime ideal $\mathfrak{p}\in\text{Spec}(R)$ such that $a_0,\cdots,a_{n-1}\in\mathfrak{p}$, $a_n\notin\mathfrak{p}$, and $a_0^2\notin\mathfrak{p}$. Then, $q(x)$ is irreducible in $\text{Frac}(R)[x]$.

Proof: By Gauss’s lemma it suffices to prove that $q(x)$ is irreducible in $R[x]$. To see this suppose that $q(x)=(b_0+\cdots+b_mx^m)(c_0+\cdots+c_kx^k)$. Consider the reduction map $\varphi:R[x]\to (R/\mathfrak{p})[x]$. We can clearly deduce (just as in the previous problem then that) $\varphi(q(x))=ux^n$ for some non-zero $u\in R/\mathfrak{p}$. We claim though that this implies $b_0=c_0=0$. Indeed, since the constant term is equal to $b_0c_0$ and is equal to zero we have that either $b_0=0$ or $c_0=0$, assume that $c_0=0$ but $b_0\ne 0$. Note then that since the linear term of the product of the polynomials is $c_0b_1+b_0c_1$ and this must be zero, we may conclude from this and the fact that $c_0\ne 0,b_0\equiv0\text{ mod }\mathfrak{p}$ that $b_1=0$ (recall that $R/\mathfrak{p}$ is an integral domain). Next, note that the quadratic coefficient of the product is $c_0b_2+c_1b_1+c_2b_0$ and applying similar logic allows us to conclude that $b_2=0$. Continuing in this way we arrive at the conclusion that $b_m\equiv0\text{ mod }\mathfrak{p}$ which is impossible since $u=b_mc_k$ and $u$ is nonzero. This is a contradiction. Thus, $c_0,b_0\equiv0\text{ mod }\mathfrak{p}$ which implies that $a_0=c_0b_0\equiv0\text{ mod }\mathfrak{p}^2$ which is a contradiction since $a_0\notin\mathfrak{p}^2$. Thus, we could not have nontrivially factored $q(x)$ in $\mathbb{Z}[x]$. $\blacksquare$

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To see an example where this theorem may be useful consider the following corollary:

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Corollary: Let $R$ be an integral domain and let $q(x,y)=y^n+xq_1(x)y^{n-1}+\cdots+xq_{n-2}(x)y+ax$. Then, $q(x,y)$ is irreducible in $R[x,y]=R[x][y]$

Proof: We know that $(x)$ is a prime ideal in $R[x]$ and $xq_1(x),\cdots,xq_{n-2}(x),x\in(x)$, $1\notin(x)$, and $x\notin (x^2)=(x)^2$ so Eisenstein’s criterion applies. $\blacksquare$

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

October 26, 2011 -

1. The first theorem should be say:
“Theorem: … Then, f(x) is REDUCIBLE in…”

Comment by Paulo | March 24, 2012 | Reply

• Dear Paulo,

Thank you very much for taking the time to point that out to me–it’s much appreciated.

Best,
Alex

Comment by Alex Youcis | March 24, 2012 | Reply

2. […] irreducible to conclude that they are, in fact, the minimal polynomials. To do this we appeal to Eisenstein’s criterion which says that since divide all but the leading term of and but don’t divide the […]

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