# Two Tricks Using Eisenstein's Criterion

/Today we're talking about Eisenstein's (not *Einstein's!) *Criterion - a well-known test to determine whether or not a polynomial is irreducible.

**Theorem (Eisenstein's Criterion)**

*Let $R$ be a UFD and $p\in R$ be a prime. If $f(x)=x^n+a_{n-1}x^{n-1}+\cdots + a_0\in R[x]$ is a monic polynomial such that $p|a_i$ for each $0\leq i \leq n-1$ and $p^2\nmid a_0$, then $f(x)$ is irreducible in $R[x]$ )and hence in $F[x]$ where $F$ is the field of fractions of $R$).*

Surely you've seen a standard type of example which makes use of this theorem:

*sometimes*the use of Eisenstein's Criterion might not be so obvious and a trick may be needed. Today we consider two such examples.

## Trick #1

**Let $p$ be a prime integer. Prove $\Phi_p(x)=\frac{x^p-1}{x-1}$ is irreducible in $\mathbb{Z}[x]$.**

$\Phi_p(x)$ is called the *cyclotomic $p$th polynomial* and is special because its roots are precisely the primitive $p$th roots of unity. That $\Phi_p(x)$ is irreducible is quite an important fact (which, for instance, turns up frequently in Galois theory), but at first glace it's not obvious that Eisenstein's Criterion is even applicable. For that we need a trick (or, a fact, really):

*$p(x)$ is irreducible $\quad \Longleftrightarrow\quad$ $p(x+1)$ is irreducible.*

for any polynomial $p(x)\in R[x]$ for any UFD $R$. (This works because any factorization of $p(x)$, say $p(x)=f(x)g(x)$, corresponds to a factorization of $p(x+1)$, namely $p(x+1)=f(x+1)g(x+1)$.)

So
$$\Phi_p(x+1)=\frac{(x+1)^p-1}{x}=\frac{ \sum_{k=0}^p\binom{p}{k}x^{p-k} \: -1 }{x}.$$
Notice that the numerator is equal to $x^p+px^{p-1}+\frac{1}{2}p(p-1)x^{p-2}+\cdots+\frac{1}{2}p(p-1)x^2+px$ and so the above becomes
$$x^{p-1}+px^{p-2}+\frac{1}{2}p(p-1)x^{p-3}+\cdots+\frac{1}{2}p(p-1)x+p.$$
This satisfies the Eisenstein criterion at the prime $p$. The key is simply that each of the binomial coefficients $\binom{p}{k}$ for $1\leq k \leq p-1$ is divisible by $p$ since the $p$ in the $p!$ of the numerator never gets canceled.

Similarly, we can show that $x^4+1$ is "Eisenstein at 2" by using the same trick. One can see this quickly by noting that fourth row of Pascal's triangle is 1-4-6-4-1 and so the coefficents of $(x+1)^4+1$ must be $1-4-6-4-2$.

## Trick #2

**Prove $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$**

Here we want to regard $\mathbb{Q}[x,y]$ as $\mathbb{Q}[x][y]$ so that our UFD is $R=\mathbb{Q}[x]$, and we want to view $x^2+y^2-1=y^2+(x+1)(x-1)$ as a monic polynomial in the variable $y$ with constant term $x^2-1\in R$. If we can show that $x+1$ (or similarly $x-1$) is a prime in $R$, then we can say $y^2+x^2-1$ is irreducible over $R$ since $x+1$ divides $(x+1)(x-1)$ but $(x+1)^2$ does not.

Indeed $x+1$ is prime. To see this, recall that it's enough to show $x+1$ is irreducible since we're in a UFD. This can be easily done with a degree argument.* *OR* if you're feeling extra fancy, you can take the high route: Notice that $$\mathbb{Q}[x]/(x+1)\cong \mathbb{Q}.$$
To see this, let $\varphi:\mathbb{Q}[x]\to\mathbb{Q}$ be the ring homomorphism given by $\varphi(p(x))=p(-1)$. Then $\varphi$ is surjective (any $q\in \mathbb{Q}$ has preimage $p(x)=x+q+1$) and $\ker\varphi$ is the ideal $(x+1)=\{(x+1)f(x):f(x)\in\mathbb{Q}[x]\}$. That $\mathbb{Q}[x]/(x+1)\cong \mathbb{Q}$ follows directly from the First Isomorphism Theorem.
Thus since $\mathbb{Q}$ is a field, so is $\mathbb{Q}[x]/(x+1)$ which implies $(x+1)$ is a maximal ideal (as $\mathbb{Q}[x]$ is a commutative ring) and therefore $x+1\in \mathbb{Q}[x]$ is irreducible. (This is "using a sledge hammer to beat a flea"!)

*Footnote:*

* Here 'tis: Suppose $x+1$ factors in $\mathbb{Q}[x]$ as $x+1=f(x)g(x)$. Then we must have $1=\deg f(x)+\deg g(x)$ which can only happen if, say, $\deg f(x)=1$ and $\deg g(x)=0$. This means $f(x)$ is linear and $g(x)=g\in \mathbb{Q}$ is a nonzero *unit* of $\mathbb{Q}[x]$ (i.e. it's a constant). But this is precisely what it means for $x+1$ to be irreducible.