# Prove the following ideal $I$ is not a principal ideal.

Prove the ideal $$I = \left \space$$of$$\space \mathbb{Z}[X]$$ is not a principal ideal.

The solution I have been given is the following:

Assume for contradiction that I were a principal ideal, i.e., $$I = \left$$ for some $$f(X) \in \mathbb{Z}[X]$$. This means that

$$f(X) = 3 \cdot g(X) + X^2 \cdot h(X), \tag{1}$$

$$3 = f(X) \cdot k(X), \tag{2}$$

$$X^2 = f(X) \cdot s(X), \tag{3}$$

for some polynomials $$g(X), h(X), k(X)$$ and $$s(X)$$ in $$\mathbb{Z}[X]$$. As $$\mathbb{Z}$$is an integral domain, the equation $$(2)$$ implies that $$0 = \deg(3) = \deg f(X) + \deg k(X)$$, whereby $$\deg f(X) = 0$$, i.e., $$f(X)$$ is a constant polynomial, say $$f(X) = n \in \mathbb{Z}$$. Next, the equation $$(1)$$ implies that $$n = f(0) = 3g(0)$$, whereby $$n \in 3 \mathbb{Z}$$. Thus, all the coefficients of $$n \cdot s(X)$$ are in $$3\mathbb{Z}$$, contradicting the fact that this is supposed to equal $$X^2$$ by the equation $$(3)$$.

Could someone help me break this down a bit please. I am struggling to come to terms with the definition of a degree. These are the current definitions I have in simple terms:

Ideal - A subring which agrees with the relevant axioms

Principal ideal - If $$I$$ is generated by a set with only one element

Apologies if this seems quite vague. TIA

• What exactly is your question? You just want the definition of the degree of a polynomial? Apr 18 '19 at 21:18
• Possible example of applying the definition of degree to which I would understand as having just come across it, and to help break down the solution to this question. Thanks Apr 18 '19 at 21:21
• Another way: $3$ and $x$ are nonassociate primes so $(x^2,3) = (f)\,\Rightarrow\, f = \gcd(x^2,3) = 1,\,$ so $\,1 = 3 g + x^2 h \,\overset{\large x=0}\Longrightarrow\, 1 = 3 g(0)\,$ in $\Bbb Z,\,$ contradiction, Apr 18 '19 at 22:40

The degree of a polynomial $$f(x)=a_nx^n+\cdots+a_0$$ (where $$a_n\neq 0$$) is by definition $$n$$, the highest exponent of $$x$$ in the expression of $$f$$. If $$f(x)=a_nx^n+\cdots+a_0$$ and $$g(x)=b_mx^m+\cdots+b_0$$ are polynomials, then we have $$f(x)g(x)=a_nb_mx^{n+m}+\cdots+a_0b_0,$$ so that $$\deg(fg)=\deg f+\deg g.$$ Observe also that the polynomials $$f$$ with $$\deg f=0$$ are exactly the constant polynomials.

Now let's take a look at the proof you've presented. Assume for a contradiction that $$(3,x^2)=I=(f(x)).$$ Since $$3,x^2\in I=(f(x)),$$ we can write $$3 = f(x)\cdot k(x)$$ and $$x^2 = f(x)\cdot s(x)$$ for some $$k(x),s(x)\in\mathbb Z[x].$$ Now, since $$3 = f(x)\cdot k(x),$$ we have $$0=\deg 3 = \deg(fk)=\deg f+\deg k.$$ This forces $$\deg f=\deg k=0,$$ since degrees are always nonnegative. [Your proof says "since $$\mathbb Z$$ is an integral domain." This may as well be good justification, but I my justification is just as good.] Hence $$f$$ is constant, and in fact equals $$\pm 3$$ since the equation $$3=f\cdot k$$ means $$f$$ divides $$3.$$ However we also have $$x^2=f(x)\cdot s(x)$$. Since $$f=\pm 3$$, this means all the coefficients of $$f(x)s(x)$$ are divisible by $$3.$$ But this is not true of $$x^2.$$ This is a contradiction, and it completes the proof.

You'll notice we never used the equation $$f(x)=3\cdot g(x)+x^2\cdot h(x).$$

Hope this helps.

• Worth emphasis: the product of the leading coef's remains nonzero because $\Bbb Z$ is a domain (otherwise the product could have lower degree, e.g. $\,(2x-3)(2-3x) = x\$ over $\Bbb Z_6 =$ integers $\bmod 6\ \$ Apr 18 '19 at 22:30
• Actually, you can only conclude that $f=\pm3$ or $f=\pm1$. You concluded correctly for the former case; but the latter case should be analyzed as well and is where $f(x)=3g(x)+x^2h(x)$ comes into play. Apr 18 '19 at 23:18
• Both good comments. I'll revise when I get home (if I remember). Apr 19 '19 at 17:15

I'll assume you are happy with the first line.

Equation (1) is saying $$f$$ must be a combination of the generators given. Similarly, equations (2) and (3) say the generators given must be recoverable from $$f$$. Thus, the equations together give equality.

The degree of a univariate polynomial in $$x$$ is simply the highest power of $$x$$ occuring (so with nonzero coefficient). This is a useful tool allowing us to make the deductions in the next paragraph. That is, if $$g = fk$$, then $$\deg g = \deg f + \deg k$$, as used in the case $$\deg g = 0$$ to force, in particular, $$\deg f = 0$$.

Finally, substituting $$X = 0$$ in equation 1 forces $$n$$ to be a multiple of 3, deriving a contradiction in equation 3.

Is it now clear? If not, what in particular could I explain better?