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Let $\mathbb Z[x]$ the ring of polynomials with integers coefficients in one variable and $I =\langle 5,x^2 + 2\rangle$, how can I prove that $I$ is maximal ideal.

I tried first see that $5$ and $x^2+2$ are both polynomial in that ring but how can i get that is maximal ? some help please.

user26857
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Rosa Maria Gtz.
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5 Answers5

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Consider the natural homomorphism from the polynomial ring to the five-element field $\mathbb{Z}/5\mathbb{Z}$, extended by $\sqrt 3$:

$$ f: \mathbb Z[x] \longrightarrow (\mathbb{Z}/5\mathbb{Z})\; (\sqrt 3)$$

generated by $f(1) = 1, f(x) = \sqrt{3}$ (you might show this is a homomorphism, if it's not clear). Notice that the kernel of this homomorphism is exactly $I$ (or show it, if it's not clear). The condition that $I$ is the kernel of a ring homomorphism onto a field is equivalent to $I$ being maximal (similarly, if $I$ is the kernel of a ring homomorphism onto an integral domain, then $I$ is prime).

Therefore $I$ is maximal.

Let's speak of the intuition here. It is often easier to show that an ideal is maximal by evidencing a homomorphism to a field with that ideal as the kernel.

Quotienting out by $5\mathbb Z$ is saying exactly that multiples of $5$ are the same as $0$. As $I$ has one generator that is $5$, this makes a lot of sense.

For $x^2 + 2$ to be in the kernel, we are saying that $f(x^2 + 2) = 0$, or rather that $f(x)^2 = -2 = 3$ (where the last equality comes from us being is $\mathbb{Z}/5\mathbb{Z}$, so we can work mod $5$ instead of with imaginary bits). Well, if $f(x)^2 = 3$, we should make $f(x) = \sqrt 3$.

And that's how we know how to make the $f$ map.

davidlowryduda
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  • By $(\mathbb{Z}/5\mathbb{Z}); (\sqrt 3)$ do you mean the field of fractions of $(\mathbb{Z}/5\mathbb{Z}); [\sqrt 3]$? – Math137 Jun 16 '14 at 17:48
  • @math137: yes, that's what I mean. – davidlowryduda Jun 16 '14 at 18:03
  • Usually $K(a)$ is the smallest subfield of something containing $K$ and $a$. What do you mean by $(\mathbb Z/5\mathbb Z)(\sqrt 3)$? Can you prove that your construction is a field? – user26857 Jun 16 '14 at 18:16
  • @user26857: I would say that usually, $K(a)$ denotes a field extension. But in this case it doesn't matter. The smallest field containing the field $\mathbb{Z}/5\mathbb{Z}$ and $\sqrt 3$ is exactly this field. Without resorting to any theory, if we define $K(\sqrt{3}) = {a + b\sqrt{3}|a,b \in K}$ for any field $K$ whose characteristic is not $2$ or $3$ and where $\sqrt 3$ is a symbol for an element whose square is $3$, then it's easy to show this is a field from only definitions. But I also think this is in a standard first semester on a course including fields as part of 'field extensions'. – davidlowryduda Jun 16 '14 at 18:31
  • @user26857: On second thought, it's very easy, and I'll tell you how to do it directly. First, show that ${a + b\sqrt 3}$ is a ring (use the definition of a ring). Then explicitly find the multiplicative inverse of every nonzero element by rationalizing the denominator of $\frac{1}{a + b\sqrt 3}$. You'll notice that you want $a^2 - 3b^2 \neq 0$, but this is equivalent to $a \neq \sqrt 3 b$, which is equivalent to $\sqrt 3 \not \in K$ (we are assuming $\sqrt 3 \not \in K$, otherwise this construction is dumb). – davidlowryduda Jun 16 '14 at 19:13
  • Finally, if you dislike $\sqrt 3$ as a symbol, use something like $j$. It will look more like $i$, which is a symbol we are all ok with using to mean something that squares to $-1$. – davidlowryduda Jun 16 '14 at 19:13
  • mixedmath: @user26857 is right about $K(a)$ - it is, by definition, the smallest subfield of a specified field containing both $K$ and $a$. Your definition as an abstract set ${a + b\sqrt{3}}$ does not make sense, as you have not defined addition. You can adjoin elements to $K$ which satisfy a certain property, but this is done by quotienting a polynomial ring over $K$ by some ideal, which a priori need not be maximal (and indeed, maximality will depend heavily on $K$) – zcn Jun 16 '14 at 19:26
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    @zcn: While it is true that I didn't explicitly define addition, I saw no reason to be pedantic. Define addition and multiplication in the obvious way, just as we define it for ${a + bi}$ for the complex numbers. – davidlowryduda Jun 16 '14 at 19:29
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    @mixedmath: This is actually not a case of being pedantic! In defining addition and multiplication as for complex numbers, you are specifying the overfield $\mathbb{C}$ in which your ring operations are taking place. But this is wrong in this case, because $\mathbb{Z}/5\mathbb{Z}$ is not a subfield of $\mathbb{C}$! Hopefully this shows why care is needed in defining field extensions – zcn Jun 16 '14 at 19:31
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    @zcn: No, I truly think this is extreme pedantry. I am not relying on being a subfield of any field. From only definitions, one can prove that ${a + b\sqrt 3}$ is a field if one uses the obvious forms of addition and multiplication. One can define addition and multiplication for ${a + bi}$ by saying you add component-wise, and multiply distributively with the extra condition that $i^2 = -1$. Everything else comes from $a,b \in K$. – davidlowryduda Jun 16 '14 at 19:33
  • Yes, you can define addition componentwise in the set ${a + b\sqrt{3}}$, but this does create ambiguity in the notation $a + b\sqrt{3}$ - the $+$ sign there is not addition in the field (you can take this as pedantry if you wish). The very fact that you are not relying on being a subfield of a given field means that more care must be taken in defining things - which is why it is much easier to just specify a larger field – zcn Jun 16 '14 at 19:37
  • This is OK, but could be cleaner. The target ring is $\Bbb F_{25}$, the quadratic extension of the prime field $\Bbb F_5$. It has four elements of period exactly eight, two of them squaring to $3$, and you choose one of them, say $\alpha$. Then you map $\Bbb Z[x]\to\Bbb F_{25}$ by sending $x$ to $\alpha$. Clearly surjective, and the kernel contains $5$ and $x^2+2$, so $\ker\supset\langle5,x^2+2\rangle$. For the reverse inclusion, let $f$ be not in your ideal, and divide by $x^2+2$, with remainder $ax+b\in\Bbb Z[x]$, with $a$ or $b$ nonzero, in fact $\notin5\Bbb Z$. Thus not in the kernel. – Lubin Jun 22 '16 at 01:35
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By the Correspondence Theorem, $(5,X^2+2)$ is maximal in ${\mathbb Z}[X] \Leftrightarrow (5,X^2+2)/(5)$ is maximal in ${\mathbb Z}[X]/(5)$. Using an isomorphism, this is the same thing as saying that $(X^2 + 2)$ is maximal in $(\mathbb{Z}/(5))[X]$. This ring is a PID, so we just need to show that $X^2 + 2$ is irreducible mod 5. Since it has degree 2, this just means checking that it has no zeros mod 5.

Jason Juett
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If you quotient by the ideal $I$, then $$5 \equiv 0 \pmod{I} \text{ and } x^2+2 \equiv 0 \pmod{I}.$$ This also suggests that $x^2 \equiv -2 \equiv 3 \pmod{I}$ This helps you get the idea that perhaps $x \mapsto \sqrt{3}$ and $5 \mapsto 0$

Anurag A
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Say you add some $f\notin\langle 5, x^2+2\rangle$, and consider the ideal $I'=I+(f)$. We want to show this is the entire ring $\mathbb{Z}[x]$.

Well, dividing by remainder, we can write $f=(x^2+2)g+h$ where $h$ is linear. Reduce $h$ mod $5$ and we write $h=ax+b$ where $0\le a, b\le 4$. By assumption that $f\notin I$, we get $h\neq 0$.

I believe there is some nice algebra to reduce the casework here, to show basically that using $h, 5$ and $x^2+2$, you can get $1$ and hence $I'=\mathbb Z[x]$, but I'm not sure off the top of my head how to prove it in a unified, clean way.

Also, if a reference would help you, in the first few pages of chapter 1, Reid Undergraduate Commutative Algebra, he proves that for any PID $A$, maximal ideals of $A[x]$ are exactly those of the form $(f,p)$ where $p\in A$ is prime and $f\in A[x]$ is irreducible mod $p$. Maybe running through that proof with your special case would be helpful?

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First look at $\mathbb Z[x]/\langle x^2+2 \rangle$ and some polynomial $p(x)\in \mathbb Z[x]$.

We can find polynomials $q(x), r(x)$ so that $p(x)=q(x)(x^2+2)+r(x)$ and $r(x)=ax+b$. Since $x^2+2$ is irreducible, it has no common factor with $r(x)$ unless $r(x)=0$ and we can find polynomials $s,t$ with $$sp+t(x^2+2)=c$$ where $c$ is an integer. This provides a multiplicative inverse for $p$ only if $c=1$, so we don't yet have a field, although we can nearly invert polynomials - it is just that $\mathbb Z$ doesn't have multiplicative inverses.

Now let's reduce the coefficients modulo $5$. $x^2+2$ remains irreducible*. We can carry through the division, but non-zero $c$ are now invertible modulo $5$, so for non-zero $p$ we can use Euclid's algorithm to find $$sp+t(x^2+2)\equiv 1 $$

So that $s(x)$ is a multiplicative inverse and every non-zero element of the ring is invertible. Hence the ring is a field and the ideal is maximal.

*Note this would not have been the case if we'd been working with $x^2+1$, or if we'd been working with $x^2+2$ modulo $11$ so we have to be careful.

Working modulo $I=\langle x^2+2,5\rangle$ we have $x^2\equiv -2\equiv 3$, so we can compute the inverse of $r=ax+b$ explicitly as $$\frac 1{ax+b}=\frac 1{ax+b}\cdot \frac{ax-b}{ax-b}\equiv\frac a{3a^2-b^2}x-\frac b{3a^2-b^2}$$ where $3a^2-b^2$ is invertible modulo $5$ unless $a,b\equiv 0$.

Working the other way about we can factor $\langle 5\rangle$ first to get the ring of polynomials with coefficients reduced modulo $5$ - so the coefficients are elements of the five element field $F_5$ and $x^2+2$ is irreducible, and we have a standard type of extension field.

Mark Bennet
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